{"id":3795,"date":"2020-02-08T20:25:25","date_gmt":"2020-02-08T20:25:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/mathforlibscoreq\/?post_type=chapter&#038;p=3795"},"modified":"2021-02-05T23:54:46","modified_gmt":"2021-02-05T23:54:46","slug":"finding-slope-given-two-points-on-a-line","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/finding-slope-given-two-points-on-a-line\/","title":{"raw":"Finding Slope Given Two Points on a Line","rendered":"Finding Slope Given Two Points on a Line"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the slope of a line given two points on the line<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p>Sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing.<\/p>\r\n<p>Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair [latex]\\left(x,y\\right)[\/latex] gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol [latex]\\left(x,y\\right)[\/latex] be used to represent two different points?<\/p>\r\n<p>Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable.<\/p>\r\n\r\n<ul id=\"fs-id1707109\">\r\n \t<li>[latex]\\left({x}_{1},{y}_{1}\\right)\\text{ read }x\\text{ sub }1,y\\text{ sub }1[\/latex]<\/li>\r\n \t<li>[latex]\\left({x}_{2},{y}_{2}\\right)\\text{ read }x\\text{ sub }2,y\\text{ sub }2[\/latex]<\/li>\r\n<\/ul>\r\nWe will use [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] to identify the first point and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] to identify the second point. If we had more than two points, we could use [latex]\\left({x}_{3},{y}_{3}\\right),\\left({x}_{4},{y}_{4}\\right)[\/latex], and so on.\r\n\r\nTo see how the rise and run relate to the coordinates of the two points, let\u2019s take another look at the slope of the line between the points [latex]\\left(2,3\\right)[\/latex] and [latex]\\left(7,6\\right)[\/latex] below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224620\/CNX_BMath_Figure_11_04_030.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6. It os labeled x sub 2 minus x sub 1, 7 minus 2, 5. \" \/>\r\nSince we have two points, we will use subscript notation.\r\n<p style=\"text-align: center\">[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(2,3\\right)}\\stackrel{{x}_{2},{y}_{2}}{\\left(7,6\\right)}[\/latex]<\/p>\r\nOn the graph, we counted the rise of [latex]3[\/latex]. The rise can also be found by subtracting the [latex]y\\text{-coordinates}[\/latex] of the points.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{y}_{2}-{y}_{1}\\\\ 6 - 3\\\\ 3\\end{array}[\/latex]<\/p>\r\nWe counted a run of [latex]5[\/latex]. The run can also be found by subtracting the [latex]x\\text{-coordinates}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{x}_{2}-{x}_{1}\\\\ 7 - 2\\\\ 5\\end{array}[\/latex]<\/p>\r\n\r\n<table id=\"eip-id1168468520883\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We know<\/td>\r\n<td>[latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>So<\/td>\r\n<td>[latex]m={\\Large\\frac{3}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We rewrite the rise and run by putting in the coordinates.<\/td>\r\n<td>[latex]m={\\Large\\frac{6 - 3}{7 - 2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>But [latex]6[\/latex] is the [latex]y[\/latex] -coordinate of the second point, [latex]{y}_{2}[\/latex]\r\n\r\nand [latex]3[\/latex] is the [latex]y[\/latex] -coordinate of the first point [latex]{y}_{1}[\/latex] .\r\n\r\nSo we can rewrite the rise using subscript notation.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{7 - 2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Also [latex]7[\/latex] is the [latex]x[\/latex] -coordinate of the second point, [latex]{x}_{2}[\/latex]\r\n\r\nand [latex]2[\/latex] is the [latex]x[\/latex] -coordinate of the first point [latex]{x}_{2}[\/latex] .\r\n\r\nSo we rewrite the run using subscript notation.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe\u2019ve shown that [latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex] is really another version of [latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]. We can use this formula to find the slope of a line when we have two points on the line.\r\n<div class=\"textbox shaded\">\r\n<h3>Slope Formula<\/h3>\r\nThe slope of the line between two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] is\r\n\r\n[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]\r\n\r\nSay the formula to yourself to help you remember it:\r\n<p style=\"text-align: center\">[latex]\\text{Slope is }y\\text{ of the second point minus }y\\text{ of the first point}[\/latex]\r\n[latex]\\text{over}[\/latex]\r\n[latex]x\\text{ of the second point minus }x\\text{ of the first point.}[\/latex]<\/p>\r\n\r\n<\/div>\r\nDoing the Manipulative Mathematics activity \"Slope of Lines Between Two Points\" will help you develop a better understanding of how to find the slope of a line between two points.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nFind the slope of the line between the points [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(4,5\\right)[\/latex].