{"id":540,"date":"2024-05-23T19:46:45","date_gmt":"2024-05-23T19:46:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/?post_type=chapter&#038;p=540"},"modified":"2026-02-05T09:50:24","modified_gmt":"2026-02-05T09:50:24","slug":"1-3-equations-of-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/1-3-equations-of-lines\/","title":{"raw":"1.3 Equations of Lines","rendered":"1.3 Equations of Lines"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Determine whether equations are linear or nonlinear.<\/li>\r\n \t<li>Graph linear functions.<\/li>\r\n \t<li>Given a linear equation, determine the slope and [latex]y[\/latex]-intercept, if they exist.<\/li>\r\n \t<li>Find the slope given a pair of points, if it exists.<\/li>\r\n \t<li>Given the slope of a line and a point on a line, write the equation of the line.<\/li>\r\n \t<li>Given two points on a line, write the equation of the line.<\/li>\r\n \t<li>Given a point on a line and the equation of a parallel line, write the equation of the line.<\/li>\r\n \t<li>Given a point on a line and the equation of a perpendicular line, write the equation of the line.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div>\r\n\r\nThis section is a review of skills and concepts from your Beginning Algebra course. Feel free to skim the explanations and try the examples, or to work on the homework and come back for more help. If you find that you need more help <a href=\"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/3-4-slope-of-a-line\/\">understanding slope<\/a>, please click on the link to learn more about slope.\r\n<h2>Determine whether equations are linear or nonlinear<\/h2>\r\nWhen you graph a linear equation, it forms a straight line on the coordinate plane. Linear equations have a constant rate of change which is called the slope. In contrast, nonlinear equations have graphs that do not form straight lines. The rate of change of a nonlinear function is not constant.\r\n<div class=\"textbox learning-objectives\">\r\n<h3>linear vs. non-linear functions<\/h3>\r\n<table style=\"border-collapse: collapse; width: 100%; border: 1px solid black; height: 60px; font-size: 110%;\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Linear<\/strong><\/th>\r\n<th style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Non-linear<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">A function that can be written in the form [latex]f(x)=mx+b[\/latex]<\/td>\r\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">A function that CANNOT be written in the form [latex]f(x)=mx+b[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">Each variable in the function\u00a0has a maximum of degree 1.<\/td>\r\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">At least one variable has<strong> a degree of 2 or more,\u00a0<\/strong>or a <strong>variable appears in the denominator<\/strong>, or a <strong>variable is inside a square root or in an exponent<\/strong>\u00a0(as a few examples)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">The function when graphed is a straight line.<\/td>\r\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">The function when graphed is not a straight line but is a curve.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Examples:<\/strong>\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]f(x)=-3x[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]7y=5x[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]f(x)=2x-9[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]3x-4y=12[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]f(x)=\\dfrac{1}{8}x[\/latex]<\/li>\r\n<\/ul>\r\n<\/td>\r\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: left;\"><strong>Examples:<\/strong>\r\n<ul>\r\n \t<li>[latex]f(x)=3x^2+4[\/latex]<\/li>\r\n \t<li>[latex]f(x)=2^x[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\dfrac{5}{x}[\/latex]<\/li>\r\n \t<li>[latex]f(x)=4x^5-3x^4-1[\/latex]<\/li>\r\n \t<li>[latex]f(x)=3\\sqrt{x}+2[\/latex]<\/li>\r\n<\/ul>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1388\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1388 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-150x150.png\" alt=\"The function graphed is a wave. It starts above the x-axis, then curves below, then back above, then below the x-axis again.\" width=\"150\" height=\"150\" \/> nonlinear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1389\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1389 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-150x150.png\" alt=\"the graph of the function first appears just to the left of the y-axis at the bottom of the image, then curves up above the x-axis and goes almost horizontal for a few units before again steeply climbing\" width=\"150\" height=\"150\" \/> nonlinear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1393\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1393 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-150x150.png\" alt=\"A decreasing linear graph.\" width=\"150\" height=\"150\" \/> linear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1392\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1392 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-150x150.png\" alt=\"The function is a U-shaped parabola curving down and then up again as you move from left to right.\" width=\"150\" height=\"150\" \/> nonlinear function[\/caption]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1391\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1391 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-150x150.png\" alt=\"Increasing curve going through origin on a coordinate plane.\" width=\"150\" height=\"150\" \/> nonlinear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1394\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1394 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-150x150.png\" alt=\"Horizontal line above the x-axis on coordinate plane.\" width=\"150\" height=\"150\" \/> linear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1390\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1390 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-150x150.png\" alt=\"Graph of curve increasing swiftly from the left, crossing the x-axis at a negative value and then slowing the increase to the right of the graph.\" width=\"150\" height=\"150\" \/> nonlinear function[\/caption]<\/td>\r\n<td style=\"width: 25%; text-align: center;\">\r\n\r\n[caption id=\"attachment_1395\" align=\"aligncenter\" width=\"150\"]<img class=\"wp-image-1395 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-150x150.png\" alt=\"An increasing linear graph.\" width=\"150\" height=\"150\" \/> linear function[\/caption]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div>\r\n<h2>Graphing Linear Functions<\/h2>\r\n<\/div>\r\nThere are multiple ways to create a graph from a linear equation. One way is to create a table of values for [latex]x[\/latex] and [latex]y[\/latex] and then plot these ordered pairs on the coordinate plane. Two points are enough to determine a line. However, it is always a good idea to plot more than two points to avoid possible errors.\r\n\r\nAfter graphing the points, draw a line through the points to show that all of the points are on the line. The arrows at each end of the graph indicate that the line continues endlessly in both directions. Every point on this line is a solution to the linear equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nGraph the linear equation [latex]y=2x-5[\/latex].\r\n\r\n[reveal-answer q=\"834421\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834421\"]\r\n\r\nEvaluate [latex]y=2x-5[\/latex]\u00a0for different values of [latex]x[\/latex], and create a table of corresponding [latex]x[\/latex] and [latex]y[\/latex] values.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]-values<\/strong><\/td>\r\n<td><strong>[latex]2x-5[\/latex]<\/strong><\/td>\r\n<td><strong>[latex]y[\/latex]-values<\/strong><\/td>\r\n<td><strong><em>Ordered Pairs<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]2(0)-5[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<td>[latex](0, -5)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]2(1)-5[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex](1, -3)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]2(2)-5[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex](2, -1)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]2(4)-5[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex](4, 3)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]2(5)-5[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex](5, 5)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the ordered pairs and draw a line through the points to obtain the graph.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183219\/image021-1.jpg\" alt=\"Coordinate plane grid with line drawn through labeled points:  (0, negative 5) (1, negative 3), (2, negative 1), (4,3) and (5,5). The line is labeled y = 2 x minus 5.\" width=\"265\" height=\"262\" \/>\r\n\r\nEvery point on the line is a solution to the equation [latex]y=2x\u20135[\/latex]. You can substitute the values from the ordered pair [latex](\\color{blue}{1},\\color{Green}{\u22123})[\/latex] into the equation and the equation will be satisfied because the point [latex](1,-3)[\/latex] is a point on the graph of [latex]y=2x-5[\/latex].\r\n\r\n[latex]\\begin{align}y&amp;=2x-5\\\\ \\color{Green}{-3}&amp;\\overset{\\large{?}}{=}2\\left(\\color{blue}{1}\\right)-5&amp;&amp; \\color{blue}{\\textsf{substitute}}\\\\-3&amp;\\overset{\\large{?}}{=}2-5\\\\-3&amp;=-3 &amp;&amp; \\color{blue}{\\textsf{true statement}}\\end{align}[\/latex]\r\n\r\nYou can also substitute ANY of the other points on the line into the equation. Every point on the line is a solution to the equation [latex]y=2x\u20135[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo determine whether an ordered pair is a solution of a linear equation, substitute the [latex]x[\/latex] and [latex]y[\/latex] values into the equation. If this results in a true statement, then the ordered pair is a solution of the linear equation. If the result is not a true statement then the ordered pair is not a solution.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nDetermine whether [latex](\u22122,4)[\/latex] is a solution to the linear equation [latex]4y+5x=3[\/latex].\r\n\r\n[reveal-answer q=\"980260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"980260\"]\r\n\r\nSubstitute [latex]\\color{Green}{x=\u22122}[\/latex]\u00a0and [latex]\\color{Blue}{y=4}[\/latex]\u00a0into the equation.\r\n\r\n[latex]\\begin{align}4y+5x&amp;=3\\\\4(\\color{Blue}{4}\\color{black}{)+5(}\\color{Green}{-2}\\color{black}{)}&amp;\\overset{\\large{?}}{=}3&amp;&amp; \\color{blue}{\\textsf{substitute}}\\\\16+(-10)&amp;\\overset{\\large{?