{"id":1786,"date":"2023-10-12T00:32:10","date_gmt":"2023-10-12T00:32:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-writing-equations-of-lines\/"},"modified":"2024-08-27T01:30:33","modified_gmt":"2024-08-27T01:30:33","slug":"introduction-writing-equations-of-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-writing-equations-of-lines\/","title":{"raw":"Linear Functions (part 1)","rendered":"Linear Functions (part 1)"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use slope-intercept form to plot and write equations of lines.<\/li>\r\n \t<li>Use point-slope form to write the equation of a line.<\/li>\r\n \t<li>Write the equation of a line in standard form.<\/li>\r\n \t<li>Recognize vertical and horizontal lines from their graphs and equations.<\/li>\r\n \t<li>Determine whether two lines are parallel or perpendicular.<\/li>\r\n \t<li>Find the equations of parallel and perpendicular lines.<\/li>\r\n \t<li>Write the equations of lines that are parallel or perpendicular to a given line.<\/li>\r\n<\/ul>\r\n<\/div>\r\nNow that we have learned how to plot points on a coordinate plane and graph linear equations, we can begin to analyze the equations of lines and evaluate\u00a0the different characteristics\u00a0of these lines. In this section, we will learn about the commonly used forms for writing linear equations and the properties\u00a0of lines that can be determined from their equations.\r\n\r\nFor example, without creating a table of values, you will be able to match each equation below to its corresponding graph. You will also be able to explain the similarities and differences of each line, how they relate to each other, and why they behave that way.\r\n\r\n&nbsp;\r\n\r\n<img class=\" wp-image-4156 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/04\/13224212\/Screen-Shot-2017-04-13-at-3.42.26-PM-300x270.png\" alt=\"6 lines graphed on a coordinate plane.\" width=\"441\" height=\"397\" \/>\r\n\r\n(a) [latex]y=3x+2[\/latex]\r\n\r\n(b) [latex]y-4=-\\frac{1}{2}(x+2)[\/latex]\r\n\r\n(c) [latex]x=5[\/latex]\r\n\r\n(d) [latex]y=-2[\/latex]\r\n\r\n(e) [latex]3x=y+1[\/latex]\r\n\r\n(f) [latex]2y-3x=6[\/latex]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<h2>Writing Equations of Lines<\/h2>\r\n<h3>Slope-Intercept Form<\/h3>\r\nPerhaps the most familiar form of a linear equation is slope-intercept form written as [latex]y=mx+b[\/latex], where [latex]m[\/latex] is the slope and [latex](0,b)[\/latex] is the [latex]y[\/latex]-intercept. Let us begin with the slope.\r\n<h3>The Slope of a Line<\/h3>\r\nThe <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.\r\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\nIf the slope is positive, the line slants upward to the right. If the slope is negative, the line slants downward to the right. As the slope increases, the line becomes steeper. Some examples are shown below. The lines indicate the following slopes: [latex]m=-3[\/latex], [latex]m=2[\/latex], and [latex]m=\\frac{1}{3}[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/11185922\/CNX_CAT_Figure_02_02_002.jpg\" alt=\"Coordinate plane with the x and y axes ranging from negative 10 to 10. Three linear functions are plotted: y = negative 3 times x minus 2; y = 2 times x plus 1; and y = x over 3 plus 2.\" width=\"487\" height=\"442\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Slope of a Line<\/h3>\r\nThe slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:\r\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\r\nFind the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].\r\n[reveal-answer q=\"688301\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688301\"]\r\n\r\nWe substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&amp;=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&amp;=\\frac{4}{-7}\\hfill \\\\ \\hfill&amp;=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\r\nThe slope is [latex]-\\frac{4}{7}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nIt does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the slope of the line that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"196055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196055\"]\r\n\r\nslope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1719&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying the Slope and <em>y-<\/em>intercept of a Line Given an Equation<\/h3>\r\nIdentify the slope and <em>y-<\/em>intercept given the equation [latex]y=-\\frac{3}{4}x - 4[\/latex].\r\n[reveal-answer q=\"757424\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"757424\"]\r\n\r\nAs the line is in [latex]y=mx+b[\/latex] form, the given line has a slope of [latex]m=-\\frac{3}{4}[\/latex]. The <em>y-<\/em>intercept is [latex]b=-4[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe <em>y<\/em>-intercept is the point at which the line crosses the <em>y-<\/em>axis. On the <em>y-<\/em>axis, [latex]x=0[\/latex]. We can always identify the <em>y-<\/em>intercept when the line is in slope-intercept form, as it will always equal <em>b.<\/em> Or, just substitute [latex]x=0[\/latex] and solve for <em>y.<\/em>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>The Point-Slope Form<\/h3>\r\nGiven the slope and one point on a line, we can find the equation of the line using point-slope form.\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\nThis is an important formula, as it will be used in other areas of College Algebra and often in Calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Point-Slope Formula<\/h3>\r\nGiven one point and the slope, using point-slope form will lead to the equation of a line:\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n[reveal-answer q=\"201330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201330\"]\r\n\r\nUsing point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\n<\/div>\r\n<div>\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<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"634647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634647\"]\r\n\r\n[latex]y=4x - 3[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110942&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Passing Through Two Given Points<\/h3>\r\nFind the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n[reveal-answer q=\"975043\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"975043\"]\r\n\r\nFirst, we calculate the slope using the slope formula and two points.