\r\n\r\nSolution\r\n<table id=\"eip-id1168468461864\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We\u2019ll call [latex]\\left(1,2\\right)[\/latex] point #1 and [latex]\\left(4,5\\right)[\/latex] point #2.<\/td>\r\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(1,2\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(4,5\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the slope formula.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute the values in the slope formula:<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{5 - 2}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{5 - 2}{4 - 1}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerator and the denominator.<\/td>\r\n<td>[latex]m={\\Large\\frac{3}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]m=1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet\u2019s confirm this by counting out the slope on the graph.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224621\/CNX_BMath_Figure_11_04_031.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -1 to 7. Two labeled points are drawn at \" \/>\r\nThe rise is [latex]3[\/latex] and the run is [latex]3[\/latex], so\r\n[latex]\\begin{array}{}\\\\ m=\\frac{\\text{rise}}{\\text{run}}\\hfill \\\\ m={\\Large\\frac{3}{3}}\\hfill \\\\ m=1\\hfill \\end{array}[\/latex]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]147021[\/ohm_question]\r\n\r\n<\/div>\r\nHow do we know which point to call #1 and which to call #2? Let\u2019s find the slope again, this time switching the names of the points to see what happens. Since we will now be counting the run from right to left, it will be negative.\r\n<table id=\"eip-id1168465988183\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We\u2019ll call [latex]\\left(4,5\\right)[\/latex] point #1 and [latex]\\left(1,2\\right)[\/latex] point #2.<\/td>\r\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(4,5\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(1,2\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the slope formula.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute the values in the slope formula:<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{2 - 5}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{2 - 5}{1 - 4}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerator and the denominator.<\/td>\r\n<td>[latex]m={\\Large\\frac{-3}{-3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]m=1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe slope is the same no matter which order we use the points.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nFind the slope of the line through the points [latex]\\left(-2,-3\\right)[\/latex] and [latex]\\left(-7,4\\right)[\/latex].\r\n[reveal-answer q=\"265532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265532\"]\r\n\r\nSolution\r\n<table id=\"eip-id1168469701671\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We\u2019ll call [latex]\\left(-2,-3\\right)[\/latex] point #1 and [latex]\\left(-7,4\\right)[\/latex] point #2.<\/td>\r\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(-2,-3\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(-7,4\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the slope formula.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute the values<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{-7-\\left(-2\\right)}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]m={\\Large\\frac{7}{-5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]m=-{\\Large\\frac{7}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet\u2019s confirm this on the graph shown.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224622\/CNX_BMath_Figure_11_04_032.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -8 to 2. The y-axis runs from -6 to 5. Two unlabeled points are drawn at \" \/>\r\n[latex]\\begin{array}{}\\\\ \\\\ \\\\ m=\\frac{\\text{rise}}{\\text{run}}\\\\ m={\\Large\\frac{-7}{5}}\\\\ m=-{\\Large\\frac{7}{5}}\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]147022[\/ohm_question]\r\n\r\n<\/div>\r\nWatch this video to see more examples of how to determine slope given two points on a line.\r\n\r\nhttps:\/\/youtu.be\/6qONExlVGgc","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the slope of a line given two points on the line<\/li>\n<\/ul>\n<\/div>\n<p>Sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing.<\/p>\n<p>Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair [latex]\\left(x,y\\right)[\/latex] gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol [latex]\\left(x,y\\right)[\/latex] be used to represent two different points?<\/p>\n<p>Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable.<\/p>\n<ul id=\"fs-id1707109\">\n<li>[latex]\\left({x}_{1},{y}_{1}\\right)\\text{ read }x\\text{ sub }1,y\\text{ sub }1[\/latex]<\/li>\n<li>[latex]\\left({x}_{2},{y}_{2}\\right)\\text{ read }x\\text{ sub }2,y\\text{ sub }2[\/latex]<\/li>\n<\/ul>\n<p>We will use [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] to identify the first point and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] to identify the second point. If we had more than two points, we could use [latex]\\left({x}_{3},{y}_{3}\\right),\\left({x}_{4},{y}_{4}\\right)[\/latex], and so on.