}}{=}3\\\\6&amp;\\color{red}{\\cancel{=}}3&amp;&amp; \\color{blue}{\\textsf{false statement}}\\\\[5pt] \\end{align}[\/latex]\r\n\r\nThe statement is not true, so [latex](\u22122,4)[\/latex] is not a solution to the equation [latex]4y+5x=3[\/latex]. Which means that the point [latex](-2,4)[\/latex] is NOT a point on the graph of [latex]4y+5x=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBelow is a video further explaining how to determine if an ordered pair is a solution to a linear equation.\r\n\r\nhttps:\/\/youtu.be\/9aWGxt7OnB8\r\n<h2 id=\"Intercepts\">Intercepts<\/h2>\r\nThe intercepts are the points where the graph of the line intersects or crosses the horizontal and vertical axes. To help you remember what \u201cintercept\u201d means, think about the word \u201cintersect.\u201d The two words sound alike and in this case mean the same thing.\r\n\r\nThe straight line on the graph below intersects the two coordinate axes. The point where the line crosses the [latex]x[\/latex]-axis is called the [latex]x[\/latex]<b>-intercept<\/b>. The [latex]y[\/latex]<b>-intercept<\/b> is the point where the line crosses the [latex]y[\/latex]-axis. For this graph, the [latex]x[\/latex]-intercept is the point [latex](\u22122,0)[\/latex] and the [latex]y[\/latex]-intercept is the point ([latex]0, 2[\/latex]). The equation of the line below is [latex]x-y=-2[\/latex]. If you substitute [latex]0[\/latex] in place of [latex]y[\/latex], you find [latex]x[\/latex] to be [latex]2[\/latex], which is the [latex]x[\/latex]-intercept of [latex](0,2)[\/latex]. If you substitute [latex]0[\/latex] in place of [latex]y[\/latex], you find [latex]x[\/latex] to be [latex]-2[\/latex], which is the [latex]y[\/latex]-intercept [latex](-2,0)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064249\/image018-1.jpg\" alt=\"A line going through two points. One point is on the x-axis and is labeled the x-intercept. The other point is on the y-axis and is labeled y-intercept.\" width=\"329\" height=\"320\" \/>\r\n\r\nNotice that the [latex]y[\/latex]-intercept always occurs where [latex]x=0[\/latex], and the [latex]x[\/latex]-intercept always occurs where [latex]y=0[\/latex]. To find the [latex]x[\/latex]- and [latex]y[\/latex]-intercepts of a linear equation, you can substitute\u00a0[latex]0[\/latex] for [latex]y[\/latex] and for [latex]x[\/latex], respectively.\r\n\r\nFor example, the linear equation [latex]3y+2x=6[\/latex]\u00a0has an [latex]x[\/latex]<i>-<\/i>intercept when [latex]\\color{green}{y=0}[\/latex]\r\n\r\n[latex]\\begin{align}3\\left(\\color{green}{0}\\right)+2x&amp;=6\\\\2x&amp;=6\\\\x&amp;=3\\end{align}[\/latex]\r\n\r\nThe [latex]x[\/latex]-intercept is [latex](3,0)[\/latex].\r\n\r\nLikewise, the [latex]y[\/latex]-intercept occurs when [latex]\\color{blue}{x=0}[\/latex]\r\n\r\n[latex]\\begin{align}3y+2\\left(\\color{blue}{0}\\right)&amp;=6\\\\3y&amp;=6\\\\y&amp;=2\\end{align}[\/latex]\r\n\r\nThe [latex]y[\/latex]-intercept is [latex](0,2)[\/latex].\r\n<h3 id=\"Using Intercepts to Graph Lines\">Using Intercepts to Graph Lines<\/h3>\r\nIntercepts can be used to graph linear equations. Once you have found the two intercepts, draw a line through them.\r\n\r\nDo this with the equation [latex]3y+2x=6[\/latex]. We found that the intercepts of the line this equation represents are [latex](0,2)[\/latex] and [latex](3,0)[\/latex]. Plot those two points and draw a line through them.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064250\/image019-1.jpg\" alt=\"A line drawn through the points (0,2) and (3,0). The point (0,2) is labeled y-intercept and the point (3,0) is labeled x-intercept. The line is labeled 3y+2x=6.\" width=\"340\" height=\"344\" \/>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nGraph [latex]5y+3x=30[\/latex]\u00a0using the [latex]x[\/latex]- and [latex]y[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"153435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153435\"]\r\n\r\nTo find the [latex]y[\/latex]-intercept, set [latex]\\color{blue}{x=0}[\/latex]\u00a0and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{align}5y+3x&amp;=30\\\\5y+3\\left(\\color{blue}{0}\\right)&amp;=30\\\\5y+0&amp;=30\\\\5y&amp;=30\\\\y&amp;=6\\\\y\\text{-intercept}\\,\\left(0,6\\right)\\end{align}[\/latex]\r\n\r\nTo find the [latex]x[\/latex]-intercept, set [latex]\\color{green}{y=0}[\/latex] and solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{align}5y+3x&amp;=30\\\\5\\left(\\color{green}{0}\\right)+3x&amp;=30\\\\0+3x&amp;=30\\\\3x&amp;=30\\\\x&amp;=10\\\\x\\text{-intercept}\\left(10,0\\right)\\end{align}[\/latex]\r\n\r\nThe graph can be seen below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064251\/image020-1.jpg\" alt=\"Line graph on coordinate plane. Line is labeled 5 y + 3 x = 30 and has negative slope, y-intercept at (0,6), and x-intercept at (10,0) and are labeled.\" width=\"425\" height=\"430\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\nThis video demonstrates graphing a linear equation using the intercepts.\r\n\r\nhttps:\/\/youtu.be\/k8r-q_T6UFk\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nGraph [latex]y=2x-4[\/latex] using the [latex]x[\/latex] and [latex]y[\/latex]-intercepts.\r\n\r\n[reveal-answer q=\"476848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"476848\"]\r\n\r\nFirst, find the [latex]y[\/latex]-intercept. Set [latex]\\color{blue}{x=0}[\/latex] equal to zero and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{align}y&amp;=2x-4\\\\y&amp;=2\\left(\\color{blue}{0}\\right)-4\\\\y&amp;=0-4\\\\y&amp;=-4\\\\y\\text{-intercept}\\left(0,-4\\right)\\end{align}[\/latex]\r\n\r\nTo find the [latex]x[\/latex]-intercept, set [latex]\\color{green}{y=0}[\/latex]\u00a0and solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{align}y&amp;=2x-4\\\\ \\color{green}{0}&amp;=2x-4\\\\4&amp;=2x\\\\x&amp;=2\\\\x\\text{-intercept}\\left(2,0\\right)\\end{align}[\/latex]\r\n\r\nThe graph can be seen below.\r\n\r\n<img class=\"aligncenter wp-image-4163 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/545\/2016\/09\/16184439\/Screen-Shot-2016-09-16-at-11.42.53-AM-300x296.png\" alt=\"Coordinate plane grid with line passing through (0, negative 4) and (2,0) and labeled y = 2 x minus 4.\" width=\"300\" height=\"296\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding slope<\/h2>\r\nGiven two values for the input, [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex], and two corresponding values for the output, [latex]{y}_{1}[\/latex]\u00a0and [latex]{y}_{2}[\/latex], represented by a set of ordered pairs, [latex]\\left({x}_{1}\\text{, }{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2}\\text{, }{y}_{2}\\right)[\/latex], we can calculate the <strong>slope<\/strong> [latex]m[\/latex] of a line through those points by\r\n\r\n[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]\r\n\r\nwhere [latex]\\Delta y[\/latex] is the vertical displacement and [latex]\\Delta x[\/latex] is the horizontal displacement.\r\n\r\nThe graph below indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2,}{y}_{2}\\right)[\/latex] is calculated. The greater the absolute value of the slope, the \"steeper\" the line is.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201021\/CNX_Precalc_Figure_02_01_005n2.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"569\" \/>\r\n\r\nThe units for slope are always [latex]\\dfrac{\\text{units for the output}}{\\text{units for the input}}[\/latex]. Think of the units as the change in output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.\r\n<div class=\"textbox shaded\">\r\n<h3>Calculating Slope<\/h3>\r\nThe slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex] is:\r\n\r\n[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]\r\n\r\n<\/div>\r\nIn the following video we show examples of how to find the slope of a line passing through two points.\r\n\r\nhttps:\/\/youtu.be\/in3NTcx11I8\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe population of a city increased from\u00a0[latex]23,400[\/latex] to\u00a0[latex]27,800[\/latex] between\u00a0[latex]2008[\/latex] and\u00a0[latex]2012[\/latex]. Find the change in population per year if we assume the change was constant from\u00a0[latex]2008[\/latex] to\u00a0[latex]2012[\/latex].\r\n\r\n[reveal-answer q=\"246268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"246268\"]\r\n\r\nThe rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.\r\n\r\n[latex]\\dfrac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ people per year}[\/latex]\r\n\r\nThe population increased by\u00a0[latex]1,100[\/latex] people per year.\r\n\r\nBecause we are told that the population increased, we expect the slope to be positive. This positive slope we calculated is therefore reasonable.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we show an example where we determine the increase in cost for producing solar panels given two data points.\r\n\r\nhttps:\/\/youtu.be\/4RbniDgEGE4\r\n<h2>The slope and y-intercept of a linear equation<\/h2>\r\nAnother way to graph a linear function is by using its slope [latex]m[\/latex] and [latex]y[\/latex]-intercept.\r\n<div class=\"textbox shaded\">\r\n<h3>SLOPE-INTERCEPT Form<\/h3>\r\nThe linear equation [latex]y=mx+b[\/latex] is in slope-intercept form.\r\n<ul>\r\n \t<li>[latex]b[\/latex] is the [latex]y[\/latex]-intercept of the graph and indicates the point [latex](0,b)[\/latex] at which the graph crosses the [latex]y[\/latex]-axis.<\/li>\r\n \t<li>[latex]m[\/latex] is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\nConsider the function [latex]f\\left(x\\right)=\\dfrac{1}{2}x+1[\/latex]. The function is in slope-intercept form, so the slope is [latex]m=\\dfrac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The [latex]y[\/latex]<em>-<\/em>intercept is the point on the graph where [latex]x=0[\/latex]. The graph of this line crosses the [latex]y[\/latex]-axis at\u00a0[latex](0, 1)[\/latex]. Begin graphing by plotting the point\u00a0[latex](0, 1)[\/latex]. The slope is [latex]m=\\dfrac{\\text{rise}}{\\text{run}}[\/latex], so for [latex]m=\\dfrac{1}{2}[\/latex], the \"rise\" is\u00a0[latex]1[\/latex] and the \"run\" from left to right is\u00a0[latex]2[\/latex]. Then starting from the [latex]y[\/latex]<em>-<\/em>intercept\u00a0[latex](0, 1)[\/latex], \"rise\" [latex]1[\/latex] unit and \"run\" [latex]2[\/latex] units to the right. (This is equivalent to \"run\" [latex]2[\/latex] units to the right and then \"rise\" [latex]1[\/latex] unit.) Repeat until you have a few points and then draw a line through the points as shown in the graph below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201048\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\dfrac{2}{3}x+5[\/latex] using the [latex]y[\/latex]<em>-<\/em>intercept and slope.