{"id":4634,"date":"2017-06-07T18:53:51","date_gmt":"2017-06-07T18:53:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-find-the-slope-of-a-line-given-two-points\/"},"modified":"2017-08-16T03:01:40","modified_gmt":"2017-08-16T03:01:40","slug":"read-find-the-slope-of-a-line-given-two-points","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-find-the-slope-of-a-line-given-two-points\/","title":{"raw":"Find the Slope of a Line Given Two Points","rendered":"Find the Slope of a Line Given Two Points"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Find the Slope from Two Points\r\n<ul>\r\n \t<li>Use the formula for slope to define the slope of a line through two points<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Finding the Slope from\u00a0Two Points on the Line<\/h2>\r\nYou\u2019ve seen that you can find the slope of a line on a graph by measuring the rise and the run. You can also find the slope of a straight line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an <i>x<\/i>-value and a <i>y<\/i>-value, written as an ordered pair (<i>x<\/i>, <i>y<\/i>). The <i>x<\/i> value tells you where a point is horizontally. The <i>y<\/i> value tells you where the point is vertically.\r\n\r\nConsider two points on a line\u2014Point 1 and Point 2. Point 1 has coordinates [latex]\\left(x_{1},y_{1}\\right)[\/latex]\u00a0and Point 2 has coordinates [latex]\\left(x_{2},y_{2}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185348\/image031.jpg\" alt=\"A line with its rise and run. The first point on the line is labeled Point 1, or (x1, y1). The second point on the line is labeled Point 2, or (x2,y2). The rise is (y2 minus y1). The run is (x2 minus X1).\" width=\"416\" height=\"401\" \/>\r\n\r\nThe rise is the vertical distance between the two points, which is the difference between their <i>y<\/i>-coordinates. That makes the rise [latex]\\left(y_{2}-y_{1}\\right)[\/latex]. The run between these two points is the difference in the <i>x<\/i>-coordinates, or [latex]\\left(x_{2}-x_{1}\\right)[\/latex].\r\n\r\nSo, [latex] \\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex] or [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex]\r\n\r\nIn the example below, you\u2019ll see that the line has two points each indicated as an ordered pair. The point [latex](0,2)[\/latex] is indicated as Point 1, and [latex](\u22122,6)[\/latex] as Point 2. So you are going to move from Point 1 to Point 2. A triangle is drawn in above the line to help illustrate the rise and run.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185350\/image032.jpg\" alt=\"A line going through Point 1, or (0,2), and Point 2, or (-2,6). The rise is 4 and the run is -2.\" width=\"410\" height=\"396\" \/>\r\n\r\nYou can see from the graph that the rise going from Point 1 to Point 2 is 4, because you are moving 4 units in a positive direction (up). The run is [latex]\u22122[\/latex], because you are then moving in a negative direction (left) 2 units. Using the slope formula,\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{-2}=-2[\/latex].<\/p>\r\nYou do not need the graph to find the slope. You can just use the coordinates, keeping careful track of which is Point 1 and which is Point 2. Let\u2019s organize the information about the two points:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Name<\/th>\r\n<th>Ordered Pair<\/th>\r\n<th>Coordinates<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Point 1<\/td>\r\n<td>[latex](0,2)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Point 2<\/td>\r\n<td>[latex](\u22122,6)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}x_{2}=-2\\\\y_{2}=6\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe slope, [latex]m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{6-2}{-2-0}=\\frac{4}{-2}=-2[\/latex]. The slope of the line, <i>m<\/i>, is [latex]\u22122[\/latex].\r\n\r\nIt doesn\u2019t matter which point is designated as Point 1 and which is Point 2. You could have called [latex](\u22122,6)[\/latex] Point 1, and [latex](0,2)[\/latex] Point 2. In that\u00a0case, putting the coordinates into the slope formula produces the equation [latex]m=\\frac{2-6}{0-\\left(-2\\right)}=\\frac{-4}{2}=-2[\/latex]. Once again, the slope is [latex]m=-2[\/latex]. That\u2019s the same slope as before. The important thing is to be consistent when you subtract: you must always subtract in the same order [latex]\\left(y_{2},y_{1}\\right)[\/latex]<sub>\u00a0<\/sub>and [latex]\\left(x_{2},x_{1}\\right)[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?\r\n\r\n[reveal-answer q=\"666697\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666697\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=4\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)=\\text{Point }1,\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{2}=5\\\\y_{2}=5\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\left(5,5\\right)=\\text{Point }2,\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{5-2}{5-4}=\\frac{3}{1}\\\\\\\\m=3\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is 3.