{"id":16025,"date":"2019-09-26T17:24:03","date_gmt":"2019-09-26T17:24:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-write-the-equation-of-a-linear-function\/"},"modified":"2024-05-02T15:54:24","modified_gmt":"2024-05-02T15:54:24","slug":"read-write-the-equation-of-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-write-the-equation-of-a-linear-function\/","title":{"raw":"Point-Slope Formula","rendered":"Point-Slope Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the equation for a linear function using the point-slope formula<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Point-Slope Form<\/h2>\r\n<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-read-or-watch-given-two-points-write-the-equation-of-a-line\/\">We previously learned how to find the equation of a line given the slope of the line and a point on the line.<\/a>\u00a0 We did this by substituting the slope and point values into the equation [latex]y=mx+b[\/latex] and then solving the equation for b.\r\n\r\nWe will now show you how you can use either use the slope and a point or just 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 can simplify it and then write the equation 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\n<p id=\"fs-id1165137911671\">Now that we know both the slope-intercept form and the point-slope for linear equations, when we are asked to find an equation for a linear function, we can choose which method to use based on the information we are given and our personal preferences. The information provided may be in the form of a graph, a point and a slope, two points, and so on. Let's look at the graph of the function <em>f<\/em>\u00a0below and determine its equation.<span id=\"fs-id1165135182766\"><\/span><\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"A linear function f which passes through the points (0, 7) and (4, 4).\" width=\"487\" height=\"347\" \/>\r\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Choose\u00a0[latex](0, 7)[\/latex]\u00a0and\u00a0[latex](4, 4)[\/latex]. We can use these points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{4 - 7}{4 - 0}\\hfill \\\\ &amp; =-\\dfrac{3}{4} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in slope-intercept form, we would do the following:<\/p>\r\n\r\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\dfrac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"The equation f of x equals mx plus b. And arrow labeled negative 3 fourths points to m, and an arrow labeled 7 points to b. The equation becomes f of x equals negative 3 fourths times x plus 7.\" width=\"487\" height=\"155\" \/>\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137769983\">If we had wanted to find the equation of the line in slope-intercept form without first using point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is\u00a0[latex]7[\/latex]. Therefore, <em>b<\/em> =\u00a0[latex]7[\/latex].\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m\u00a0<\/em>so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into slope-intercept form of a line.<span id=\"fs-id1165137548391\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137705273\">The function is [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\dfrac{3}{4}x+7[\/latex].<\/p>\r\nLet's take a look at another example like the one above.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite an equation for a linear function\u00a0<em>f<\/em> given its graph shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/>\r\n[reveal-answer q=\"728685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"728685\"]\r\nIdentify two points on the line such as\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](\u20132, \u20134)[\/latex]. Use the points to calculate the slope.\r\n\r\n[latex]\\begin{array}{l}m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{-4 - 2}{-2 - 0}\\hfill \\\\ &amp; =\\dfrac{-6}{-2}\\hfill \\\\ &amp; =3\\hfill \\end{array}[\/latex]\r\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\r\n[latex]\\begin{array}{l}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ y=3x+2 \\hfill \\end{array}[\/latex]\r\n\r\n&nbsp;\r\n\r\nThis makes sense because we can see from the graph above that the line crosses the\u00a0<em>y<\/em>-axis at the point [latex](0,2)[\/latex], which is the\u00a0<em>y<\/em>-intercept, so [latex]b=2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn our next 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\nNow, let's look at an example in which we 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\r\nThis next video shows another example of writing the equation of a line given two points on the line.\r\n\r\nhttps:\/\/youtu.be\/ndRpJxdmZJI\r\n\r\nNow that we have seen some examples of how to use the point-slope form of a linear equation, let us look briefly at where the point-slope form comes from.\r\n\r\nRecall that the formula for the slope of a line is defined as\u00a0[latex]m=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].\u00a0 For any two points on the same line, the slope formula will always give us the same value for [latex]m[\/latex].\u00a0 With this in mind, let us rewrite the slope equation as [latex]m=\\dfrac{y-{y}_{1}}{x-{x}_{1}}[\/latex] where [latex](x_1, y_1)[\/latex] represents a given point on the line and [latex](x, y)[\/latex] represents any other point on our line.\u00a0 Now, if we multiply both sides of this slope equation by [latex]x-x_1,[\/latex] we get the following result:\r\n<p style=\"text-align: center;\">[latex]m\\left(x-{x}_{1}\\right)=y-{y}_{1}[\/latex]<\/p>\r\nWhich is equivalent to our point-slope form:\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nWe will conclude this section by looking at an example that provides the points of the function using function notation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.\r\n[reveal-answer q=\"834495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834495\"]\r\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\r\n\r\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}{l}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array}{l} m &amp; =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ &amp; =\\dfrac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ &amp; =\\dfrac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\r\n\r\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y+2=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\dfrac{3}{5}x-\\dfrac{9}{5}\\hfill \\\\ \\text{ }y=\\dfrac{3}{5}x-\\dfrac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this last video, we show an example of how to write a linear function given two points written with function notation.