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ m=\\frac{-3 - 4}{0 - 3}\\hfill \\\\ =\\frac{-7}{-3}\\hfill \\\\ =\\frac{7}{3}\\hfill \\end{array}[\/latex]<\/div>\r\nNext, we use point-slope form with the slope of [latex]\\frac{7}{3}[\/latex] and either point. Let\u2019s pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=\\frac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\frac{7}{3}x - 7\\hfill&amp; \\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\r\nIn slope-intercept form, the equation is written as [latex]y=\\frac{7}{3}x - 3[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\n<div>\r\n\r\nTo prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\frac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\frac{7}{3}x\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\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<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Standard Form of a Line<\/h3>\r\nAnother way that we can represent the equation of a line is in <strong>standard form<\/strong>. Standard form is given as\r\n<div style=\"text-align: center;\">[latex]Ax+By=C[\/latex]<\/div>\r\nwhere [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x\u00a0<\/em>and <em>y-<\/em>terms are on one side of the equal sign and the constant term is on the other side.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line and Writing It in Standard Form<\/h3>\r\nFind the equation of the line with [latex]m=-6[\/latex] and passing through the point [latex]\\left(\\frac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.\r\n[reveal-answer q=\"111657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111657\"]\r\n\r\nWe begin by using point-slope form.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\frac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nFrom here, we multiply through by 2 as no fractions are permitted in standard form. Then we move both variables to the left aside of the equal sign and move the constants to the right.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\frac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/div>\r\nThis equation is now written in standard form.\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\nFind the equation of the line in standard form with slope [latex]m=-\\frac{1}{3}[\/latex] which passes through the point [latex]\\left(1,\\frac{1}{3}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"3712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3712\"][latex]x+3y=2[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110946&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Vertical and Horizontal Lines<\/h3>\r\nThe equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a <strong>vertical line<\/strong> is given as\r\n<div style=\"text-align: center;\">[latex]x=c[\/latex]<\/div>\r\nwhere <em>c <\/em>is a constant. The slope of a vertical line is undefined, and regardless of the <em>y-<\/em>value of any point on the line, the <em>x-<\/em>coordinate of the point will be <em>c<\/em>.\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.\r\n<div style=\"text-align: center;\">[latex]m=\\frac{5 - 3}{-3-\\left(-3\\right)}=\\frac{2}{0}[\/latex]<\/div>\r\nZero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the <em>x-<\/em>coordinates are the same and we find a vertical line through [latex]x=-3[\/latex].\r\n\r\nThe equation of a <strong>horizontal line<\/strong> is given as\r\n<div style=\"text-align: center;\">[latex]y=c[\/latex]<\/div>\r\nwhere <em>c <\/em>is a constant. The slope of a horizontal line is zero, and for any <em>x-<\/em>value of a point on the line, the <em>y-<\/em>coordinate will be <em>c<\/em>.\r\n\r\nSuppose we want to find the equation of a line that contains the following set of 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. First, we find the slope using any two points on the line.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m=\\frac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ =\\frac{0}{2}\\hfill \\\\ =0\\hfill \\end{array}[\/latex]<\/div>\r\nUse any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in the formula, or use the <em>y<\/em>-intercept.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/div>\r\nThe graph is a horizontal line through [latex]y=-2[\/latex]. Notice that all of the <em>y-<\/em>coordinates are the same.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/11185925\/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 <i>x<\/i> = \u22123 is a vertical line. The line <i>y<\/i> = \u22122 is a horizontal line.[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Passing Through the Given Points<\/h3>\r\nFind the equation of the line passing through the given points: [latex]\\left(1,-3\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].\r\n[reveal-answer q=\"122244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"122244\"]\r\n\r\nThe <em>x-<\/em>coordinate of both points is 1. Therefore, we have a vertical line, [latex]x=1[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the equation of the line passing through [latex]\\left(-5,2\\right)[\/latex] and [latex]\\left(2,2\\right)[\/latex].\r\n\r\n[reveal-answer q=\"864962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"864962\"]\r\n\r\nHorizontal line: [latex]y=2[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110951&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110952&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Parallel and Perpendicular Lines<\/h2>\r\nParallel lines have the same slope and different <em>y-<\/em>intercepts. Lines that are <strong>parallel<\/strong> to each other will never intersect. For example, the figure below\u00a0shows the graphs of various lines with the same slope, [latex]m=2[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232102\/CNX_CAT_Figure_02_02_004.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.\" width=\"487\" height=\"593\" \/> Parallel lines have slopes that are the same.[\/caption]\r\n\r\nAll of the lines shown in the graph are parallel because they have the same slope and different <em>y-<\/em>intercepts.\r\n\r\nLines that are <strong>perpendicular<\/strong> intersect to form a [latex]{90}^{\\circ }[\/latex] angle. The slope of one line is the negative <strong>reciprocal<\/strong> of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\\cdot {m}_{2}=-1[\/latex]. For example, the figure below shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\\frac{1}{3}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{m}_{1}\\cdot {m}_{2}=-1\\hfill \\\\ \\text{ }3\\cdot \\left(-\\frac{1}{3}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232106\/CNX_CAT_Figure_02_02_005.