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe example below shows the solution when you reverse the order of the points, calling [latex](5,5)[\/latex] Point 1 and [latex](4,2)[\/latex] Point 2.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?\r\n\r\n[reveal-answer q=\"76013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"76013\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=5\\\\y_{1}=5\\end{array}[\/latex]<\/td>\r\n<td>[latex](5,5)=\\text{Point }1[\/latex], [latex]\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](4,2)=\\text{Point }2[\/latex], [latex]\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{{x_2}-x_{1}}\\\\\\\\m=\\frac{2-5}{4-5}=\\frac{-3}{-1}=3\\\\\\\\m=3\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is 3.[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that regardless of which ordered pair is named Point 1 and which is named Point 2, the slope is still 3.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nWhat is the slope of the line that contains the points [latex](3,-6.25)[\/latex] and [latex](-1,8.5)[\/latex]?\r\n\r\n[reveal-answer q=\"291649\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"291649\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=3\\\\y_{1}=-6.25\\end{array}[\/latex]<\/td>\r\n<td>[latex](3,-6.25)=\\text{Point }1[\/latex], [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-1\\\\{{y}_{2}}=8.5\\end{array}[\/latex]<\/td>\r\n<td>[latex](-1,8.5)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{8.5-(-6.25)}{-1-3}\\\\\\\\m=\\frac{14.75}{-4}\\\\\\\\m=-3.6875\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is [latex]-3.6875[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nLet\u2019s consider a horizontal line on a graph. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. The equation for this line is [latex]y=3[\/latex]. The equation can also be written as [latex]y=\\left(0\\right)x+3[\/latex].\r\n\r\nThe following video shows more examples of finding the slope of a line given two points on the line:\r\n\r\nhttps:\/\/youtu.be\/ZW7rQa8SJSU","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Find the Slope from Two Points\n<ul>\n<li>Use the formula for slope to define the slope of a line through two points<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Finding the Slope from\u00a0Two Points on the Line<\/h2>\n<p>You\u2019ve seen that you can find the slope of a line on a graph by measuring the rise and the run. You can also find the slope of a straight line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an <i>x<\/i>-value and a <i>y<\/i>-value, written as an ordered pair (<i>x<\/i>, <i>y<\/i>). The <i>x<\/i> value tells you where a point is horizontally. The <i>y<\/i> value tells you where the point is vertically.<\/p>\n<p>Consider two points on a line\u2014Point 1 and Point 2. Point 1 has coordinates [latex]\\left(x_{1},y_{1}\\right)[\/latex]\u00a0and Point 2 has coordinates [latex]\\left(x_{2},y_{2}\\right)[\/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\/2011\/2017\/06\/07185348\/image031.jpg\" alt=\"A line with its rise and run. The first point on the line is labeled Point 1, or (x1, y1). The second point on the line is labeled Point 2, or (x2,y2). The rise is (y2 minus y1). The run is (x2 minus X1).\" width=\"416\" height=\"401\" \/><\/p>\n<p>The rise is the vertical distance between the two points, which is the difference between their <i>y<\/i>-coordinates. That makes the rise [latex]\\left(y_{2}-y_{1}\\right)[\/latex]. The run between these two points is the difference in the <i>x<\/i>-coordinates, or [latex]\\left(x_{2}-x_{1}\\right)[\/latex].<\/p>\n<p>So, [latex]\\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex] or [latex]\\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex]<\/p>\n<p>In the example below, you\u2019ll see that the line has two points each indicated as an ordered pair. The point [latex](0,2)[\/latex] is indicated as Point 1, and [latex](\u22122,6)[\/latex] as Point 2. So you are going to move from Point 1 to Point 2. A triangle is drawn in above the line to help illustrate the rise and run.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185350\/image032.jpg\" alt=\"A line going through Point 1, or (0,2), and Point 2, or (-2,6). The rise is 4 and the run is -2.\" width=\"410\" height=\"396\" \/><\/p>\n<p>You can see from the graph that the rise going from Point 1 to Point 2 is 4, because you are moving 4 units in a positive direction (up). The run is [latex]\u22122[\/latex], because you are then moving in a negative direction (left) 2 units. Using the slope formula,<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{-2}=-2[\/latex].<\/p>\n<p>You do not need the graph to find the slope. You can just use the coordinates, keeping careful track of which is Point 1 and which is Point 2. Let\u2019s organize the information about the two points:<\/p>\n<table>\n<thead>\n<tr>\n<th>Name<\/th>\n<th>Ordered Pair<\/th>\n<th>Coordinates<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Point 1<\/td>\n<td>[latex](0,2)[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=2\\end{array}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Point 2<\/td>\n<td>[latex](\u22122,6)[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}x_{2}=-2\\\\y_{2}=6\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The slope, [latex]m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{6-2}{-2-0}=\\frac{4}{-2}=-2[\/latex]. The slope of the line, <i>m<\/i>, is [latex]\u22122[\/latex].