\r\n\r\n[reveal-answer q=\"421669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421669\"]\r\n\r\nEvaluate the function at [latex]x=0[\/latex]\u00a0to find the [latex]y[\/latex]<em>-<\/em>intercept. The output value when [latex]x=0[\/latex] is\u00a0[latex]5[\/latex], so the graph will cross the [latex]y[\/latex]-axis at\u00a0[latex](0, 5)[\/latex].\r\n\r\nAccording to the equation for the function, the slope of the line is [latex]-\\dfrac{2}{3}[\/latex]. This tells us that for each vertical decrease in the \"rise\" of [latex]\u20132[\/latex] units, the \"run\" increases by\u00a0[latex]3[\/latex] units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in the graph below. From the initial value\u00a0[latex](0, 5)[\/latex], we move down\u00a0[latex]2[\/latex] units and to the right\u00a0[latex]3[\/latex] units. (Which is equivalent to moving up [latex]2[\/latex] units and left [latex]3[\/latex] units.) We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201050\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/>\r\n\r\nThe graph slants downward from left to right, which means it has a negative slope as expected.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]288727[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video we show another example of how to graph a linear function given the y-intercepts and the slope.\r\n\r\nhttps:\/\/youtu.be\/N6lEPh11gk8\r\n<div class=\"textbox exercises\"><\/div>\r\nIn the following video, we show an example of how to write the equation of a line given its graph.\r\n\r\nhttps:\/\/youtu.be\/mmWf_oLTNSQ\r\n<h2>Point-Slope Form<\/h2>\r\nIf we have the slope of a line and a point on the line, we can write the equation of the line.\r\n<div class=\"textbox shaded\">\r\n<h3>Point-Slope Form<\/h3>\r\nGiven a point [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] on a line and the slope of the line, [latex]m[\/latex], point-slope form will give the following equation of a line: [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]\r\n\r\n<\/div>\r\nIn this example, we will start with a given slope.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] that passes through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"524449\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"524449\"]\r\n\r\nUsing point-slope form, substitute [latex]\\color{green}{-3}[\/latex] for [latex]m[\/latex] and the point [latex]\\color{blue}{\\left(4,8\\right)}[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n\r\n[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\ y - \\color{blue}{8}&amp;=\\color{green}{-3}\\left(x - \\color{blue}{4}\\right)\\\\ y - 8&amp;=-3x+12\\\\ y&amp;=-3x+20\\end{align}[\/latex]\r\n\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of how to write the equation for a line given its slope and a point on the line.\r\n\r\nhttps:\/\/youtu.be\/vut5b2fRQQ0\r\n<h2>Finding the Equation of a Line Given Two Points<\/h2>\r\nIf we start with two points, we can first find the slope of a line through those two points, and then find the equation of the line that passes through them.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite an equation for a linear function [latex]f[\/latex] given its graph shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (negative 2, negative 4) and (0,2).\" width=\"369\" height=\"378\" \/>\r\n[reveal-answer q=\"728685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728685\"]\r\n<p id=\"fs-id1165135536538\">Identify two points on the line such as\u00a0[latex]\\color{Green}{(0, 2)}[\/latex] and\u00a0[latex]\\color{Blue}{(\u20132, \u20134)}[\/latex]. Use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\" style=\"text-align: left;\">[latex]\\begin{align}m&amp;=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\&amp;=\\dfrac{\\color{Blue}{-4} - \\color{Green}{2}}{\\color{Blue}{-2} - \\color{Green}{0}}\\\\&amp;=\\dfrac{-6}{-2}\\\\&amp;=3\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\y-\\left(\\color{Blue}{-4}\\right)&amp;=3\\left(x-\\left(\\color{Blue}{-2}\\right)\\right)\\\\ y+4&amp;=3\\left(x+2\\right)\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\r\n[latex]\\begin{align}y+4&amp;=3\\left(x+2\\right)\\\\y+4&amp;=3x+6\\\\ y&amp;=3x+2\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\nExamining the graph above, we see that the line crosses the [latex]y[\/latex]-axis at the point [latex](0,2)[\/latex] (the [latex]y[\/latex]-intercept), so [latex]b=2[\/latex] in the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the equation of the line that passes through the points [latex]\\left(\\color{blue}{3,4}\\right)[\/latex] and [latex]\\left(\\color{green}{0,-3}\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"249539\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249539\"]\r\n\r\nFirst, we calculate the slope using the slope formula and two points.\r\n\r\n[latex]\\begin{align}m&amp;=\\dfrac{\\color{green}{-3} - \\color{blue}{4}}{\\color{green}{0} - \\color{blue}{3}}\\\\ &amp;=\\dfrac{-7}{-3}\\\\ &amp;=\\dfrac{7}{3}\\end{align}[\/latex]\r\n\r\nNext, we use point-slope form with the slope of [latex]\\dfrac{7}{3}[\/latex] and either point. Let us pick the point [latex]\\left(\\color{blue}{3,4}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n\r\n[latex]\\begin{align}y - \\color{blue}{4}&amp;=\\dfrac{7}{3}\\left(x - \\color{blue}{3}\\right)\\\\ y - 4&amp;=\\dfrac{7}{3}x - 7\\\\ y&amp;=\\dfrac{7}{3}x - 3\\end{align}[\/latex]\r\n\r\nIn slope-intercept form, the equation is written as [latex]y=\\dfrac{7}{3}x - 3[\/latex].\r\n\r\nTo prove that either point can be used, let us use the second point [latex]\\left(\\color{green}{0,-3}\\right)[\/latex] and see if we get the same equation.\r\n\r\n[latex]\r\n\r\n\\begin{align}\r\n\r\ny-\\left(\\color{green}{-3}\\right)&amp;=\\dfrac{7}{3}\\left(x - \\color{green}{0}\\right)\\\\y+3&amp;=\\dfrac{7}{3}x\\\\y&amp;=\\dfrac{7}{3}x - 3\\end{align}[\/latex]\r\n\r\nWe see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next video shows another example of writing the equation of a line given two points on the line.\r\n\r\nhttps:\/\/youtu.be\/ndRpJxdmZJI\r\n<h2>Vertical and Horizontal Lines<\/h2>\r\n<h4>Vertical Lines<\/h4>\r\nThe equation of a <strong>vertical line<\/strong> is given as [latex]x=c[\/latex] where [latex]c[\/latex] is a constant. The slope of a vertical line is undefined, and regardless of the [latex]y[\/latex]<em>-<\/em>value of any point on the line, the [latex]x[\/latex]<em>-<\/em>coordinate of the point will be [latex]c[\/latex].\r\n\r\nSuppose that we want to find the equation of a line containing the following points: [latex]\\left(-3,-5\\right),\\left(-3,1\\right),\\left(-3,3\\right)[\/latex], and [latex]\\left(-3,5\\right)[\/latex]. First, we will find the slope, choosing two of the given points. In this case, we are choosing the points [latex](-3,3)[\/latex] and [latex](-3,5)[\/latex].\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{5 - 3}{-3-\\left(-3\\right)}=\\dfrac{2}{0}[\/latex]<\/p>\r\nDividing by zero is undefined so the slope of the line through these points is undefined. Using point-slope form is not possible. Notice that all of the [latex]x[\/latex]<em>-<\/em>coordinates in the given ordered pairs are the same. Then a vertical line through these points is [latex]x=-3[\/latex]. The line is on the graph below.\r\n<h4>Horizontal Lines<\/h4>\r\nThe equation of a <strong>horizontal line<\/strong> is given as [latex]y=c[\/latex] where [latex]c[\/latex] is a constant. The slope of a horizontal line is zero, and for any [latex]x[\/latex]<em>-<\/em>value of a point on the line, the corresponding [latex]y[\/latex]<em>-<\/em>coordinate will be [latex]c[\/latex].\r\n\r\nSuppose we want to find the equation of a line that contains the following points: [latex]\\left(-2,-2\\right),\\left(0,-2\\right),\\left(3,-2\\right)[\/latex], and [latex]\\left(5,-2\\right)[\/latex]. We can use point-slope form to find the equation of the line. First, we find the slope using any two points on the line. In this case, we are choosing [latex](-2,-2)[\/latex] and [latex](0,-2)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}m&amp;=\\dfrac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\\\&amp;=\\dfrac{0}{2}\\\\&amp;=0\\end{align}[\/latex]<\/p>\r\nUse any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in point-slope form.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y-\\left(-2\\right)&amp;=0\\left(x - 3\\right)\\\\y+2&amp;=0\\\\ y&amp;=-2\\end{align}[\/latex]<\/p>\r\nThe graph (below) is a horizontal line [latex]y=-2[\/latex]. Notice that all of the [latex]y[\/latex]<em>-<\/em>coordinates are the same in the list of ordered pairs.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/> The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.[\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line that passes through the points [latex]\\left(1,-3\\right)[\/latex] and [latex]\\left(1,4\\right).[\/latex]\r\n\r\n[reveal-answer q=\"346281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346281\"]\r\n\r\nThe [latex]x[\/latex]<em>-<\/em>coordinate of both points is\u00a0[latex]1.[\/latex] Therefore, we have a vertical line; [latex]x=1.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graphing a Linear Function: Another Approach<\/h2>\r\nAnother option for graphing linear functions is to use a vertical shift of the function.\r\n<h3>Vertical Shift<\/h3>\r\nIn a linear function [latex]f\\left(x\\right)=mx+b[\/latex], the [latex]b[\/latex] acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice in the figure below\u00a0that adding a value of [latex]b[\/latex] to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of [latex]f[\/latex] a total of [latex]b[\/latex] units up if [latex]b[\/latex] is positive and [latex]|b|[\/latex] units down if [latex]b[\/latex] is negative.\u00a0The graph below illustrates vertical shifts of the linear function [latex]f\\left(x\\right)=x[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201052\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"Coordinate system of 5 parallel lines labeled: y = x , y = x plus 2, y = x plus 4, y = x minus 2, and y = x minus 4.\" width=\"900\" height=\"759\" \/>\r\n\r\nIn this next example, we will start with the function [latex]f(x)=\\dfrac{1}{2}x[\/latex] and use a vertical shift to graph [latex]f\\left(x\\right)=\\dfrac{1}{2}x - 3[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f\\left(x\\right)=\\dfrac{1}{2}x - 3[\/latex].\r\n\r\n[reveal-answer q=\"208708\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"208708\"]\r\n\r\nFirst graph [latex]y=\\dfrac{1}{2}x[\/latex] by starting at the [latex]y[\/latex]-intercept of [latex](0,0)[\/latex] (this is also the [latex]x[\/latex]-intercept). Then \"rise\" [latex]1[\/latex] unit and \"run\" [latex]2[\/latex] units to the right to plot a point at [latex](2,1)[\/latex]. Draw a line through those points.