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x\/3 minus 2. Their intersection is marked by a box to show that it is a right angle.\" width=\"487\" height=\"329\" \/> Perpendicular lines have slopes that are negative reciprocals of each other.[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing Two Equations and Determining Whether the Lines are Parallel, Perpendicular, or Neither<\/h3>\r\nGraph the equations of the given lines and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[\/latex] and [latex]3x - 4y=8[\/latex].\r\n[reveal-answer q=\"130903\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130903\"]\r\n\r\nThe first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.\r\n\r\nFirst equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=-4x+3\\hfill \\\\ y=-\\frac{4}{3}x+1\\hfill \\end{array}[\/latex]<\/p>\r\nSecond equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x - 4y=8\\hfill \\\\ -4y=-3x+8\\hfill \\\\ y=\\frac{3}{4}x - 2\\hfill \\end{array}[\/latex]<\/p>\r\nSee the graph of both lines in the graph below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232109\/CNX_CAT_Figure_02_02_006.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x\/3 plus 1 and y = 3 times x\/4 minus 2. A box is placed at the intersection to note that it forms a right angle.\" width=\"487\" height=\"366\" \/>\r\n\r\nFrom the graph, we can see that the lines appear perpendicular, but we must compare the slopes.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=-\\frac{4}{3}\\hfill \\\\ {m}_{2}=\\frac{3}{4}\\hfill \\\\ {m}_{1}\\cdot {m}_{2}=\\left(-\\frac{4}{3}\\right)\\left(\\frac{3}{4}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\r\nThe slopes are negative reciprocals of each other confirming that the lines are perpendicular.\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\nGraph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[\/latex] and [latex]2y=x+4[\/latex].\r\n\r\n[reveal-answer q=\"727314\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"727314\"]\r\n\r\nParallel lines. Write the equations in slope-intercept form.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200328\/CNX_CAT_Figure_02_02_007.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x\/2 plus 5 and y = x\/2 plus 2. The lines do not cross.\" \/>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110960&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"750\"><\/iframe>\r\n\r\n<\/div>\r\nIf we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.\r\n<h3>Writing Equations of Parallel Lines<\/h3>\r\nSuppose we are given the following equation:\r\n<p style=\"text-align: center;\">[latex]y=3x+1[\/latex]<\/p>\r\nWe know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to [latex]y=3x+1[\/latex]. So all of the following lines will be parallel to the given line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+6\\hfill &amp; \\\\ y=3x+1\\hfill &amp; \\\\ y=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nSuppose then we want to write the equation of a line that is parallel to [latex]y=3x+6[\/latex]\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with point-slope form of a line and then rewrite it in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill &amp; \\\\ y - 7=3\\left(x - 1\\right)\\hfill &amp; \\\\ y - 7=3x - 3\\hfill &amp; \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]y=3x+4[\/latex] is parallel to [latex]y=3x+1[\/latex] and passes through the point (1, 7).\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a LINE, write the equation of a line parallel to the given line that passes through A given point<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the line.<\/li>\r\n \t<li>Substitute the given values into either point-slope form or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\r\nFind a line parallel to the graph of [latex]y=3x+6[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"672987\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"672987\"]\r\n\r\nThe slope of the given line is 3. If we choose to use slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and y\u00a0= 0 into slope-intercept form to find the <em>y-<\/em>intercept.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+b\\hfill &amp; \\\\ \\text{}0=3\\left(3\\right)+b\\hfill &amp; \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\r\nThe line parallel to\u00a0[latex]y=3x+6[\/latex]\u00a0that passes through (3, 0) is [latex]y=3x - 9[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/>\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<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110970&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Writing Equations of Perpendicular Lines<\/h3>\r\nWe can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following line:\r\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\r\nThe slope of the line is 2 and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to [latex]y=2x+4[\/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=-\\frac{1}{2}x+4\\hfill &amp; \\\\ y=-\\frac{1}{2}x+2\\hfill &amp; \\\\ y=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nAs before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to [latex]y=2x+4[\/latex]\u00a0and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into slope-intercept form and solving for <em>b<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=mx+b\\hfill &amp; \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill &amp; \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill &amp; \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]y=-\\frac{1}{2}x+2[\/latex].\r\n\r\nSo [latex]y=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]y=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong>\r\n\r\n<em>No. For two perpendicular linear functions, the product of their slopes is \u20131. As you will learn later, a vertical line is not a function so the definition is not contradicted.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a LINE, write the equation of a line Perpendicular to the given line that passes through A given point<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the given line.