{"id":2006,"date":"2016-06-30T22:19:09","date_gmt":"2016-06-30T22:19:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2006"},"modified":"2019-08-06T18:43:47","modified_gmt":"2019-08-06T18:43:47","slug":"read-write-the-equation-of-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-write-the-equation-of-a-linear-function\/","title":{"raw":"Write the Equation of a Linear Function Part I","rendered":"Write the Equation of a Linear Function Part I"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define the equation of a line given the slope and a point<\/li>\r\n \t<li>Define the equation of a line given two points<\/li>\r\n \t<li>Write a linear function in standard form<\/li>\r\n \t<li>Write the equation for linear functions whose graphs are horizontal and vertical lines<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Point-Slope Form<\/h2>\r\nWe have seen that we can define the slope of a line given two points on the line and use that information along with the y-intercept to graph the line. \u00a0If you do not know the y-intercept or the equation for the line, you can use two points to define\u00a0the equation of the line using point-slope form.\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nThis is an important formula, as it will be used in other algebra courses 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 shaded\">\r\n<h3>Point-Slope Form<\/h3>\r\nGiven a point [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and slope <em>m<\/em>, point-slope form will give the following equation of a line:\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn our first example, we will start with the slope. Then, we will show how to find the equation of a line when the slope is not given.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] that passes through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"524449\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"524449\"]\r\n\r\nUsing point-slope form, substitute [latex]-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<p 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]<\/p>\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we will start with two points and find the equation of the line that passes through them.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the equation of the line that passes through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"249539\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249539\"]\r\n\r\nFirst, we calculate the slope using the slope formula and two points.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill &amp; =\\dfrac{-3 - 4}{0 - 3}\\hfill \\\\ \\hfill &amp; =\\dfrac{-7}{-3}\\hfill \\\\ \\hfill &amp; =\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nNext, we use point-slope form with the slope of [latex]\\dfrac{7}{3}[\/latex] and either point. Let us pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=\\dfrac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\dfrac{7}{3}x - 7\\hfill&amp;\\text{Distribute the }\\dfrac{7}{3}.\\hfill \\\\ y=\\dfrac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\r\nIn slope-intercept form, the equation is written as [latex]y=\\dfrac{7}{3}x - 3[\/latex].\r\n\r\nTo prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\dfrac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\dfrac{7}{3}x\\hfill \\\\ y=\\dfrac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\r\nWe see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows how to write the equation for a line given its slope and a point on the line.\r\n\r\nhttps:\/\/youtu.be\/vut5b2fRQQ0\r\n<h2>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<p style=\"text-align: center;\">[latex]Ax+By=C[\/latex]<\/p>\r\nwhere [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x\u00a0<\/em>and <em>y\u00a0<\/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<\/h3>\r\nFind the equation of the line with [latex]m=-6[\/latex] that passes through the point [latex]\\left(\\dfrac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.\r\n\r\n[reveal-answer q=\"526704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"526704\"]\r\n\r\nWe begin by using point-slope form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\dfrac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\dfrac{3}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nFrom here, we multiply through by\u00a0[latex]2[\/latex], as no fractions are permitted in standard form. Then, move both variables to the left side of the equal sign and constants to the right side of the equal sign.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\dfrac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/p>\r\nThis equation is now written in standard form.\r\n\r\n[\/hidden-answer]\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<p style=\"text-align: center;\">[latex]x=c[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]m=\\dfrac{5 - 3}{-3-\\left(-3\\right)}=\\dfrac{2}{0}[\/latex]<\/p>\r\nZero in the denominator means that the slope is undefined which makes using point-slope form impossible. 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<p style=\"text-align: center;\">[latex]y=c[\/latex]<\/p>\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 corresponding\u00a0<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<p style=\"text-align: center;\">[latex]\\begin{array}{l}m &amp; =\\dfrac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ &amp; =\\dfrac{0}{2}\\hfill \\\\ &amp; =0\\hfill \\end{array}[\/latex]<\/p>\r\nUse any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in point-slope form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/p>\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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/> The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.[\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\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\r\n[reveal-answer q=\"346281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346281\"]\r\n\r\nThe <em>x-<\/em>coordinate of both points is\u00a0[latex]1[\/latex]. Therefore, we have a vertical line, [latex]x=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur last video shows another example of writing the equation of a line given two points on the line.\r\n\r\nhttps:\/\/youtu.be\/ndRpJxdmZJI\r\n<h2>Summary<\/h2>\r\n<ul>\r\n \t<li>The <strong>slope<\/strong> of a line\u00a0indicates the direction which it slants as well as its steepness. Slope is\u00a0defined algebraically as:[latex]m=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/li>\r\n \t<li>Given the slope and one point on a line, we can find the equation of the line using point-slope form.