<\/p>\n<p>To see how the rise and run relate to the coordinates of the two points, let\u2019s take another look at the slope of the line between the points [latex]\\left(2,3\\right)[\/latex] and [latex]\\left(7,6\\right)[\/latex] below.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224620\/CNX_BMath_Figure_11_04_030.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6. It os labeled x sub 2 minus x sub 1, 7 minus 2, 5.\" \/><br \/>\nSince we have two points, we will use subscript notation.<\/p>\n<p style=\"text-align: center\">[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(2,3\\right)}\\stackrel{{x}_{2},{y}_{2}}{\\left(7,6\\right)}[\/latex]<\/p>\n<p>On the graph, we counted the rise of [latex]3[\/latex]. The rise can also be found by subtracting the [latex]y\\text{-coordinates}[\/latex] of the points.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{y}_{2}-{y}_{1}\\\\ 6 - 3\\\\ 3\\end{array}[\/latex]<\/p>\n<p>We counted a run of [latex]5[\/latex]. The run can also be found by subtracting the [latex]x\\text{-coordinates}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{x}_{2}-{x}_{1}\\\\ 7 - 2\\\\ 5\\end{array}[\/latex]<\/p>\n<table id=\"eip-id1168468520883\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We know<\/td>\n<td>[latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>So<\/td>\n<td>[latex]m={\\Large\\frac{3}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>We rewrite the rise and run by putting in the coordinates.<\/td>\n<td>[latex]m={\\Large\\frac{6 - 3}{7 - 2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>But [latex]6[\/latex] is the [latex]y[\/latex] -coordinate of the second point, [latex]{y}_{2}[\/latex]<\/p>\n<p>and [latex]3[\/latex] is the [latex]y[\/latex] -coordinate of the first point [latex]{y}_{1}[\/latex] .<\/p>\n<p>So we can rewrite the rise using subscript notation.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{7 - 2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Also [latex]7[\/latex] is the [latex]x[\/latex] -coordinate of the second point, [latex]{x}_{2}[\/latex]<\/p>\n<p>and [latex]2[\/latex] is the [latex]x[\/latex] -coordinate of the first point [latex]{x}_{2}[\/latex] .<\/p>\n<p>So we rewrite the run using subscript notation.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We\u2019ve shown that [latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex] is really another version of [latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]. We can use this formula to find the slope of a line when we have two points on the line.<\/p>\n<div class=\"textbox shaded\">\n<h3>Slope Formula<\/h3>\n<p>The slope of the line between two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] is<\/p>\n<p>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/p>\n<p>Say the formula to yourself to help you remember it:<\/p>\n<p style=\"text-align: center\">[latex]\\text{Slope is }y\\text{ of the second point minus }y\\text{ of the first point}[\/latex]<br \/>\n[latex]\\text{over}[\/latex]<br \/>\n[latex]x\\text{ of the second point minus }x\\text{ of the first point.}[\/latex]<\/p>\n<\/div>\n<p>Doing the Manipulative Mathematics activity &#8220;Slope of Lines Between Two Points&#8221; will help you develop a better understanding of how to find the slope of a line between two points.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Find the slope of the line between the points [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(4,5\\right)[\/latex].<\/p>\n<p>Solution<\/p>\n<table id=\"eip-id1168468461864\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We\u2019ll call [latex]\\left(1,2\\right)[\/latex] point #1 and [latex]\\left(4,5\\right)[\/latex] point #2.<\/td>\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(1,2\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(4,5\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the slope formula.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute the values in the slope formula:<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{5 - 2}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{5 - 2}{4 - 1}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerator and the denominator.<\/td>\n<td>[latex]m={\\Large\\frac{3}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]m=1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let\u2019s confirm this by counting out the slope on the graph.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224621\/CNX_BMath_Figure_11_04_031.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -1 to 7. Two labeled points are drawn at\" \/><br \/>\nThe rise is [latex]3[\/latex] and the run is [latex]3[\/latex], so<br \/>\n[latex]\\begin{array}{}\\\\ m=\\frac{\\text{rise}}{\\text{run}}\\hfill \\\\ m={\\Large\\frac{3}{3}}\\hfill \\\\ m=1\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm147021\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=147021&theme=oea&iframe_resize_id=ohm147021&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>How do we know which point to call #1 and which to call #2? Let\u2019s find the slope again, this time switching the names of the points to see what happens. Since we will now be counting the run from right to left, it will be negative.