<\/li>\r\n \t<li>Determine the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\r\nFind the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"566588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"566588\"]\r\n\r\nThe original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}y=-\\frac{1}{3}x+b\\hfill &amp; \\\\ \\text{}0=-\\frac{1}{3}\\left(3\\right)+b\\hfill &amp; \\\\ \\text{}1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{array}[\/latex]<\/p>\r\nThe line perpendicular to [latex]y=3x+3[\/latex]\u00a0that passes through (3, 0) is [latex]y=-\\frac{1}{3}x+1[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the two lines is shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/>\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\nGiven the line [latex]y=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is\r\n<ol>\r\n \t<li>parallel to\u00a0<em>y<\/em><\/li>\r\n \t<li>perpendicular to\u00a0<em>y<\/em><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"924175\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924175\"]\r\n<ol>\r\n \t<li>[latex]y=2x[\/latex] is parallel<\/li>\r\n \t<li>[latex]y=-\\frac{1}{2}x[\/latex] is perpendicular<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110971&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two points on a line, write the equation of A perpendicular line that passes through A Third point<\/h3>\r\n<ol>\r\n \t<li>Determine the slope of the line passing through the points.<\/li>\r\n \t<li>Find the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3 style=\"text-align: center;\">Example: Finding the Equation of a Perpendicular Line<\/h3>\r\nA line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).\r\n\r\n[reveal-answer q=\"119051\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"119051\"]\r\nFrom the two points of the given line, we can calculate the slope of that line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill &amp; =\\frac{-1}{6}\\hfill &amp; =-\\frac{1}{6}\\hfill \\end{array}[\/latex]<\/p>\r\nFind the negative reciprocal of the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill &amp; =-1\\left(-\\frac{6}{1}\\right)\\hfill &amp; =6\\hfill \\end{array}[\/latex]<\/p>\r\nWe can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is:\r\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA line passes through the points (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point (6, 4).\r\n\r\n[reveal-answer q=\"69699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"69699\"]\r\n\r\n[latex]y=-\\frac{1}{3}x+6[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Writing the Equations of Lines Parallel or Perpendicular to a Given Line<\/h3>\r\nAs we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use\u00a0<strong>point-slope form<\/strong>\u00a0to write the equation of the new line.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Equation of a Line Parallel to a Given Line<\/h3>\r\nWrite the equation of line parallel to a [latex]5x+3y=1[\/latex] which passes through the point [latex]\\left(3,5\\right)[\/latex].\r\n[reveal-answer q=\"72460\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"72460\"]\r\n\r\nFirst, we will write the equation in slope-intercept form to find the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x+3y=1\\hfill \\\\ 3y=-5x+1\\hfill \\\\ y=-\\frac{5}{3}x+\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe slope is [latex]m=-\\frac{5}{3}[\/latex]. The <em>y-<\/em>intercept is [latex]\\frac{1}{3}[\/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope.\r\n\r\nThe one exception is that if the <em>y-<\/em>intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point in point-slope form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 5=-\\frac{5}{3}\\left(x - 3\\right)\\hfill \\\\ y - 5=-\\frac{5}{3}x+5\\hfill \\\\ y=-\\frac{5}{3}x+10\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation of the line is [latex]y=-\\frac{5}{3}x+10[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232111\/CNX_CAT_Figure_02_02_008.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 2 to 12 in intervals of 2. Two functions are graphed on the same plot: y = negative 5 times x\/3 plus 1\/3 and y = negative 5 times x\/3 plus 10. The lines do not cross.\" width=\"487\" height=\"329\" \/>\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\nFind the equation of the line parallel to [latex]5x=7+y[\/latex] which passes through the point [latex]\\left(-1,-2\\right)[\/latex].\r\n\r\n[reveal-answer q=\"985300\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"985300\"]\r\n\r\n[latex]y=5x+3[\/latex][\/hidden-answer]\r\n<iframe id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1436&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\r\nFind the equation of the line perpendicular to [latex]5x - 3y+4=0[\/latex] which goes through the point [latex]\\left(-4,1\\right)[\/latex].\r\n[reveal-answer q=\"408679\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"408679\"]\r\n\r\nThe first step is to write the equation in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x - 3y+4=0\\hfill \\\\ -3y=-5x - 4\\hfill \\\\ y=\\frac{5}{3}x+\\frac{4}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nWe see that the slope is [latex]m=\\frac{5}{3}[\/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal or [latex]-\\frac{3}{5}[\/latex]. Next, we use point-slope form with this new slope and the given point.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=-\\frac{3}{5}\\left(x-\\left(-4\\right)\\right)\\hfill \\\\ y - 1=-\\frac{3}{5}x-\\frac{12}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{12}{5}+\\frac{5}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{7}{5}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Given two points, we can find the slope of a line using the slope formula.<\/li>\r\n \t<li>We can identify the slope and <em>y<\/em>-intercept of an equation in slope-intercept form.<\/li>\r\n \t<li>We can find the equation of a line given the slope and a point.<\/li>\r\n \t<li>We can also find the equation of a line given two points. Find the slope and use point-slope form.<\/li>\r\n \t<li>The standard form of a line has no fractions.<\/li>\r\n \t<li>Horizontal lines have a slope of zero and are defined as [latex]y=c[\/latex], where <em>c <\/em>is a constant.<\/li>\r\n \t<li>Vertical lines have an undefined slope (zero in the denominator) and are defined as [latex]x=c[\/latex], where <em>c <\/em>is a constant.<\/li>\r\n \t<li>Parallel lines have the same slope and different <em>y-<\/em>intercepts.<\/li>\r\n \t<li>Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165131990658\" class=\"definition\">\r\n \t<dt><strong>slope<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165131990661\">the change in <em>y-<\/em>values over the change in <em>x-<\/em>values<\/dd>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use slope-intercept form to plot and write equations of lines.