\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/li>\r\n \t<li>Standard form of a line is given as\u00a0[latex]Ax+By=C[\/latex]<\/li>\r\n \t<li>The equation of a <strong>vertical line<\/strong> is given as\u00a0[latex]x=c[\/latex]<\/li>\r\n \t<li>The equation of a <strong>horizontal line<\/strong> is given as\u00a0[latex]y=c[\/latex]<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define the equation of a line given the slope and a point<\/li>\n<li>Define the equation of a line given two points<\/li>\n<li>Write a linear function in standard form<\/li>\n<li>Write the equation for linear functions whose graphs are horizontal and vertical lines<\/li>\n<\/ul>\n<\/div>\n<h2>Point-Slope Form<\/h2>\n<p>We have seen that we can define the slope of a line given two points on the line and use that information along with the y-intercept to graph the line. \u00a0If you do not know the y-intercept or the equation for the line, you can use two points to define\u00a0the equation of the line using point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>This is an important formula, as it will be used in other algebra courses 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 shaded\">\n<h3>Point-Slope Form<\/h3>\n<p>Given a point [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and slope <em>m<\/em>, point-slope form will give the following equation of a line:<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<\/div>\n<p>In our first example, we will start with the slope. Then, we will show how to find the equation of a line when the slope is not given.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] that passes through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q524449\">Show Solution<\/span><\/p>\n<div id=\"q524449\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using point-slope form, substitute [latex]-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<p 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]<\/p>\n<p>Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we will start with two points and find the equation of the line that passes through them.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the equation of the line that passes through the points [latex]\\left(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=\"q249539\">Show Solution<\/span><\/p>\n<div id=\"q249539\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we calculate the slope using the slope formula and two points.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill & =\\dfrac{-3 - 4}{0 - 3}\\hfill \\\\ \\hfill & =\\dfrac{-7}{-3}\\hfill \\\\ \\hfill & =\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Next, we use point-slope form with the slope of [latex]\\dfrac{7}{3}[\/latex] and either point. Let us pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=\\dfrac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\dfrac{7}{3}x - 7\\hfill&\\text{Distribute the }\\dfrac{7}{3}.\\hfill \\\\ y=\\dfrac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\n<p>In slope-intercept form, the equation is written as [latex]y=\\dfrac{7}{3}x - 3[\/latex].<\/p>\n<p>To prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\dfrac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\dfrac{7}{3}x\\hfill \\\\ y=\\dfrac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\n<p>We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows how to write the equation for a line given its slope and a point on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vut5b2fRQQ0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>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<p style=\"text-align: center;\">[latex]Ax+By=C[\/latex]<\/p>\n<p>where [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x\u00a0<\/em>and <em>y\u00a0<\/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<\/h3>\n<p>Find the equation of the line with [latex]m=-6[\/latex] that passes through the point [latex]\\left(\\dfrac{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=\"q526704\">Show Solution<\/span><\/p>\n<div id=\"q526704\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by using point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\dfrac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\dfrac{3}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>From here, we multiply through by\u00a0[latex]2[\/latex], as no fractions are permitted in standard form. Then, move both variables to the left side of the equal sign and constants to the right side of the equal sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\dfrac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/p>\n<p>This equation is now written in standard form.<\/p>\n<\/div>\n<\/div>\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<p style=\"text-align: center;\">[latex]x=c[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]m=\\dfrac{5 - 3}{-3-\\left(-3\\right)}=\\dfrac{2}{0}[\/latex]<\/p>\n<p>Zero in the denominator means that the slope is undefined which makes using point-slope form impossible. 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<p style=\"text-align: center;\">[latex]y=c[\/latex]<\/p>\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 corresponding\u00a0<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<p style=\"text-align: center;\">[latex]\\begin{array}{l}m & =\\dfrac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ & =\\dfrac{0}{2}\\hfill \\\\ & =0\\hfill \\end{array}[\/latex]<\/p>\n<p>Use any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/p>\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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\">The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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=\"q346281\">Show Solution<\/span><\/p>\n<div id=\"q346281\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <em>x-<\/em>coordinate of both points is\u00a0[latex]1[\/latex]. Therefore, we have a vertical line, [latex]x=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Our last video shows another example of writing the equation of a line given two points on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ndRpJxdmZJI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<ul>\n<li>The <strong>slope<\/strong> of a line\u00a0indicates the direction which it slants as well as its steepness. Slope is\u00a0defined algebraically as:[latex]m=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/li>\n<li>Given the slope and one point on a line, we can find the equation of the line using point-slope form.\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/li>\n<li>Standard form of a line is given as\u00a0[latex]Ax+By=C[\/latex]<\/li>\n<li>The equation of a <strong>vertical line<\/strong> is given as\u00a0[latex]x=c[\/latex]<\/li>\n<li>The equation of a <strong>horizontal line<\/strong> is given as\u00a0[latex]y=c[\/latex]<\/li>\n<\/ul>\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-2006\">\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>Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/vut5b2fRQQ0\">https:\/\/youtu.be\/vut5b2fRQQ0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ndRpJxdmZJI\">https:\/\/youtu.be\/ndRpJxdmZJI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>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>. <strong>License Terms<\/strong>: Download for free:  http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/vut5b2fRQQ0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ndRpJxdmZJI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College 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