\r\n\r\nhttps:\/\/youtu.be\/TSIcryAtCmY","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the equation for a linear function using the point-slope formula<\/li>\n<\/ul>\n<\/div>\n<h2>Point-Slope Form<\/h2>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-read-or-watch-given-two-points-write-the-equation-of-a-line\/\">We previously learned how to find the equation of a line given the slope of the line and a point on the line.<\/a>\u00a0 We did this by substituting the slope and point values into the equation [latex]y=mx+b[\/latex] and then solving the equation for b.<\/p>\n<p>We will now show you how you can use either use the slope and a point or just 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 can simplify it and then write the equation 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 id=\"fs-id1165137911671\">Now that we know both the slope-intercept form and the point-slope for linear equations, when we are asked to find an equation for a linear function, we can choose which method to use based on the information we are given and our personal preferences. The information provided may be in the form of a graph, a point and a slope, two points, and so on. Let&#8217;s look at the graph of the function <em>f<\/em>\u00a0below and determine its equation.<span id=\"fs-id1165135182766\"><\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"A linear function f which passes through the points (0, 7) and (4, 4).\" width=\"487\" height=\"347\" \/><\/p>\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Choose\u00a0[latex](0, 7)[\/latex]\u00a0and\u00a0[latex](4, 4)[\/latex]. We can use these points to calculate the slope.<\/p>\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l} m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{4 - 7}{4 - 0}\\hfill \\\\ & =-\\dfrac{3}{4} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in slope-intercept form, we would do the following:<\/p>\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{l}y - 4=-\\dfrac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\dfrac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\dfrac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"The equation f of x equals mx plus b. And arrow labeled negative 3 fourths points to m, and an arrow labeled 7 points to b. The equation becomes f of x equals negative 3 fourths times x plus 7.\" width=\"487\" height=\"155\" \/><\/p>\n<\/div>\n<p id=\"fs-id1165137769983\">If we had wanted to find the equation of the line in slope-intercept form without first using point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is\u00a0[latex]7[\/latex]. Therefore, <em>b<\/em> =\u00a0[latex]7[\/latex].\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m\u00a0<\/em>so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into slope-intercept form of a line.<span id=\"fs-id1165137548391\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137705273\">The function is [latex]f\\left(x\\right)=-\\dfrac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\dfrac{3}{4}x+7[\/latex].<\/p>\n<p>Let&#8217;s take a look at another example like the one above.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write an equation for a linear function\u00a0<em>f<\/em> given its graph shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728685\">Show Solution<\/span><\/p>\n<div id=\"q728685\" class=\"hidden-answer\" style=\"display: none\">\nIdentify two points on the line such as\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](\u20132, \u20134)[\/latex]. Use the points to calculate the slope.<\/p>\n<p>[latex]\\begin{array}{l}m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{-4 - 2}{-2 - 0}\\hfill \\\\ & =\\dfrac{-6}{-2}\\hfill \\\\ & =3\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<p>[latex]\\begin{array}{l}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ y=3x+2 \\hfill \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>This makes sense because we can see from the graph above that the line crosses the\u00a0<em>y<\/em>-axis at the point [latex](0,2)[\/latex], which is the\u00a0<em>y<\/em>-intercept, so [latex]b=2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In our next 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>Now, let&#8217;s look at an example in which we 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<p>This next video shows another example of writing the equation of a line given two points on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-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<p>Now that we have seen some examples of how to use the point-slope form of a linear equation, let us look briefly at where the point-slope form comes from.<\/p>\n<p>Recall that the formula for the slope of a line is defined as\u00a0[latex]m=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex].\u00a0 For any two points on the same line, the slope formula will always give us the same value for [latex]m[\/latex].\u00a0 With this in mind, let us rewrite the slope equation as [latex]m=\\dfrac{y-{y}_{1}}{x-{x}_{1}}[\/latex] where [latex](x_1, y_1)[\/latex] represents a given point on the line and [latex](x, y)[\/latex] represents any other point on our line.\u00a0 Now, if we multiply both sides of this slope equation by [latex]x-x_1,[\/latex] we get the following result:<\/p>\n<p style=\"text-align: center;\">[latex]m\\left(x-{x}_{1}\\right)=y-{y}_{1}[\/latex]<\/p>\n<p>Which is equivalent to our point-slope form:<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>We will conclude this section by looking at an example that provides the points of the function using function notation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834495\">Show Solution<\/span><\/p>\n<div id=\"q834495\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}{l}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array}{l} m & =\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ & =\\dfrac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ & =\\dfrac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into point-slope form.<\/p>\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}{l}y+2=\\dfrac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\dfrac{3}{5}x-\\dfrac{9}{5}\\hfill \\\\ \\text{ }y=\\dfrac{3}{5}x-\\dfrac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In this last video, we show an example of how to write a linear function given two points written with function notation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Find the Linear Function Given Two Function Values in Function Notation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSIcryAtCmY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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-16025\">\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":169554,"menu_order":10,"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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