<\/li>\n<li>Use point-slope form to write the equation of a line.<\/li>\n<li>Write the equation of a line in standard form.<\/li>\n<li>Recognize vertical and horizontal lines from their graphs and equations.<\/li>\n<li>Determine whether two lines are parallel or perpendicular.<\/li>\n<li>Find the equations of parallel and perpendicular lines.<\/li>\n<li>Write the equations of lines that are parallel or perpendicular to a given line.<\/li>\n<\/ul>\n<\/div>\n<p>Now that we have learned how to plot points on a coordinate plane and graph linear equations, we can begin to analyze the equations of lines and evaluate\u00a0the different characteristics\u00a0of these lines. In this section, we will learn about the commonly used forms for writing linear equations and the properties\u00a0of lines that can be determined from their equations.<\/p>\n<p>For example, without creating a table of values, you will be able to match each equation below to its corresponding graph. You will also be able to explain the similarities and differences of each line, how they relate to each other, and why they behave that way.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4156 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/04\/13224212\/Screen-Shot-2017-04-13-at-3.42.26-PM-300x270.png\" alt=\"6 lines graphed on a coordinate plane.\" width=\"441\" height=\"397\" \/><\/p>\n<p>(a) [latex]y=3x+2[\/latex]<\/p>\n<p>(b) [latex]y-4=-\\frac{1}{2}(x+2)[\/latex]<\/p>\n<p>(c) [latex]x=5[\/latex]<\/p>\n<p>(d) [latex]y=-2[\/latex]<\/p>\n<p>(e) [latex]3x=y+1[\/latex]<\/p>\n<p>(f) [latex]2y-3x=6[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<h2>Writing Equations of Lines<\/h2>\n<h3>Slope-Intercept Form<\/h3>\n<p>Perhaps the most familiar form of a linear equation is slope-intercept form written as [latex]y=mx+b[\/latex], where [latex]m[\/latex] is the slope and [latex](0,b)[\/latex] is the [latex]y[\/latex]-intercept. Let us begin with the slope.<\/p>\n<h3>The Slope of a Line<\/h3>\n<p>The <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.<\/p>\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<p>If the slope is positive, the line slants upward to the right. If the slope is negative, the line slants downward to the right. As the slope increases, the line becomes steeper. Some examples are shown below. The lines indicate the following slopes: [latex]m=-3[\/latex], [latex]m=2[\/latex], and [latex]m=\\frac{1}{3}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/11185922\/CNX_CAT_Figure_02_02_002.jpg\" alt=\"Coordinate plane with the x and y axes ranging from negative 10 to 10. Three linear functions are plotted: y = negative 3 times x minus 2; y = 2 times x plus 1; and y = x over 3 plus 2.\" width=\"487\" height=\"442\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Slope of a Line<\/h3>\n<p>The slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:<\/p>\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\n<p>Find the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688301\">Show Solution<\/span><\/p>\n<div id=\"q688301\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&=\\frac{4}{-7}\\hfill \\\\ \\hfill&=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\n<p>The slope is [latex]-\\frac{4}{7}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>It does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the slope of the line that passes through the points [latex]\\left(-2,6\\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=\"q196055\">Show Solution<\/span><\/p>\n<div id=\"q196055\" class=\"hidden-answer\" style=\"display: none\">\n<p>slope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1719&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying the Slope and <em>y-<\/em>intercept of a Line Given an Equation<\/h3>\n<p>Identify the slope and <em>y-<\/em>intercept given the equation [latex]y=-\\frac{3}{4}x - 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q757424\">Show Solution<\/span><\/p>\n<div id=\"q757424\" class=\"hidden-answer\" style=\"display: none\">\n<p>As the line is in [latex]y=mx+b[\/latex] form, the given line has a slope of [latex]m=-\\frac{3}{4}[\/latex]. The <em>y-<\/em>intercept is [latex]b=-4[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The <em>y<\/em>-intercept is the point at which the line crosses the <em>y-<\/em>axis. On the <em>y-<\/em>axis, [latex]x=0[\/latex]. We can always identify the <em>y-<\/em>intercept when the line is in slope-intercept form, as it will always equal <em>b.<\/em> Or, just substitute [latex]x=0[\/latex] and solve for <em>y.<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>The Point-Slope Form<\/h3>\n<p>Given the slope and one point on a line, we can find the equation of the line using point-slope form.<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<p>This is an important formula, as it will be used in other areas of College Algebra and often in Calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Point-Slope Formula<\/h3>\n<p>Given one point and the slope, using point-slope form will lead to the equation of a line:<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<\/div>\n<div style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] and passing 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=\"q201330\">Show Solution<\/span><\/p>\n<div id=\"q201330\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<\/div>\n<div>\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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634647\">Show Solution<\/span><\/p>\n<div id=\"q634647\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=4x - 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110942&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Passing Through Two Given Points<\/h3>\n<p>Find the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(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=\"q975043\">Show Solution<\/span><\/p>\n<div id=\"q975043\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we calculate the slope using the slope formula and two points.