{"id":136,"date":"2016-10-17T23:54:14","date_gmt":"2016-10-17T23:54:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=136"},"modified":"2017-04-10T20:16:57","modified_gmt":"2017-04-10T20:16:57","slug":"methods-for-plotting-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/methods-for-plotting-linear-equations\/","title":{"raw":"Writing Equations of Lines","rendered":"Writing Equations of Lines"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Use the slope intercept form to plot and write equations of lines<\/li>\r\n \t<li>Use the point-slope formula 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<\/ul>\r\n<\/div>\r\n<h2>Slope-Intercept Form<\/h2>\r\nPerhaps the most familiar form of a linear equation is the slope-intercept form, written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. 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 to the right. If the slope is negative, the line slants to the left. 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\" data-media-type=\"image\/jpg\" \/>\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\"]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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"196055\"][latex]x=-5[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"757424\"]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].\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><\/h3>\r\n<h2>The Point-Slope Formula<\/h2>\r\nGiven the slope and one point on a line, we can find the equation of the line using the point-slope formula.\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, the point-slope formula 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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"201330\"]\r\n\r\nUsing the point-slope formula, 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><\/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\"]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:\/\/www.myopenmath.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\"]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 the point-slope formula 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;\\text{Distribute the }\\frac{7}{3}.\\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<h2>Standard Form of a Line<\/h2>\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- <\/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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"111657\"]\r\n\r\nWe begin using the point-slope formula.\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, and then 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] and passing through the point [latex]\\left(1,\\frac{1}{3}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"3712\"]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:\/\/www.myopenmath.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<h2>Vertical and Horizontal Lines<\/h2>\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 the point-slope formula. 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 the point-slope formula. 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\" data-media-type=\"image\/jpg\" \/> 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 key-takeaways\">\r\n<h3>try it now<\/h3>\r\nUse the Desmos calculator to graph the following:\r\n<ol>\r\n \t<li>A horizontal line that passes through the point (-5,2)<\/li>\r\n \t<li>A vertical line that passes through the point (3,3)<\/li>\r\n<\/ol>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n\r\n<\/div>\r\n&nbsp;\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"122244\"]The <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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"864962\"]Horizontal line: [latex]y=2[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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:\/\/www.myopenmath.com\/multiembedq.php?id=110952&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Use the slope intercept form to plot and write equations of lines<\/li>\n<li>Use the point-slope formula 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<\/ul>\n<\/div>\n<h2>Slope-Intercept Form<\/h2>\n<p>Perhaps the most familiar form of a linear equation is the slope-intercept form, written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. 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 to the right. If the slope is negative, the line slants to the left. 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\" data-media-type=\"image\/jpg\" \/><\/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\">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\">Solution<\/span><\/p>\n<div id=\"q196055\" class=\"hidden-answer\" style=\"display: none\">[latex]x=-5[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\">Solution<\/span><\/p>\n<div id=\"q757424\" class=\"hidden-answer\" style=\"display: none\">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><\/h3>\n<h2>The Point-Slope Formula<\/h2>\n<p>Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.<\/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, the point-slope formula 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\">Solution<\/span><\/p>\n<div id=\"q201330\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the point-slope formula, 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><\/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\">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:\/\/www.myopenmath.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\">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 the point-slope formula 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&\\text{Distribute the }\\frac{7}{3}.\\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<h2>Standard Form of a Line<\/h2>\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- <\/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\">Solution<\/span><\/p>\n<div id=\"q111657\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin using the point-slope formula.<\/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, and then 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] and passing 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\">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:\/\/www.myopenmath.com\/multiembedq.php?id=110946&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Vertical and Horizontal Lines<\/h2>\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 the point-slope formula. 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 the point-slope formula. 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\" data-media-type=\"image\/jpg\" \/><\/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 key-takeaways\">\n<h3>try it now<\/h3>\n<p>Use the Desmos calculator to graph the following:<\/p>\n<ol>\n<li>A horizontal line that passes through the point (-5,2)<\/li>\n<li>A vertical line that passes through the point (3,3)<\/li>\n<\/ol>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\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\">Solution<\/span><\/p>\n<div id=\"q122244\" class=\"hidden-answer\" style=\"display: none\">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\">Solution<\/span><\/p>\n<div id=\"q864962\" class=\"hidden-answer\" style=\"display: none\">Horizontal line: [latex]y=2[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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:\/\/www.myopenmath.com\/multiembedq.php?id=110952&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\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-136\">\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. <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><\/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 <\/section>","protected":false},"author":21,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College 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