<\/p>\n<table id=\"eip-id1168465988183\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We\u2019ll call [latex]\\left(4,5\\right)[\/latex] point #1 and [latex]\\left(1,2\\right)[\/latex] point #2.<\/td>\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(4,5\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(1,2\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the slope formula.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute the values in the slope formula:<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{2 - 5}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{2 - 5}{1 - 4}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerator and the denominator.<\/td>\n<td>[latex]m={\\Large\\frac{-3}{-3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]m=1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The slope is the same no matter which order we use the points.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Find the slope of the line through the points [latex]\\left(-2,-3\\right)[\/latex] and [latex]\\left(-7,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265532\">Show Solution<\/span><\/p>\n<div id=\"q265532\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<\/p>\n<table id=\"eip-id1168469701671\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We\u2019ll call [latex]\\left(-2,-3\\right)[\/latex] point #1 and [latex]\\left(-7,4\\right)[\/latex] point #2.<\/td>\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(-2,-3\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(-7,4\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the slope formula.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute the values<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{-7-\\left(-2\\right)}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]m={\\Large\\frac{7}{-5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]m=-{\\Large\\frac{7}{5}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let\u2019s confirm this on the graph shown.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224622\/CNX_BMath_Figure_11_04_032.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -8 to 2. The y-axis runs from -6 to 5. Two unlabeled points are drawn at\" \/><br \/>\n[latex]\\begin{array}{}\\\\ \\\\ \\\\ m=\\frac{\\text{rise}}{\\text{run}}\\\\ m={\\Large\\frac{-7}{5}}\\\\ m=-{\\Large\\frac{7}{5}}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm147022\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=147022&theme=oea&iframe_resize_id=ohm147022&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Watch this video to see more examples of how to determine slope given two points on a line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Determine the Slope a Line Given Two Points on a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6qONExlVGgc?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-3795\">\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>Question ID 147022, 147021, 147020. <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><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Determine the Slope a Line Given Two Points on a Line. <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6qONExlVGgc\">https:\/\/youtu.be\/6qONExlVGgc<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <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\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757<\/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":25777,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"},{\"type\":\"original\",\"description\":\"Question ID 147022, 147021, 147020\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Determine the Slope a Line Given Two Points on a Line\",\"author\":\"James Sousa (mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/6qONExlVGgc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3795","chapter","type-chapter","status-web-only","hentry"],"part":1040,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3795","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/users\/25777"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3795\/revisions"}],"predecessor-version":[{"id":3796,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3795\/revisions\/3796"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/parts\/1040"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/3795\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/media?parent=3795"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=3795"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/contributor?post=3795"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/license?post=3795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}