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ m=\\frac{-3 - 4}{0 - 3}\\hfill \\\\ =\\frac{-7}{-3}\\hfill \\\\ =\\frac{7}{3}\\hfill \\end{array}[\/latex]<\/div>\n<p>Next, we use point-slope form with the slope of [latex]\\frac{7}{3}[\/latex] and either point. Let\u2019s pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=\\frac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\frac{7}{3}x - 7\\hfill& \\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\n<p>In slope-intercept form, the equation is written as [latex]y=\\frac{7}{3}x - 3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<div>\n<p>To prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\frac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\frac{7}{3}x\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\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<\/div>\n<h3>Standard Form of a Line<\/h3>\n<p>Another way that we can represent the equation of a line is in <strong>standard form<\/strong>. Standard form is given as<\/p>\n<div style=\"text-align: center;\">[latex]Ax+By=C[\/latex]<\/div>\n<p>where [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x\u00a0<\/em>and <em>y-<\/em>terms are on one side of the equal sign and the constant term is on the other side.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line and Writing It in Standard Form<\/h3>\n<p>Find the equation of the line with [latex]m=-6[\/latex] and passing through the point [latex]\\left(\\frac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q111657\">Show Solution<\/span><\/p>\n<div id=\"q111657\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by using point-slope form.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\frac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>From here, we multiply through by 2 as no fractions are permitted in standard form. Then we move both variables to the left aside of the equal sign and move the constants to the right.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\frac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/div>\n<p>This equation is now written in standard form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the equation of the line in standard form with slope [latex]m=-\\frac{1}{3}[\/latex] which passes through the point [latex]\\left(1,\\frac{1}{3}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3712\">Show Solution<\/span><\/p>\n<div id=\"q3712\" class=\"hidden-answer\" style=\"display: none\">[latex]x+3y=2[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110946&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Vertical and Horizontal Lines<\/h3>\n<p>The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a <strong>vertical line<\/strong> is given as<\/p>\n<div style=\"text-align: center;\">[latex]x=c[\/latex]<\/div>\n<p>where <em>c <\/em>is a constant. The slope of a vertical line is undefined, and regardless of the <em>y-<\/em>value of any point on the line, the <em>x-<\/em>coordinate of the point will be <em>c<\/em>.<\/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.<\/p>\n<div style=\"text-align: center;\">[latex]m=\\frac{5 - 3}{-3-\\left(-3\\right)}=\\frac{2}{0}[\/latex]<\/div>\n<p>Zero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the <em>x-<\/em>coordinates are the same and we find a vertical line through [latex]x=-3[\/latex].<\/p>\n<p>The equation of a <strong>horizontal line<\/strong> is given as<\/p>\n<div style=\"text-align: center;\">[latex]y=c[\/latex]<\/div>\n<p>where <em>c <\/em>is a constant. The slope of a horizontal line is zero, and for any <em>x-<\/em>value of a point on the line, the <em>y-<\/em>coordinate will be <em>c<\/em>.<\/p>\n<p>Suppose we want to find the equation of a line that contains the following set of 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. First, we find the slope using any two points on the line.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m=\\frac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ =\\frac{0}{2}\\hfill \\\\ =0\\hfill \\end{array}[\/latex]<\/div>\n<p>Use any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in the formula, or use the <em>y<\/em>-intercept.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/div>\n<p>The graph is a horizontal line through [latex]y=-2[\/latex]. Notice that all of the <em>y-<\/em>coordinates are the same.<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/11185925\/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 <i>x<\/i> = \u22123 is a vertical line. The line <i>y<\/i> = \u22122 is a horizontal line.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Passing Through the Given Points<\/h3>\n<p>Find the equation of the line passing through the given 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=\"q122244\">Show Solution<\/span><\/p>\n<div id=\"q122244\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <em>x-<\/em>coordinate of both points is 1. Therefore, we have a vertical line, [latex]x=1[\/latex].<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the equation of the line passing through [latex]\\left(-5,2\\right)[\/latex] and [latex]\\left(2,2\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864962\">Show Solution<\/span><\/p>\n<div id=\"q864962\" class=\"hidden-answer\" style=\"display: none\">\n<p>Horizontal line: [latex]y=2[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110951&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110952&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Parallel and Perpendicular Lines<\/h2>\n<p>Parallel lines have the same slope and different <em>y-<\/em>intercepts. Lines that are <strong>parallel<\/strong> to each other will never intersect. For example, the figure below\u00a0shows the graphs of various lines with the same slope, [latex]m=2[\/latex].<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/17232102\/CNX_CAT_Figure_02_02_004.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.\" width=\"487\" height=\"593\" \/><\/p>\n<p class=\"wp-caption-text\">Parallel lines have slopes that are the same.<\/p>\n<\/div>\n<p>All of the lines shown in the graph are parallel because they have the same slope and different <em>y-<\/em>intercepts.<\/p>\n<p>Lines that are <strong>perpendicular<\/strong> intersect to form a [latex]{90}^{\\circ }[\/latex] angle. The slope of one line is the negative <strong>reciprocal<\/strong> of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\\cdot {m}_{2}=-1[\/latex]. For example, the figure below shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\\frac{1}{3}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{m}_{1}\\cdot {m}_{2}=-1\\hfill \\\\ \\text{ }3\\cdot \\left(-\\frac{1}{3}\\right)=-1\\hfill \\end{array}[\/latex]<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/17232106\/CNX_CAT_Figure_02_02_005.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x\/3 minus 2. Their intersection is marked by a box to show that it is a right angle.