<\/p>\n<p>It doesn\u2019t matter which point is designated as Point 1 and which is Point 2. You could have called [latex](\u22122,6)[\/latex] Point 1, and [latex](0,2)[\/latex] Point 2. In that\u00a0case, putting the coordinates into the slope formula produces the equation [latex]m=\\frac{2-6}{0-\\left(-2\\right)}=\\frac{-4}{2}=-2[\/latex]. Once again, the slope is [latex]m=-2[\/latex]. That\u2019s the same slope as before. The important thing is to be consistent when you subtract: you must always subtract in the same order [latex]\\left(y_{2},y_{1}\\right)[\/latex]<sub>\u00a0<\/sub>and [latex]\\left(x_{2},x_{1}\\right)[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q666697\">Show Solution<\/span><\/p>\n<div id=\"q666697\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=4\\\\y_{1}=2\\end{array}[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)=\\text{Point }1,\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}x_{2}=5\\\\y_{2}=5\\end{array}[\/latex]<\/td>\n<td>[latex]\\left(5,5\\right)=\\text{Point }2,\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{5-2}{5-4}=\\frac{3}{1}\\\\\\\\m=3\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is 3.<\/p><\/div>\n<\/div>\n<\/div>\n<p>The example below shows the solution when you reverse the order of the points, calling [latex](5,5)[\/latex] Point 1 and [latex](4,2)[\/latex] Point 2.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q76013\">Show Solution<\/span><\/p>\n<div id=\"q76013\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=5\\\\y_{1}=5\\end{array}[\/latex]<\/td>\n<td>[latex](5,5)=\\text{Point }1[\/latex], [latex]\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=2\\end{array}[\/latex]<\/td>\n<td>[latex](4,2)=\\text{Point }2[\/latex], [latex]\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{{x_2}-x_{1}}\\\\\\\\m=\\frac{2-5}{4-5}=\\frac{-3}{-1}=3\\\\\\\\m=3\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is 3.<\/p><\/div>\n<\/div>\n<\/div>\n<p>Notice that regardless of which ordered pair is named Point 1 and which is named Point 2, the slope is still 3.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>What is the slope of the line that contains the points [latex](3,-6.25)[\/latex] and [latex](-1,8.5)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q291649\">Show Solution<\/span><\/p>\n<div id=\"q291649\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=3\\\\y_{1}=-6.25\\end{array}[\/latex]<\/td>\n<td>[latex](3,-6.25)=\\text{Point }1[\/latex], [latex]\\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{2}}=-1\\\\{{y}_{2}}=8.5\\end{array}[\/latex]<\/td>\n<td>[latex](-1,8.5)=\\text{Point }2[\/latex], [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{8.5-(-6.25)}{-1-3}\\\\\\\\m=\\frac{14.75}{-4}\\\\\\\\m=-3.6875\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is [latex]-3.6875[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<p>Let\u2019s consider a horizontal line on a graph. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. The equation for this line is [latex]y=3[\/latex]. The equation can also be written as [latex]y=\\left(0\\right)x+3[\/latex].<\/p>\n<p>The following video shows more examples of finding the slope of a line given two points on the line:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Determine the Slope a Line Given Two Points on a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ZW7rQa8SJSU?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-4634\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Ex 1: Determine the Slope a Line Given Two Points on a Line. <strong>Authored by<\/strong>: mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ZW7rQa8SJSU\">https:\/\/youtu.be\/ZW7rQa8SJSU<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Ex 1: Determine the Slope a Line Given Two Points on a Line\",\"author\":\"mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ZW7rQa8SJSU\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"e204ef99-80d0-4dfc-b543-385f2dc38b28","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4634","chapter","type-chapter","status-publish","hentry"],"part":4587,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4634","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4634\/revisions"}],"predecessor-version":[{"id":5079,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4634\/revisions\/5079"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/parts\/4587"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4634\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/media?parent=4634"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapter-type?post=4634"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/contributor?post=4634"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/license?post=4634"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}