\r\n\r\nNext, shift the line vertically down [latex]3[\/latex] units to graph the line [latex]y=\\dfrac{1}{2}x-3[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201055\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Coordinate system of 2 parallel lines. An arrow is drawn from the line labeled y = one half x to the line labeled y = one half x minus 3.  \" width=\"487\" height=\"377\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"Write the equations of parallel and perpendicular lines\">Write the equation of a line given a point and a parallel or perpendicular line<\/h2>\r\nThe relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.\r\n\r\nLet\u2019s start with an example involving parallel lines. <strong>Parallel lines<\/strong> have the same slope.\r\n<h3>Write the equation of a line given a point and a parallel line<\/h3>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]x\u2013y=5[\/latex] and passes through the point [latex](\u22122,1)[\/latex].\r\n\r\n[reveal-answer q=\"763534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"763534\"]\r\n\r\nRewrite the line you want to be parallel to into the\u00a0[latex]y=mx+b[\/latex] form, if needed.\r\n\r\n[latex]\\begin{align}x\u2013y&amp;=5\\\\\u2212y&amp;=\u2212x+5\\\\y&amp;=x\u20135\\end{align}[\/latex]\r\n\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=1[\/latex].\u00a0 Therefore, the slope is [latex]1[\/latex].\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is [latex]1[\/latex].\r\n\r\nUse the method for writing an equation from the slope and a point on the line. We will use the Point-Slope Form and substitute 1 for [latex]m[\/latex], and the point [latex]\\left(\\color{green}{\u22122,1}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] .\r\n\r\n&nbsp;\r\n\r\n[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\y \\color{green}{-1}&amp;=1\\left(x-(\\color{green}{-2})\\right)\\\\y - 1&amp;=1\\left(x + 2\\right)\\\\y-1&amp;=x+2\\\\y&amp;=x+3\\end{align}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThe answer is\u00a0[latex]y=x+3.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 class=\"yt watch-title-container\"><\/h2>\r\nhttps:\/\/youtu.be\/TQKz2XHI09E\r\n<h3>Write the equation of a line given a point and a perpendicular line<\/h3>\r\nWhen you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are <strong>perpendicular<\/strong> if the slope of one is the <strong>opposite\u00a0reciprocal<\/strong> of the slope of the other.\u00a0To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal. \u00a0In other words, flip it and change the sign.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that passes through the point [latex](1,5)[\/latex] and is perpendicular to the line [latex]y=2x\u2013 6[\/latex].\r\n\r\n[reveal-answer q=\"604282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604282\"]\r\n\r\nIdentify the slope of the line you want to be perpendicular to.\r\n\r\nThe given line is written in [latex]y=mx+b[\/latex] form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope is 2.\r\n\r\nTo find the slope of a perpendicular line, find the reciprocal, [latex] \\displaystyle \\frac{1}{2}[\/latex], then the opposite, [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nThe slope of the perpendicular line is [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nUse the method for writing an equation from the slope and a point on the line. We will use the Point-Slope Form and substitute [latex]-\\frac{1}{2}[\/latex] for [latex]m[\/latex], and the point [latex]\\left(\\color{green}{1,5}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] .\r\n\r\n[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\y -\\color{green}{5}&amp;=-\\frac{1}{2}\\left(x-\\color{green}{1}\\right)\\\\y - 5&amp;= -\\frac{1}{2}\\left(x -1\\right)\\\\y-5&amp;=-\\frac{1}{2}x+\\frac{1}{2}\\\\y&amp;=-\\dfrac{1}{2}x+\\frac{11}{2}\\end{align}[\/latex]\r\n\r\nFor practice, this time, we will use the alternate method for finding the equation of a line. Substitute [latex] \\displaystyle -\\dfrac{1}{2}[\/latex] for [latex]m[\/latex], and the point [latex]\\color{green}{(1,5)}[\/latex] for [latex]x[\/latex] and [latex]y[\/latex] into slope-intercept form.\r\n\r\n[latex] \\displaystyle \\begin{align}y&amp;=mx+b\\\\\\color{green}{5}&amp;=-\\frac{1}{2}(\\color{green}{1})+b\\end{align}[\/latex]\r\n\r\nSolve for [latex]b[\/latex].\r\n\r\n[latex] \\displaystyle \\begin{align}5&amp;=-\\frac{1}{2}+b\\\\\\frac{11}{2}&amp;=b\\end{align}[\/latex]\r\n\r\nWrite the equation using the new slope for [latex]m[\/latex] and the [latex]b[\/latex] you just found.\r\n\r\n[latex]y=mx+b[\/latex]\r\n\r\n[latex]y=-\\dfrac{1}{2}x+\\dfrac{11}{2}[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]y=-\\dfrac{1}{2}x+\\dfrac{11}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Video: Write the equation of a line given a point and a perpendicular line (Alternate Method)<\/h3>\r\nhttps:\/\/youtu.be\/QtvtzKjtowA\r\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Write the equations of lines parallel and perpendicular to horizontal and vertical lines<\/span><\/h2>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]y=4[\/latex] and passes through the point [latex](0,10)[\/latex].\r\n\r\n[reveal-answer q=\"426450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426450\"]\r\n\r\nRewrite the line into [latex]y=mx+b[\/latex]\u00a0form, if needed.\r\n\r\nYou may notice without doing this that [latex]y=4[\/latex]\u00a0is a horizontal line 4 units above the [latex]x[\/latex]-axis. Because it is horizontal, you know its slope is zero.\r\n\r\n[latex]\\begin{align}y&amp;=4\\\\y&amp;=0x+4\\end{align}[\/latex]\r\n\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=4[\/latex].\r\n\r\nSince [latex]m=0[\/latex], the slope is [latex]0[\/latex]. This is a horizontal line.\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is also [latex]0[\/latex].\r\n\r\nSince the parallel line will be a horizontal line, its form is\r\n\r\n[latex]y=\\text{a constant}[\/latex]\r\n\r\nSince we want this new line to pass through the point [latex](0,10)[\/latex], we will need to write the equation of the new line as:\r\n\r\n[latex]y=10[\/latex]\r\n\r\nThis line is parallel to [latex]y=4[\/latex]\u00a0and passes through [latex](0,10)[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=10[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is perpendicular to the line [latex]y=-3[\/latex] and passes through the point [latex](-2,5)[\/latex].\r\n\r\n[reveal-answer q=\"426550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426550\"]\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=-3[\/latex].\r\n\r\nA perpendicular line will have a slope that is the negative reciprocal of the slope of\u00a0[latex]y=-3[\/latex], but\u00a0what does that mean in this case?\r\n\r\nThe reciprocal of [latex]0[\/latex] is [latex]\\frac{1}{0}[\/latex], but we know that dividing by [latex]0[\/latex] is undefined.\r\n\r\nThis means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.\r\n\r\nThe form of a vertical line is [latex]x=\\text{a constant}[\/latex], where every <em>x<\/em>-value on the line is equal to some constant. \u00a0Since we are looking for a line that goes through the point [latex](-2,5)[\/latex], all of the <em>x<\/em>-values on this line must be [latex]-2[\/latex].\r\n\r\nThe equation of a line passing through [latex](-2,5)[\/latex] that is perpendicular to the horizontal line\u00a0[latex]y=-3[\/latex] is therefore,\r\n\r\n[latex]x=-2[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]x=-2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 class=\"yt watch-title-container\"><\/h2>\r\nhttps:\/\/youtu.be\/Qpn3f3wMeIs\r\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Summary<\/span><\/h2>\r\nThe slope-intercept form of a linear equation is written as [latex]y=mx+b[\/latex], where [latex]m[\/latex] is the slope and [latex]y[\/latex] is the value of [latex]y[\/latex] at the [latex]y[\/latex]-intercept, which can be written as [latex](0,b)[\/latex]. When you know the slope and the [latex]y[\/latex]-intercept of a line you can use the slope-intercept form to immediately write the equation of that line. The point-slope form, [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex], can be used to write the equation of a line when you know the slope and a point on the line or when you know two points on the line.\r\n\r\nWhen lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a 90\u00b0 angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[\/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Determine whether equations are linear or nonlinear.<\/li>\n<li>Graph linear functions.<\/li>\n<li>Given a linear equation, determine the slope and [latex]y[\/latex]-intercept, if they exist.<\/li>\n<li>Find the slope given a pair of points, if it exists.<\/li>\n<li>Given the slope of a line and a point on a line, write the equation of the line.<\/li>\n<li>Given two points on a line, write the equation of the line.<\/li>\n<li>Given a point on a line and the equation of a parallel line, write the equation of the line.<\/li>\n<li>Given a point on a line and the equation of a perpendicular line, write the equation of the line.<\/li>\n<\/ul>\n<\/div>\n<div>\n<p>This section is a review of skills and concepts from your Beginning Algebra course. Feel free to skim the explanations and try the examples, or to work on the homework and come back for more help. If you find that you need more help <a href=\"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/3-4-slope-of-a-line\/\">understanding slope<\/a>, please click on the link to learn more about slope.<\/p>\n<h2>Determine whether equations are linear or nonlinear<\/h2>\n<p>When you graph a linear equation, it forms a straight line on the coordinate plane. Linear equations have a constant rate of change which is called the slope. In contrast, nonlinear equations have graphs that do not form straight lines. The rate of change of a nonlinear function is not constant.<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>linear vs. non-linear functions<\/h3>\n<table style=\"border-collapse: collapse; width: 100%; border: 1px solid black; height: 60px; font-size: 110%;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Linear<\/strong><\/th>\n<th style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Non-linear<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">A function that can be written in the form [latex]f(x)=mx+b[\/latex]<\/td>\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">A function that CANNOT be written in the form [latex]f(x)=mx+b[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">Each variable in the function\u00a0has a maximum of degree 1.