\" width=\"487\" height=\"329\" \/><\/p>\n<p class=\"wp-caption-text\">Perpendicular lines have slopes that are negative reciprocals of each other.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing Two Equations and Determining Whether the Lines are Parallel, Perpendicular, or Neither<\/h3>\n<p>Graph the equations of the given lines and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[\/latex] and [latex]3x - 4y=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q130903\">Show Solution<\/span><\/p>\n<div id=\"q130903\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.<\/p>\n<p>First equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=-4x+3\\hfill \\\\ y=-\\frac{4}{3}x+1\\hfill \\end{array}[\/latex]<\/p>\n<p>Second equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x - 4y=8\\hfill \\\\ -4y=-3x+8\\hfill \\\\ y=\\frac{3}{4}x - 2\\hfill \\end{array}[\/latex]<\/p>\n<p>See the graph of both lines in the graph below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232109\/CNX_CAT_Figure_02_02_006.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x\/3 plus 1 and y = 3 times x\/4 minus 2. A box is placed at the intersection to note that it forms a right angle.\" width=\"487\" height=\"366\" \/><\/p>\n<p>From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=-\\frac{4}{3}\\hfill \\\\ {m}_{2}=\\frac{3}{4}\\hfill \\\\ {m}_{1}\\cdot {m}_{2}=\\left(-\\frac{4}{3}\\right)\\left(\\frac{3}{4}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\n<p>The slopes are negative reciprocals of each other confirming that the lines are perpendicular.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Graph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[\/latex] and [latex]2y=x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q727314\">Show Solution<\/span><\/p>\n<div id=\"q727314\" class=\"hidden-answer\" style=\"display: none\">\n<p>Parallel lines. Write the equations in slope-intercept form.<br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200328\/CNX_CAT_Figure_02_02_007.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x\/2 plus 5 and y = x\/2 plus 2. The lines do not cross.\" \/><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110960&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"750\"><\/iframe><\/p>\n<\/div>\n<p>If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n<h3>Writing Equations of Parallel Lines<\/h3>\n<p>Suppose we are given the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]y=3x+1[\/latex]<\/p>\n<p>We know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to [latex]y=3x+1[\/latex]. So all of the following lines will be parallel to the given line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+6\\hfill & \\\\ y=3x+1\\hfill & \\\\ y=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Suppose then we want to write the equation of a line that is parallel to [latex]y=3x+6[\/latex]\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with point-slope form of a line and then rewrite it in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\\\ y - 7=3\\left(x - 1\\right)\\hfill & \\\\ y - 7=3x - 3\\hfill & \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]y=3x+4[\/latex] is parallel to [latex]y=3x+1[\/latex] and passes through the point (1, 7).<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a LINE, write the equation of a line parallel to the given line that passes through A given point<\/h3>\n<ol>\n<li>Find the slope of the line.<\/li>\n<li>Substitute the given values into either point-slope form or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\n<p>Find a line parallel to the graph of [latex]y=3x+6[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q672987\">Show Solution<\/span><\/p>\n<div id=\"q672987\" class=\"hidden-answer\" style=\"display: none\">\n<p>The slope of the given line is 3. If we choose to use slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and y\u00a0= 0 into slope-intercept form to find the <em>y-<\/em>intercept.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+b\\hfill & \\\\ \\text{}0=3\\left(3\\right)+b\\hfill & \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\n<p>The line parallel to\u00a0[latex]y=3x+6[\/latex]\u00a0that passes through (3, 0) is [latex]y=3x - 9[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110970&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Writing Equations of Perpendicular Lines<\/h3>\n<p>We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following line:<\/p>\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\n<p>The slope of the line is 2 and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to [latex]y=2x+4[\/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=-\\frac{1}{2}x+4\\hfill & \\\\ y=-\\frac{1}{2}x+2\\hfill & \\\\ y=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to [latex]y=2x+4[\/latex]\u00a0and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into slope-intercept form and solving for <em>b<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=mx+b\\hfill & \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill & \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill & \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]y=-\\frac{1}{2}x+2[\/latex].<\/p>\n<p>So [latex]y=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]y=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n<p><em>No. For two perpendicular linear functions, the product of their slopes is \u20131. As you will learn later, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a LINE, write the equation of a line Perpendicular to the given line that passes through A given point<\/h3>\n<ol>\n<li>Find the slope of the given line.<\/li>\n<li>Determine the negative reciprocal of the slope.<\/li>\n<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\n<p>Find the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q566588\">Show Solution<\/span><\/p>\n<div id=\"q566588\" class=\"hidden-answer\" style=\"display: none\">\n<p>The original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}y=-\\frac{1}{3}x+b\\hfill & \\\\ \\text{}0=-\\frac{1}{3}\\left(3\\right)+b\\hfill & \\\\ \\text{}1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{array}[\/latex]<\/p>\n<p>The line perpendicular to [latex]y=3x+3[\/latex]\u00a0that passes through (3, 0) is [latex]y=-\\frac{1}{3}x+1[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the two lines is shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the line [latex]y=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\n<ol>\n<li>parallel to\u00a0<em>y<\/em><\/li>\n<li>perpendicular to\u00a0<em>y<\/em><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924175\">Show Solution<\/span><\/p>\n<div id=\"q924175\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]y=2x[\/latex] is parallel<\/li>\n<li>[latex]y=-\\frac{1}{2}x[\/latex] is perpendicular<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110971&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two points on a line, write the equation of A perpendicular line that passes through A Third point<\/h3>\n<ol>\n<li>Determine the slope of the line passing through the points.