{"id":16515,"date":"2019-10-03T17:08:40","date_gmt":"2019-10-03T17:08:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-read-or-watch-given-two-points-write-the-equation-of-a-line\/"},"modified":"2024-04-30T23:16:07","modified_gmt":"2024-04-30T23:16:07","slug":"read-or-watch-read-or-watch-given-two-points-write-the-equation-of-a-line","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-read-or-watch-given-two-points-write-the-equation-of-a-line\/","title":{"raw":"Writing Equations Using Point-Slope","rendered":"Writing Equations Using Point-Slope"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write the equation of a line using slope and a point on the line<\/li>\r\n \t<li>Write the equation of a line using two points on the line<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"Find the Equation of a Line Given the Slope and a Point on the Line\">Find the Equation of a Line Given the Slope and a Point on the Line<\/h2>\r\nUsing the slope-intercept equation of a line is possible when you know both the slope (m) and the [latex]y[\/latex]-intercept (b), but what if you know the slope and just any point on the line, not specifically the [latex]y[\/latex]-intercept? Can you still write the equation? The answer is yes, but you will need to put in a little more thought and work than you did previously.\r\n\r\nRecall that a point is an [latex](x, y)[\/latex] coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don\u2019t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[\/latex].\r\n\r\nYou do know the slope (m), but you just don\u2019t know the value of the y-intercept (b). Since point \u00a0[latex](x, y)[\/latex] is a solution to the equation, you can substitute its coordinates for [latex]x[\/latex] and [latex]y[\/latex] in [latex]y=mx+b[\/latex]\u00a0and solve to find b!\r\n\r\nThis may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line that has a slope of [latex]3[\/latex] and contains the point [latex](1,4)[\/latex].\r\n\r\n[reveal-answer q=\"161353\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161353\"]\r\n\r\nSubstitute the slope (m) into\u00a0[latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center;\">[latex]y=3x+b[\/latex]<\/p>\r\nSubstitute the point [latex](1,4)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]4=3\\left(1\\right)+b[\/latex]<\/p>\r\nSolve for b.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4=3+b\\\\1=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=3[\/latex]\u00a0and [latex]b=1[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=3x+1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nTo confirm our algebra, you can check by graphing the equation [latex]y=3x+1[\/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064327\/image045.jpg\" alt=\"An uphill line passes through the y-intercept of (0,1) and the point (1,4). The rise is 3 and the run is 1.\" width=\"348\" height=\"349\" \/>\r\n\r\n&nbsp;\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nWrite the equation of the line that has a slope of [latex]\\frac{7}{8}[\/latex]\u00a0and contains the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"31452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"31452\"]\r\n\r\nSubstitute the slope (m) into [latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=mx+b\\\\\\\\y=\\frac{7}{8}x+b\\end{array}[\/latex]<\/p>\r\nSubstitute the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex]\u00a0for x and y.\r\n<p style=\"text-align: center;\">[latex]\\frac{5}{4}=\\frac{7}{8}\\left(4\\right)+b[\/latex]<\/p>\r\nSolve for b.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=\\frac{28}{8}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=\\frac{14}{4}+b\\\\\\\\\\frac{5}{4}-\\frac{14}{4}=\\frac{14}{4}-\\frac{14}{4}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\frac{9}{4}=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex] with [latex] \\displaystyle m=\\frac{7}{8}[\/latex] and [latex] \\displaystyle b=-\\frac{9}{4}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=\\frac{7}{8}x-\\frac{9}{4}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"video2\">Watch the video below for another example of how to find the equation given the slope and a point on the line.<\/p>\r\nhttps:\/\/youtu.be\/URYnKqEctgc\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]31488[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 id=\"Find the Equation of a Line Given Two Points on the Line\">Find the Equation of a Line Given Two Points on the Line<\/h2>\r\nLet\u2019s suppose you don\u2019t know either the slope or the [latex]y[\/latex]-intercept, but you do know the location of two points on the line. It is more challenging, but you can find the equation of the line that would pass through those two points. You will again use slope-intercept form to help you.\r\n\r\nThe slope of a linear equation is always the same, no matter which two points you use to find the slope. Since you have two points, you can use those points to find the slope (m). Now you have the slope and a point on the line! You can now substitute values for [latex]m, x[\/latex], and [latex]y[\/latex] into the equation [latex]y=mx+b[\/latex] and find b.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line that passes through the points [latex](2,1)[\/latex] and [latex](\u22121,\u22125)[\/latex].\r\n\r\n[reveal-answer q=\"333536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333536\"]\r\n\r\nFind the slope using the given points.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{1-(-5)}{2-(-1)}=\\frac{6}{3}=2[\/latex]<\/p>\r\nSubstitute the slope (m) into [latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center;\">[latex]y=2x+b[\/latex]<\/p>\r\nSubstitute the coordinates of either point for [latex]x[\/latex] and [latex]y[\/latex]\u2013 this example uses [latex](2, 1)[\/latex].