<\/td>\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">At least one variable has<strong> a degree of 2 or more,\u00a0<\/strong>or a <strong>variable appears in the denominator<\/strong>, or a <strong>variable is inside a square root or in an exponent<\/strong>\u00a0(as a few examples)<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\">The function when graphed is a straight line.<\/td>\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: center;\">The function when graphed is not a straight line but is a curve.<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 50.036629%; border: 1px solid #999999; height: 12px; text-align: center;\"><strong>Examples:<\/strong><\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]f(x)=-3x[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]7y=5x[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]f(x)=2x-9[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]3x-4y=12[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]f(x)=\\dfrac{1}{8}x[\/latex]<\/li>\n<\/ul>\n<\/td>\n<td style=\"width: 89.963371%; border: 1px solid #999999; height: 12px; text-align: left;\"><strong>Examples:<\/strong><\/p>\n<ul>\n<li>[latex]f(x)=3x^2+4[\/latex]<\/li>\n<li>[latex]f(x)=2^x[\/latex]<\/li>\n<li>[latex]f(x)=\\dfrac{5}{x}[\/latex]<\/li>\n<li>[latex]f(x)=4x^5-3x^4-1[\/latex]<\/li>\n<li>[latex]f(x)=3\\sqrt{x}+2[\/latex]<\/li>\n<\/ul>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1388\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1388\" class=\"wp-image-1388 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-150x150.png\" alt=\"The function graphed is a wave. It starts above the x-axis, then curves below, then back above, then below the x-axis again.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-sine.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1388\" class=\"wp-caption-text\">nonlinear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1389\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1389\" class=\"wp-image-1389 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-150x150.png\" alt=\"the graph of the function first appears just to the left of the y-axis at the bottom of the image, then curves up above the x-axis and goes almost horizontal for a few units before again steeply climbing\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-cubic.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1389\" class=\"wp-caption-text\">nonlinear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1393\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1393\" class=\"wp-image-1393 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-150x150.png\" alt=\"A decreasing linear graph.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-3.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1393\" class=\"wp-caption-text\">linear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1392\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1392\" class=\"wp-image-1392 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-150x150.png\" alt=\"The function is a U-shaped parabola curving down and then up again as you move from left to right.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-quadratic.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1392\" class=\"wp-caption-text\">nonlinear function<\/p>\n<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1391\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1391\" class=\"wp-image-1391 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-150x150.png\" alt=\"Increasing curve going through origin on a coordinate plane.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-exponential.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1391\" class=\"wp-caption-text\">nonlinear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1394\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1394\" class=\"wp-image-1394 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-150x150.png\" alt=\"Horizontal line above the x-axis on coordinate plane.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-2.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1394\" class=\"wp-caption-text\">linear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1390\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1390\" class=\"wp-image-1390 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-150x150.png\" alt=\"Graph of curve increasing swiftly from the left, crossing the x-axis at a negative value and then slowing the increase to the right of the graph.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-logarithmic.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1390\" class=\"wp-caption-text\">nonlinear function<\/p>\n<\/div>\n<\/td>\n<td style=\"width: 25%; text-align: center;\">\n<div id=\"attachment_1395\" style=\"width: 160px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1395\" class=\"wp-image-1395 size-thumbnail\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-150x150.png\" alt=\"An increasing linear graph.\" width=\"150\" height=\"150\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-150x150.png 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-300x300.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-768x768.png 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-65x65.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-225x225.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1-350x350.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/05\/M1010-1.3-linear-1.png 800w\" sizes=\"auto, (max-width: 150px) 100vw, 150px\" \/><\/p>\n<p id=\"caption-attachment-1395\" class=\"wp-caption-text\">linear function<\/p>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div>\n<h2>Graphing Linear Functions<\/h2>\n<\/div>\n<p>There are multiple ways to create a graph from a linear equation. One way is to create a table of values for [latex]x[\/latex] and [latex]y[\/latex] and then plot these ordered pairs on the coordinate plane. Two points are enough to determine a line. However, it is always a good idea to plot more than two points to avoid possible errors.<\/p>\n<p>After graphing the points, draw a line through the points to show that all of the points are on the line. The arrows at each end of the graph indicate that the line continues endlessly in both directions. Every point on this line is a solution to the linear equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Graph the linear equation [latex]y=2x-5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834421\">Show Solution<\/span><\/p>\n<div id=\"q834421\" class=\"hidden-answer\" style=\"display: none\">\n<p>Evaluate [latex]y=2x-5[\/latex]\u00a0for different values of [latex]x[\/latex], and create a table of corresponding [latex]x[\/latex] and [latex]y[\/latex] values.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]-values<\/strong><\/td>\n<td><strong>[latex]2x-5[\/latex]<\/strong><\/td>\n<td><strong>[latex]y[\/latex]-values<\/strong><\/td>\n<td><strong><em>Ordered Pairs<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]2(0)-5[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<td>[latex](0, -5)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]2(1)-5[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex](1, -3)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]2(2)-5[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex](2, -1)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]2(4)-5[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex](4, 3)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]2(5)-5[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex](5, 5)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the ordered pairs and draw a line through the points to obtain the graph.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/06\/01183219\/image021-1.jpg\" alt=\"Coordinate plane grid with line drawn through labeled points:  (0, negative 5) (1, negative 3), (2, negative 1), (4,3) and (5,5). The line is labeled y = 2 x minus 5.\" width=\"265\" height=\"262\" \/><\/p>\n<p>Every point on the line is a solution to the equation [latex]y=2x\u20135[\/latex]. You can substitute the values from the ordered pair [latex](\\color{blue}{1},\\color{Green}{\u22123})[\/latex] into the equation and the equation will be satisfied because the point [latex](1,-3)[\/latex] is a point on the graph of [latex]y=2x-5[\/latex].<\/p>\n<p>[latex]\\begin{align}y&=2x-5\\\\ \\color{Green}{-3}&\\overset{\\large{?}}{=}2\\left(\\color{blue}{1}\\right)-5&& \\color{blue}{\\textsf{substitute}}\\\\-3&\\overset{\\large{?}}{=}2-5\\\\-3&=-3 && \\color{blue}{\\textsf{true statement}}\\end{align}[\/latex]<\/p>\n<p>You can also substitute ANY of the other points on the line into the equation. Every point on the line is a solution to the equation [latex]y=2x\u20135[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>To determine whether an ordered pair is a solution of a linear equation, substitute the [latex]x[\/latex] and [latex]y[\/latex] values into the equation. If this results in a true statement, then the ordered pair is a solution of the linear equation. If the result is not a true statement then the ordered pair is not a solution.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Determine whether [latex](\u22122,4)[\/latex] is a solution to the linear equation [latex]4y+5x=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q980260\">Show Solution<\/span><\/p>\n<div id=\"q980260\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]\\color{Green}{x=\u22122}[\/latex]\u00a0and [latex]\\color{Blue}{y=4}[\/latex]\u00a0into the equation.<\/p>\n<p>[latex]\\begin{align}4y+5x&=3\\\\4(\\color{Blue}{4}\\color{black}{)+5(}\\color{Green}{-2}\\color{black}{)}&\\overset{\\large{?}}{=}3&& \\color{blue}{\\textsf{substitute}}\\\\16+(-10)&\\overset{\\large{?}}{=}3\\\\6&\\color{red}{\\cancel{=}}3&& \\color{blue}{\\textsf{false statement}}\\\\[5pt] \\end{align}[\/latex]<\/p>\n<p>The statement is not true, so [latex](\u22122,4)[\/latex] is not a solution to the equation [latex]4y+5x=3[\/latex]. Which means that the point [latex](-2,4)[\/latex] is NOT a point on the graph of [latex]4y+5x=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Below is a video further explaining how to determine if an ordered pair is a solution to a linear equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine If an Ordered Pair is a Solution to a Linear Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9aWGxt7OnB8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 id=\"Intercepts\">Intercepts<\/h2>\n<p>The intercepts are the points where the graph of the line intersects or crosses the horizontal and vertical axes. To help you remember what \u201cintercept\u201d means, think about the word \u201cintersect.