<\/li>\n<li>Find the negative reciprocal of the slope.<\/li>\n<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3 style=\"text-align: center;\">Example: Finding the Equation of a Perpendicular Line<\/h3>\n<p>A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q119051\">Show Solution<\/span><\/p>\n<div id=\"q119051\" class=\"hidden-answer\" style=\"display: none\">\nFrom the two points of the given line, we can calculate the slope of that line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill & =\\frac{-1}{6}\\hfill & =-\\frac{1}{6}\\hfill \\end{array}[\/latex]<\/p>\n<p>Find the negative reciprocal of the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill & =-1\\left(-\\frac{6}{1}\\right)\\hfill & =6\\hfill \\end{array}[\/latex]<\/p>\n<p>We can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is:<\/p>\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A line passes through the points (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point (6, 4).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q69699\">Show Solution<\/span><\/p>\n<div id=\"q69699\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=-\\frac{1}{3}x+6[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h3>Writing the Equations of Lines Parallel or Perpendicular to a Given Line<\/h3>\n<p>As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use\u00a0<strong>point-slope form<\/strong>\u00a0to write the equation of the new line.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Equation of a Line Parallel to a Given Line<\/h3>\n<p>Write the equation of line parallel to a [latex]5x+3y=1[\/latex] which passes through the point [latex]\\left(3,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q72460\">Show Solution<\/span><\/p>\n<div id=\"q72460\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will write the equation in slope-intercept form to find the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x+3y=1\\hfill \\\\ 3y=-5x+1\\hfill \\\\ y=-\\frac{5}{3}x+\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The slope is [latex]m=-\\frac{5}{3}[\/latex]. The <em>y-<\/em>intercept is [latex]\\frac{1}{3}[\/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope.<\/p>\n<p>The one exception is that if the <em>y-<\/em>intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point in point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 5=-\\frac{5}{3}\\left(x - 3\\right)\\hfill \\\\ y - 5=-\\frac{5}{3}x+5\\hfill \\\\ y=-\\frac{5}{3}x+10\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation of the line is [latex]y=-\\frac{5}{3}x+10[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232111\/CNX_CAT_Figure_02_02_008.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 2 to 12 in intervals of 2. Two functions are graphed on the same plot: y = negative 5 times x\/3 plus 1\/3 and y = negative 5 times x\/3 plus 10. The lines do not cross.\" width=\"487\" height=\"329\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the equation of the line parallel to [latex]5x=7+y[\/latex] which passes through the point [latex]\\left(-1,-2\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q985300\">Show Solution<\/span><\/p>\n<div id=\"q985300\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=5x+3[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1436&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\n<p>Find the equation of the line perpendicular to [latex]5x - 3y+4=0[\/latex] which goes through the point [latex]\\left(-4,1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q408679\">Show Solution<\/span><\/p>\n<div id=\"q408679\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first step is to write the equation in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x - 3y+4=0\\hfill \\\\ -3y=-5x - 4\\hfill \\\\ y=\\frac{5}{3}x+\\frac{4}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>We see that the slope is [latex]m=\\frac{5}{3}[\/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal or [latex]-\\frac{3}{5}[\/latex]. Next, we use point-slope form with this new slope and the given point.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=-\\frac{3}{5}\\left(x-\\left(-4\\right)\\right)\\hfill \\\\ y - 1=-\\frac{3}{5}x-\\frac{12}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{12}{5}+\\frac{5}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{7}{5}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Given two points, we can find the slope of a line using the slope formula.<\/li>\n<li>We can identify the slope and <em>y<\/em>-intercept of an equation in slope-intercept form.<\/li>\n<li>We can find the equation of a line given the slope and a point.<\/li>\n<li>We can also find the equation of a line given two points. Find the slope and use point-slope form.<\/li>\n<li>The standard form of a line has no fractions.<\/li>\n<li>Horizontal lines have a slope of zero and are defined as [latex]y=c[\/latex], where <em>c <\/em>is a constant.<\/li>\n<li>Vertical lines have an undefined slope (zero in the denominator) and are defined as [latex]x=c[\/latex], where <em>c <\/em>is a constant.<\/li>\n<li>Parallel lines have the same slope and different <em>y-<\/em>intercepts.<\/li>\n<li>Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165131990658\" class=\"definition\">\n<dt><strong>slope<\/strong><\/dt>\n<dd id=\"fs-id1165131990661\">the change in <em>y-<\/em>values over the change in <em>x-<\/em>values<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1786\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 1719. <strong>Authored by<\/strong>: Barbara Goldner. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 110942, 110946, 110951, 110952, 110960, 110970, 110971. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 1436. <strong>Authored by<\/strong>: WebWork-Rochester. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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