\r\n<p style=\"text-align: center;\">[latex]1=2(2)+b[\/latex]<\/p>\r\nSolve for b.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,1=4+b\\\\\u22123=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=2[\/latex] and [latex]b=-3[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]\\begin{array}{l}y=2x+\\left(-3\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\\\y=2x-3\\end{array}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that is doesn\u2019t matter which point you use when you substitute and solve for b\u2014you get the same result for b either way. In the example above, you substituted the coordinates of the point [latex](2, 1)[\/latex] in the equation [latex]y=2x+b[\/latex]. Let\u2019s start with the same equation, [latex]y=2x+b[\/latex], but substitute in [latex](\u22121,\u22125)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,y=2x+b\\\\-5=2\\left(-1\\right)+b\\\\-5=-2+b\\\\-3=b\\end{array}[\/latex]<\/p>\r\nThe final equation is the same: [latex]y=2x\u20133[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nWrite the equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex].\r\n\r\n[reveal-answer q=\"347882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"347882\"]\r\n\r\nFind the slope using the given points.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\frac{7.6-6.45}{1.15-(-4.6)}=\\frac{1.15}{5.75}=0.2[\/latex]<\/p>\r\nSubstitute the slope (m) into [latex] \\displaystyle y=mx+b[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\displaystyle y=0.2x+b[\/latex]<\/p>\r\nSubstitute either point for [latex]x[\/latex] and [latex]y[\/latex]\u2014this example uses [latex](1.15,7.6)[\/latex]. Then solve for b.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.2(1.15)+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\underline{-0.23\\,\\,\\,\\,-0.23}\\\\\\,\\,\\,\\,\\,7.37\\,=\\,\\,b\\end{array}[\/latex]<\/p>\r\nRewrite [latex] \\displaystyle y=mx+b[\/latex] with [latex]m=0.2[\/latex] and [latex]b=7.37[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\displaystyle y=0.2x+7.37[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex] is [latex]y=0.2x+7.37[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/P1ex_a6iYDo\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]152704[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write the equation of a line using slope and a point on the line<\/li>\n<li>Write the equation of a line using two points on the line<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"Find the Equation of a Line Given the Slope and a Point on the Line\">Find the Equation of a Line Given the Slope and a Point on the Line<\/h2>\n<p>Using the slope-intercept equation of a line is possible when you know both the slope (m) and the [latex]y[\/latex]-intercept (b), but what if you know the slope and just any point on the line, not specifically the [latex]y[\/latex]-intercept? Can you still write the equation? The answer is yes, but you will need to put in a little more thought and work than you did previously.<\/p>\n<p>Recall that a point is an [latex](x, y)[\/latex] coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don\u2019t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[\/latex].<\/p>\n<p>You do know the slope (m), but you just don\u2019t know the value of the y-intercept (b). Since point \u00a0[latex](x, y)[\/latex] is a solution to the equation, you can substitute its coordinates for [latex]x[\/latex] and [latex]y[\/latex] in [latex]y=mx+b[\/latex]\u00a0and solve to find b!<\/p>\n<p>This may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of the line that has a slope of [latex]3[\/latex] and contains the point [latex](1,4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161353\">Show Solution<\/span><\/p>\n<div id=\"q161353\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the slope (m) into\u00a0[latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]y=3x+b[\/latex]<\/p>\n<p>Substitute the point [latex](1,4)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]4=3\\left(1\\right)+b[\/latex]<\/p>\n<p>Solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}4=3+b\\\\1=b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=3[\/latex]\u00a0and [latex]b=1[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=3x+1[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>To confirm our algebra, you can check by graphing the equation [latex]y=3x+1[\/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[\/latex].<\/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\/1468\/2016\/02\/04064327\/image045.jpg\" alt=\"An uphill line passes through the y-intercept of (0,1) and the point (1,4). The rise is 3 and the run is 1.\" width=\"348\" height=\"349\" \/><\/p>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>Write the equation of the line that has a slope of [latex]\\frac{7}{8}[\/latex]\u00a0and contains the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q31452\">Show Solution<\/span><\/p>\n<div id=\"q31452\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the slope (m) into [latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=mx+b\\\\\\\\y=\\frac{7}{8}x+b\\end{array}[\/latex]<\/p>\n<p>Substitute the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex]\u00a0for x and y.