\u201d The two words sound alike and in this case mean the same thing.<\/p>\n<p>The straight line on the graph below intersects the two coordinate axes. The point where the line crosses the [latex]x[\/latex]-axis is called the [latex]x[\/latex]<b>-intercept<\/b>. The [latex]y[\/latex]<b>-intercept<\/b> is the point where the line crosses the [latex]y[\/latex]-axis. For this graph, the [latex]x[\/latex]-intercept is the point [latex](\u22122,0)[\/latex] and the [latex]y[\/latex]-intercept is the point ([latex]0, 2[\/latex]). The equation of the line below is [latex]x-y=-2[\/latex]. If you substitute [latex]0[\/latex] in place of [latex]y[\/latex], you find [latex]x[\/latex] to be [latex]2[\/latex], which is the [latex]x[\/latex]-intercept of [latex](0,2)[\/latex]. If you substitute [latex]0[\/latex] in place of [latex]y[\/latex], you find [latex]x[\/latex] to be [latex]-2[\/latex], which is the [latex]y[\/latex]-intercept [latex](-2,0)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064249\/image018-1.jpg\" alt=\"A line going through two points. One point is on the x-axis and is labeled the x-intercept. The other point is on the y-axis and is labeled y-intercept.\" width=\"329\" height=\"320\" \/><\/p>\n<p>Notice that the [latex]y[\/latex]-intercept always occurs where [latex]x=0[\/latex], and the [latex]x[\/latex]-intercept always occurs where [latex]y=0[\/latex]. To find the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts of a linear equation, you can substitute\u00a0[latex]0[\/latex] for [latex]y[\/latex] and for [latex]x[\/latex], respectively.<\/p>\n<p>For example, the linear equation [latex]3y+2x=6[\/latex]\u00a0has an [latex]x[\/latex]<i>&#8211;<\/i>intercept when [latex]\\color{green}{y=0}[\/latex]<\/p>\n<p>[latex]\\begin{align}3\\left(\\color{green}{0}\\right)+2x&=6\\\\2x&=6\\\\x&=3\\end{align}[\/latex]<\/p>\n<p>The [latex]x[\/latex]-intercept is [latex](3,0)[\/latex].<\/p>\n<p>Likewise, the [latex]y[\/latex]-intercept occurs when [latex]\\color{blue}{x=0}[\/latex]<\/p>\n<p>[latex]\\begin{align}3y+2\\left(\\color{blue}{0}\\right)&=6\\\\3y&=6\\\\y&=2\\end{align}[\/latex]<\/p>\n<p>The [latex]y[\/latex]-intercept is [latex](0,2)[\/latex].<\/p>\n<h3 id=\"Using Intercepts to Graph Lines\">Using Intercepts to Graph Lines<\/h3>\n<p>Intercepts can be used to graph linear equations. Once you have found the two intercepts, draw a line through them.<\/p>\n<p>Do this with the equation [latex]3y+2x=6[\/latex]. We found that the intercepts of the line this equation represents are [latex](0,2)[\/latex] and [latex](3,0)[\/latex]. Plot those two points and draw a line through them.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064250\/image019-1.jpg\" alt=\"A line drawn through the points (0,2) and (3,0). The point (0,2) is labeled y-intercept and the point (3,0) is labeled x-intercept. The line is labeled 3y+2x=6.\" width=\"340\" height=\"344\" \/><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Graph [latex]5y+3x=30[\/latex]\u00a0using the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q153435\">Show Solution<\/span><\/p>\n<div id=\"q153435\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the [latex]y[\/latex]-intercept, set [latex]\\color{blue}{x=0}[\/latex]\u00a0and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{align}5y+3x&=30\\\\5y+3\\left(\\color{blue}{0}\\right)&=30\\\\5y+0&=30\\\\5y&=30\\\\y&=6\\\\y\\text{-intercept}\\,\\left(0,6\\right)\\end{align}[\/latex]<\/p>\n<p>To find the [latex]x[\/latex]-intercept, set [latex]\\color{green}{y=0}[\/latex] and solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{align}5y+3x&=30\\\\5\\left(\\color{green}{0}\\right)+3x&=30\\\\0+3x&=30\\\\3x&=30\\\\x&=10\\\\x\\text{-intercept}\\left(10,0\\right)\\end{align}[\/latex]<\/p>\n<p>The graph can be seen below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064251\/image020-1.jpg\" alt=\"Line graph on coordinate plane. Line is labeled 5 y + 3 x = 30 and has negative slope, y-intercept at (0,6), and x-intercept at (10,0) and are labeled.\" width=\"425\" height=\"430\" \/><\/div>\n<\/div>\n<\/div>\n<p>This video demonstrates graphing a linear equation using the intercepts.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Graph Linear Equations Using Intercepts\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/k8r-q_T6UFk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Graph [latex]y=2x-4[\/latex] using the [latex]x[\/latex] and [latex]y[\/latex]-intercepts.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q476848\">Show Solution<\/span><\/p>\n<div id=\"q476848\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the [latex]y[\/latex]-intercept. Set [latex]\\color{blue}{x=0}[\/latex] equal to zero and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{align}y&=2x-4\\\\y&=2\\left(\\color{blue}{0}\\right)-4\\\\y&=0-4\\\\y&=-4\\\\y\\text{-intercept}\\left(0,-4\\right)\\end{align}[\/latex]<\/p>\n<p>To find the [latex]x[\/latex]-intercept, set [latex]\\color{green}{y=0}[\/latex]\u00a0and solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{align}y&=2x-4\\\\ \\color{green}{0}&=2x-4\\\\4&=2x\\\\x&=2\\\\x\\text{-intercept}\\left(2,0\\right)\\end{align}[\/latex]<\/p>\n<p>The graph can be seen below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-4163 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/545\/2016\/09\/16184439\/Screen-Shot-2016-09-16-at-11.42.53-AM-300x296.png\" alt=\"Coordinate plane grid with line passing through (0, negative 4) and (2,0) and labeled y = 2 x minus 4.\" width=\"300\" height=\"296\" \/><\/div>\n<\/div>\n<\/div>\n<h2>Finding slope<\/h2>\n<p>Given two values for the input, [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex], and two corresponding values for the output, [latex]{y}_{1}[\/latex]\u00a0and [latex]{y}_{2}[\/latex], represented by a set of ordered pairs, [latex]\\left({x}_{1}\\text{, }{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2}\\text{, }{y}_{2}\\right)[\/latex], we can calculate the <strong>slope<\/strong> [latex]m[\/latex] of a line through those points by<\/p>\n<p>[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<p>where [latex]\\Delta y[\/latex] is the vertical displacement and [latex]\\Delta x[\/latex] is the horizontal displacement.<\/p>\n<p>The graph below indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2,}{y}_{2}\\right)[\/latex] is calculated. The greater the absolute value of the slope, the &#8220;steeper&#8221; the line is.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201021\/CNX_Precalc_Figure_02_01_005n2.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"569\" \/><\/p>\n<p>The units for slope are always [latex]\\dfrac{\\text{units for the output}}{\\text{units for the input}}[\/latex]. Think of the units as the change in output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.<\/p>\n<div class=\"textbox shaded\">\n<h3>Calculating Slope<\/h3>\n<p>The slope [latex]m[\/latex], or rate of change of a linear function between two points [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex] is:<\/p>\n<p>[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<\/div>\n<p>In the following video we show examples of how to find the slope of a line passing through two points.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Find the Slope Given Two Points and Describe the Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/in3NTcx11I8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The population of a city increased from\u00a0[latex]23,400[\/latex] to\u00a0[latex]27,800[\/latex] between\u00a0[latex]2008[\/latex] and\u00a0[latex]2012[\/latex]. Find the change in population per year if we assume the change was constant from\u00a0[latex]2008[\/latex] to\u00a0[latex]2012[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q246268\">Show Solution<\/span><\/p>\n<div id=\"q246268\" class=\"hidden-answer\" style=\"display: none\">\n<p>The rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.<\/p>\n<p>[latex]\\dfrac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ people per year}[\/latex]<\/p>\n<p>The population increased by\u00a0[latex]1,100[\/latex] people per year.<\/p>\n<p>Because we are told that the population increased, we expect the slope to be positive. This positive slope we calculated is therefore reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, we show an example where we determine the increase in cost for producing solar panels given two data points.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Slope Application Involving Production Costs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4RbniDgEGE4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>The slope and y-intercept of a linear equation<\/h2>\n<p>Another way to graph a linear function is by using its slope [latex]m[\/latex] and [latex]y[\/latex]-intercept.<\/p>\n<div class=\"textbox shaded\">\n<h3>SLOPE-INTERCEPT Form<\/h3>\n<p>The linear equation [latex]y=mx+b[\/latex] is in slope-intercept form.<\/p>\n<ul>\n<li>[latex]b[\/latex] is the [latex]y[\/latex]-intercept of the graph and indicates the point [latex](0,b)[\/latex] at which the graph crosses the [latex]y[\/latex]-axis.<\/li>\n<li>[latex]m[\/latex] is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<\/div>\n<p>Consider the function [latex]f\\left(x\\right)=\\dfrac{1}{2}x+1[\/latex]. The function is in slope-intercept form, so the slope is [latex]m=\\dfrac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The [latex]y[\/latex]<em>&#8211;<\/em>intercept is the point on the graph where [latex]x=0[\/latex]. The graph of this line crosses the [latex]y[\/latex]-axis at\u00a0[latex](0, 1)[\/latex]. Begin graphing by plotting the point\u00a0[latex](0, 1)[\/latex]. The slope is [latex]m=\\dfrac{\\text{rise}}{\\text{run}}[\/latex], so for [latex]m=\\dfrac{1}{2}[\/latex], the &#8220;rise&#8221; is\u00a0[latex]1[\/latex] and the &#8220;run&#8221; from left to right is\u00a0[latex]2[\/latex]. Then starting from the [latex]y[\/latex]<em>&#8211;<\/em>intercept\u00a0[latex](0, 1)[\/latex], &#8220;rise&#8221; [latex]1[\/latex] unit and &#8220;run&#8221; [latex]2[\/latex] units to the right. (This is equivalent to &#8220;run&#8221; [latex]2[\/latex] units to the right and then &#8220;rise&#8221; [latex]1[\/latex] unit.) Repeat until you have a few points and then draw a line through the points as shown in the graph below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201048\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f\\left(x\\right)=-\\dfrac{2}{3}x+5[\/latex] using the [latex]y[\/latex]<em>&#8211;<\/em>intercept and slope.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421669\">Show Solution<\/span><\/p>\n<div id=\"q421669\" class=\"hidden-answer\" style=\"display: none\">\n<p>Evaluate the function at [latex]x=0[\/latex]\u00a0to find the [latex]y[\/latex]<em>&#8211;<\/em>intercept. The output value when [latex]x=0[\/latex] is\u00a0[latex]5[\/latex], so the graph will cross the [latex]y[\/latex]-axis at\u00a0[latex](0, 5)[\/latex].<\/p>\n<p>According to the equation for the function, the slope of the line is [latex]-\\dfrac{2}{3}[\/latex]. This tells us that for each vertical decrease in the &#8220;rise&#8221; of [latex]\u20132[\/latex] units, the &#8220;run&#8221; increases by\u00a0[latex]3[\/latex] units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in the graph below. From the initial value\u00a0[latex](0, 5)[\/latex], we move down\u00a0[latex]2[\/latex] units and to the right\u00a0[latex]3[\/latex] units. (Which is equivalent to moving up [latex]2[\/latex] units and left [latex]3[\/latex] units.) We can extend the line to the left and right by using this relationship to plot additional points and then drawing a line through the points.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201050\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/><\/p>\n<p>The graph slants downward from left to right, which means it has a negative slope as expected.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm288727\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288727&theme=oea&iframe_resize_id=ohm288727&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show another example of how to graph a linear function given the y-intercepts and the slope.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex: Graph a Line and ID the Slope and Intercepts (Fraction Slope)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/N6lEPh11gk8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\"><\/div>\n<p>In the following video, we show an example of how to write the equation of a line given its graph.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex 1:  Find the Equation of a Line in Slope Intercept Form Given the Graph of a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mmWf_oLTNSQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Point-Slope Form<\/h2>\n<p>If we have the slope of a line and a point on the line, we can write the equation of the line.<\/p>\n<div class=\"textbox shaded\">\n<h3>Point-Slope Form<\/h3>\n<p>Given a point [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] on a line and the slope of the line, [latex]m[\/latex], point-slope form will give the following equation of a line: [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<\/div>\n<p>In this example, we will start with a given slope.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] that passes through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q524449\">Show Solution<\/span><\/p>\n<div id=\"q524449\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using point-slope form, substitute [latex]\\color{green}{-3}[\/latex] for [latex]m[\/latex] and the point [latex]\\color{blue}{\\left(4,8\\right)}[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\ y - \\color{blue}{8}&=\\color{green}{-3}\\left(x - \\color{blue}{4}\\right)\\\\ y - 8&=-3x+12\\\\ y&=-3x+20\\end{align}[\/latex]<\/p>\n<p>Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows an example of how to write the equation for a line given its slope and a point on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vut5b2fRQQ0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Finding the Equation of a Line Given Two Points<\/h2>\n<p>If we start with two points, we can first find the slope of a line through those two points, and then find the equation of the line that passes through them.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write an equation for a linear function [latex]f[\/latex] given its graph shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (negative 2, negative 4) and (0,2).\" width=\"369\" height=\"378\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728685\">Show Solution<\/span><\/p>\n<div id=\"q728685\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135536538\">Identify two points on the line such as\u00a0[latex]\\color{Green}{(0, 2)}[\/latex] and\u00a0[latex]\\color{Blue}{(\u20132, \u20134)}[\/latex]. Use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\" style=\"text-align: left;\">[latex]\\begin{align}m&=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\&=\\dfrac{\\color{Blue}{-4} - \\color{Green}{2}}{\\color{Blue}{-2} - \\color{Green}{0}}\\\\&=\\dfrac{-6}{-2}\\\\&=3\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\y-\\left(\\color{Blue}{-4}\\right)&=3\\left(x-\\left(\\color{Blue}{-2}\\right)\\right)\\\\ y+4&=3\\left(x+2\\right)\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<p>[latex]\\begin{align}y+4&=3\\left(x+2\\right)\\\\y+4&=3x+6\\\\ y&=3x+2\\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Examining the graph above, we see that the line crosses the [latex]y[\/latex]-axis at the point [latex](0,2)[\/latex] (the [latex]y[\/latex]-intercept), so [latex]b=2[\/latex] in the equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the equation of the line that passes through the points [latex]\\left(\\color{blue}{3,4}\\right)[\/latex] and [latex]\\left(\\color{green}{0,-3}\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q249539\">Show Solution<\/span><\/p>\n<div id=\"q249539\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we calculate the slope using the slope formula and two points.<\/p>\n<p>[latex]\\begin{align}m&=\\dfrac{\\color{green}{-3} - \\color{blue}{4}}{\\color{green}{0} - \\color{blue}{3}}\\\\ &=\\dfrac{-7}{-3}\\\\ &=\\dfrac{7}{3}\\end{align}[\/latex]<\/p>\n<p>Next, we use point-slope form with the slope of [latex]\\dfrac{7}{3}[\/latex] and either point. Let us pick the point [latex]\\left(\\color{blue}{3,4}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}y - \\color{blue}{4}&=\\dfrac{7}{3}\\left(x - \\color{blue}{3}\\right)\\\\ y - 4&=\\dfrac{7}{3}x - 7\\\\ y&=\\dfrac{7}{3}x - 3\\end{align}[\/latex]<\/p>\n<p>In slope-intercept form, the equation is written as [latex]y=\\dfrac{7}{3}x - 3[\/latex].<\/p>\n<p>To prove that either point can be used, let us use the second point [latex]\\left(\\color{green}{0,-3}\\right)[\/latex] and see if we get the same equation.<\/p>\n<p>[latex]\\begin{align}    y-\\left(\\color{green}{-3}\\right)&=\\dfrac{7}{3}\\left(x - \\color{green}{0}\\right)\\\\y+3&=\\dfrac{7}{3}x\\\\y&=\\dfrac{7}{3}x - 3\\end{align}[\/latex]<\/p>\n<p>We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next video shows another example of writing the equation of a line given two points on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Ex:  Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ndRpJxdmZJI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Vertical and Horizontal Lines<\/h2>\n<h4>Vertical Lines<\/h4>\n<p>The equation of a <strong>vertical line<\/strong> is given as [latex]x=c[\/latex] where [latex]c[\/latex] is a constant. The slope of a vertical line is undefined, and regardless of the [latex]y[\/latex]<em>&#8211;<\/em>value of any point on the line, the [latex]x[\/latex]<em>&#8211;<\/em>coordinate of the point will be [latex]c[\/latex].<\/p>\n<p>Suppose that we want to find the equation of a line containing the following points: [latex]\\left(-3,-5\\right),\\left(-3,1\\right),\\left(-3,3\\right)[\/latex], and [latex]\\left(-3,5\\right)[\/latex]. First, we will find the slope, choosing two of the given points. In this case, we are choosing the points [latex](-3,3)[\/latex] and [latex](-3,5)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{5 - 3}{-3-\\left(-3\\right)}=\\dfrac{2}{0}[\/latex]<\/p>\n<p>Dividing by zero is undefined so the slope of the line through these points is undefined. Using point-slope form is not possible. Notice that all of the [latex]x[\/latex]<em>&#8211;<\/em>coordinates in the given ordered pairs are the same. Then a vertical line through these points is [latex]x=-3[\/latex]. The line is on the graph below.<\/p>\n<h4>Horizontal Lines<\/h4>\n<p>The equation of a <strong>horizontal line<\/strong> is given as [latex]y=c[\/latex] where [latex]c[\/latex] is a constant. The slope of a horizontal line is zero, and for any [latex]x[\/latex]<em>&#8211;<\/em>value of a point on the line, the corresponding [latex]y[\/latex]<em>&#8211;<\/em>coordinate will be [latex]c[\/latex].<\/p>\n<p>Suppose we want to find the equation of a line that contains the following points: [latex]\\left(-2,-2\\right),\\left(0,-2\\right),\\left(3,-2\\right)[\/latex], and [latex]\\left(5,-2\\right)[\/latex]. We can use point-slope form to find the equation of the line. First, we find the slope using any two points on the line. In this case, we are choosing [latex](-2,-2)[\/latex] and [latex](0,-2)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}m&=\\dfrac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\\\&=\\dfrac{0}{2}\\\\&=0\\end{align}[\/latex]<\/p>\n<p>Use any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y-\\left(-2\\right)&=0\\left(x - 3\\right)\\\\y+2&=0\\\\ y&=-2\\end{align}[\/latex]<\/p>\n<p>The graph (below) is a horizontal line [latex]y=-2[\/latex]. Notice that all of the [latex]y[\/latex]<em>&#8211;<\/em>coordinates are the same in the list of ordered pairs.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\">The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write the equation of the line that passes through the points [latex]\\left(1,-3\\right)[\/latex] and [latex]\\left(1,4\\right).[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346281\">Show Solution<\/span><\/p>\n<div id=\"q346281\" class=\"hidden-answer\" style=\"display: none\">\n<p>The [latex]x[\/latex]<em>&#8211;<\/em>coordinate of both points is\u00a0[latex]1.[\/latex] Therefore, we have a vertical line; [latex]x=1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graphing a Linear Function: Another Approach<\/h2>\n<p>Another option for graphing linear functions is to use a vertical shift of the function.<\/p>\n<h3>Vertical Shift<\/h3>\n<p>In a linear function [latex]f\\left(x\\right)=mx+b[\/latex], the [latex]b[\/latex] acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice in the figure below\u00a0that adding a value of [latex]b[\/latex] to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of [latex]f[\/latex] a total of [latex]b[\/latex] units up if [latex]b[\/latex] is positive and [latex]|b|[\/latex] units down if [latex]b[\/latex] is negative.\u00a0The graph below illustrates vertical shifts of the linear function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201052\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"Coordinate system of 5 parallel lines labeled: y = x , y = x plus 2, y = x plus 4, y = x minus 2, and y = x minus 4.