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{5}{4}=\\frac{7}{8}\\left(4\\right)+b[\/latex]<\/p>\n<p>Solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=\\frac{28}{8}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=\\frac{14}{4}+b\\\\\\\\\\frac{5}{4}-\\frac{14}{4}=\\frac{14}{4}-\\frac{14}{4}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-\\frac{9}{4}=b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]y=mx+b[\/latex] with [latex]\\displaystyle m=\\frac{7}{8}[\/latex] and [latex]\\displaystyle b=-\\frac{9}{4}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=\\frac{7}{8}x-\\frac{9}{4}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p id=\"video2\">Watch the video below for another example of how to find the equation given the slope and a point on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Determine a Linear Equation Given Slope and a Point (Slope-Intercept Form) (09x-32)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/URYnKqEctgc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm31488\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=31488&theme=oea&iframe_resize_id=ohm31488&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 id=\"Find the Equation of a Line Given Two Points on the Line\">Find the Equation of a Line Given Two Points on the Line<\/h2>\n<p>Let\u2019s suppose you don\u2019t know either the slope or the [latex]y[\/latex]-intercept, but you do know the location of two points on the line. It is more challenging, but you can find the equation of the line that would pass through those two points. You will again use slope-intercept form to help you.<\/p>\n<p>The slope of a linear equation is always the same, no matter which two points you use to find the slope. Since you have two points, you can use those points to find the slope (m). Now you have the slope and a point on the line! You can now substitute values for [latex]m, x[\/latex], and [latex]y[\/latex] into the equation [latex]y=mx+b[\/latex] and find b.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of the line that passes through the points [latex](2,1)[\/latex] and [latex](\u22121,\u22125)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333536\">Show Solution<\/span><\/p>\n<div id=\"q333536\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the slope using the given points.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{1-(-5)}{2-(-1)}=\\frac{6}{3}=2[\/latex]<\/p>\n<p>Substitute the slope (m) into [latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]y=2x+b[\/latex]<\/p>\n<p>Substitute the coordinates of either point for [latex]x[\/latex] and [latex]y[\/latex]\u2013 this example uses [latex](2, 1)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]1=2(2)+b[\/latex]<\/p>\n<p>Solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,1=4+b\\\\\u22123=b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=2[\/latex] and [latex]b=-3[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\begin{array}{l}y=2x+\\left(-3\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\\\y=2x-3\\end{array}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Notice that is doesn\u2019t matter which point you use when you substitute and solve for b\u2014you get the same result for b either way. In the example above, you substituted the coordinates of the point [latex](2, 1)[\/latex] in the equation [latex]y=2x+b[\/latex]. Let\u2019s start with the same equation, [latex]y=2x+b[\/latex], but substitute in [latex](\u22121,\u22125)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,y=2x+b\\\\-5=2\\left(-1\\right)+b\\\\-5=-2+b\\\\-3=b\\end{array}[\/latex]<\/p>\n<p>The final equation is the same: [latex]y=2x\u20133[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>Write the equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q347882\">Show Solution<\/span><\/p>\n<div id=\"q347882\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the slope using the given points.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\frac{7.6-6.45}{1.15-(-4.6)}=\\frac{1.15}{5.75}=0.2[\/latex]<\/p>\n<p>Substitute the slope (m) into [latex]\\displaystyle y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle y=0.2x+b[\/latex]<\/p>\n<p>Substitute either point for [latex]x[\/latex] and [latex]y[\/latex]\u2014this example uses [latex](1.15,7.6)[\/latex]. Then solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.2(1.15)+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\underline{-0.23\\,\\,\\,\\,-0.23}\\\\\\,\\,\\,\\,\\,7.37\\,=\\,\\,b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]\\displaystyle y=mx+b[\/latex] with [latex]m=0.2[\/latex] and [latex]b=7.37[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle y=0.2x+7.37[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex] is [latex]y=0.2x+7.37[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Find the Equation of a Line in Slope Intercept Form Given Two Points\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/P1ex_a6iYDo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm152704\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=152704&theme=oea&iframe_resize_id=ohm152704&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n","protected":false},"author":169554,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"33769246025745d88f2fea56ee77e766, a2157c5160df49a0ae5539c98ac9c76d, 528e3b104ae2431190ef940d08adf28d","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16515","chapter","type-chapter","status-publish","hentry"],"part":8524,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16515","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16515\/revisions"}],"predecessor-version":[{"id":20472,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16515\/revisions\/20472"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/8524"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16515\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16515"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16515"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16515"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}