\" width=\"900\" height=\"759\" \/><\/p>\n<p>In this next example, we will start with the function [latex]f(x)=\\dfrac{1}{2}x[\/latex] and use a vertical shift to graph [latex]f\\left(x\\right)=\\dfrac{1}{2}x - 3[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f\\left(x\\right)=\\dfrac{1}{2}x - 3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q208708\">Show Solution<\/span><\/p>\n<div id=\"q208708\" class=\"hidden-answer\" style=\"display: none\">\n<p>First graph [latex]y=\\dfrac{1}{2}x[\/latex] by starting at the [latex]y[\/latex]-intercept of [latex](0,0)[\/latex] (this is also the [latex]x[\/latex]-intercept). Then &#8220;rise&#8221; [latex]1[\/latex] unit and &#8220;run&#8221; [latex]2[\/latex] units to the right to plot a point at [latex](2,1)[\/latex]. Draw a line through those points.<\/p>\n<p>Next, shift the line vertically down [latex]3[\/latex] units to graph the line [latex]y=\\dfrac{1}{2}x-3[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201055\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Coordinate system of 2 parallel lines. An arrow is drawn from the line labeled y = one half x to the line labeled y = one half x minus 3.\" width=\"487\" height=\"377\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"Write the equations of parallel and perpendicular lines\">Write the equation of a line given a point and a parallel or perpendicular line<\/h2>\n<p>The relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.<\/p>\n<p>Let\u2019s start with an example involving parallel lines. <strong>Parallel lines<\/strong> have the same slope.<\/p>\n<h3>Write the equation of a line given a point and a parallel line<\/h3>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is parallel to the line [latex]x\u2013y=5[\/latex] and passes through the point [latex](\u22122,1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q763534\">Show Solution<\/span><\/p>\n<div id=\"q763534\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the line you want to be parallel to into the\u00a0[latex]y=mx+b[\/latex] form, if needed.<\/p>\n<p>[latex]\\begin{align}x\u2013y&=5\\\\\u2212y&=\u2212x+5\\\\y&=x\u20135\\end{align}[\/latex]<\/p>\n<p>Identify the slope of the given line.<\/p>\n<p>In the equation above, [latex]m=1[\/latex].\u00a0 Therefore, the slope is [latex]1[\/latex].<\/p>\n<p>To find the slope of a parallel line, use the same slope.<\/p>\n<p>The slope of the parallel line is [latex]1[\/latex].<\/p>\n<p>Use the method for writing an equation from the slope and a point on the line. We will use the Point-Slope Form and substitute 1 for [latex]m[\/latex], and the point [latex]\\left(\\color{green}{\u22122,1}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] .<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\y \\color{green}{-1}&=1\\left(x-(\\color{green}{-2})\\right)\\\\y - 1&=1\\left(x + 2\\right)\\\\y-1&=x+2\\\\y&=x+3\\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The answer is\u00a0[latex]y=x+3.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 class=\"yt watch-title-container\"><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Determine the Equation of a Line Parallel to a Line in General Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TQKz2XHI09E?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Write the equation of a line given a point and a perpendicular line<\/h3>\n<p>When you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are <strong>perpendicular<\/strong> if the slope of one is the <strong>opposite\u00a0reciprocal<\/strong> of the slope of the other.\u00a0To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal. \u00a0In other words, flip it and change the sign.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that passes through the point [latex](1,5)[\/latex] and is perpendicular to the line [latex]y=2x\u2013 6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604282\">Show Solution<\/span><\/p>\n<div id=\"q604282\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify the slope of the line you want to be perpendicular to.<\/p>\n<p>The given line is written in [latex]y=mx+b[\/latex] form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope is 2.<\/p>\n<p>To find the slope of a perpendicular line, find the reciprocal, [latex]\\displaystyle \\frac{1}{2}[\/latex], then the opposite, [latex]\\displaystyle -\\frac{1}{2}[\/latex].<\/p>\n<p>The slope of the perpendicular line is [latex]\\displaystyle -\\frac{1}{2}[\/latex].<\/p>\n<p>Use the method for writing an equation from the slope and a point on the line. We will use the Point-Slope Form and substitute [latex]-\\frac{1}{2}[\/latex] for [latex]m[\/latex], and the point [latex]\\left(\\color{green}{1,5}\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] .<\/p>\n<p>[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\y -\\color{green}{5}&=-\\frac{1}{2}\\left(x-\\color{green}{1}\\right)\\\\y - 5&= -\\frac{1}{2}\\left(x -1\\right)\\\\y-5&=-\\frac{1}{2}x+\\frac{1}{2}\\\\y&=-\\dfrac{1}{2}x+\\frac{11}{2}\\end{align}[\/latex]<\/p>\n<p>For practice, this time, we will use the alternate method for finding the equation of a line. Substitute [latex]\\displaystyle -\\dfrac{1}{2}[\/latex] for [latex]m[\/latex], and the point [latex]\\color{green}{(1,5)}[\/latex] for [latex]x[\/latex] and [latex]y[\/latex] into slope-intercept form.<\/p>\n<p>[latex]\\displaystyle \\begin{align}y&=mx+b\\\\\\color{green}{5}&=-\\frac{1}{2}(\\color{green}{1})+b\\end{align}[\/latex]<\/p>\n<p>Solve for [latex]b[\/latex].<\/p>\n<p>[latex]\\displaystyle \\begin{align}5&=-\\frac{1}{2}+b\\\\\\frac{11}{2}&=b\\end{align}[\/latex]<\/p>\n<p>Write the equation using the new slope for [latex]m[\/latex] and the [latex]b[\/latex] you just found.<\/p>\n<p>[latex]y=mx+b[\/latex]<\/p>\n<p>[latex]y=-\\dfrac{1}{2}x+\\dfrac{11}{2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=-\\dfrac{1}{2}x+\\dfrac{11}{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h3>Video: Write the equation of a line given a point and a perpendicular line (Alternate Method)<\/h3>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QtvtzKjtowA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Write the equations of lines parallel and perpendicular to horizontal and vertical lines<\/span><\/h2>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is parallel to the line [latex]y=4[\/latex] and passes through the point [latex](0,10)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426450\">Show Solution<\/span><\/p>\n<div id=\"q426450\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the line into [latex]y=mx+b[\/latex]\u00a0form, if needed.<\/p>\n<p>You may notice without doing this that [latex]y=4[\/latex]\u00a0is a horizontal line 4 units above the [latex]x[\/latex]-axis. Because it is horizontal, you know its slope is zero.<\/p>\n<p>[latex]\\begin{align}y&=4\\\\y&=0x+4\\end{align}[\/latex]<\/p>\n<p>Identify the slope of the given line.<\/p>\n<p>In the equation above, [latex]m=0[\/latex] and [latex]b=4[\/latex].<\/p>\n<p>Since [latex]m=0[\/latex], the slope is [latex]0[\/latex]. This is a horizontal line.<\/p>\n<p>To find the slope of a parallel line, use the same slope.<\/p>\n<p>The slope of the parallel line is also [latex]0[\/latex].<\/p>\n<p>Since the parallel line will be a horizontal line, its form is<\/p>\n<p>[latex]y=\\text{a constant}[\/latex]<\/p>\n<p>Since we want this new line to pass through the point [latex](0,10)[\/latex], we will need to write the equation of the new line as:<\/p>\n<p>[latex]y=10[\/latex]<\/p>\n<p>This line is parallel to [latex]y=4[\/latex]\u00a0and passes through [latex](0,10)[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=10[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is perpendicular to the line [latex]y=-3[\/latex] and passes through the point [latex](-2,5)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426550\">Show Solution<\/span><\/p>\n<div id=\"q426550\" class=\"hidden-answer\" style=\"display: none\">\n<p>In the equation above, [latex]m=0[\/latex] and [latex]b=-3[\/latex].<\/p>\n<p>A perpendicular line will have a slope that is the negative reciprocal of the slope of\u00a0[latex]y=-3[\/latex], but\u00a0what does that mean in this case?<\/p>\n<p>The reciprocal of [latex]0[\/latex] is [latex]\\frac{1}{0}[\/latex], but we know that dividing by [latex]0[\/latex] is undefined.<\/p>\n<p>This means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.<\/p>\n<p>The form of a vertical line is [latex]x=\\text{a constant}[\/latex], where every <em>x<\/em>-value on the line is equal to some constant. \u00a0Since we are looking for a line that goes through the point [latex](-2,5)[\/latex], all of the <em>x<\/em>-values on this line must be [latex]-2[\/latex].<\/p>\n<p>The equation of a line passing through [latex](-2,5)[\/latex] that is perpendicular to the horizontal line\u00a0[latex]y=-3[\/latex] is therefore,<\/p>\n<p>[latex]x=-2[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-2[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2 class=\"yt watch-title-container\"><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Qpn3f3wMeIs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Summary<\/span><\/h2>\n<p>The slope-intercept form of a linear equation is written as [latex]y=mx+b[\/latex], where [latex]m[\/latex] is the slope and [latex]y[\/latex] is the value of [latex]y[\/latex] at the [latex]y[\/latex]-intercept, which can be written as [latex](0,b)[\/latex]. When you know the slope and the [latex]y[\/latex]-intercept of a line you can use the slope-intercept form to immediately write the equation of that line. The point-slope form, [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex], can be used to write the equation of a line when you know the slope and a point on the line or when you know two points on the line.<\/p>\n<p>When lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a 90\u00b0 angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[\/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.<\/p>\n","protected":false},"author":773621,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-540","chapter","type-chapter","status-publish","hentry"],"part":24,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/540","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/773621"}],"version-history":[{"count":39,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/540\/revisions"}],"predecessor-version":[{"id":2104,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/540\/revisions\/2104"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/24"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/540\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=540"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=540"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=540"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=540"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}