{"id":1677,"date":"2024-06-03T18:49:09","date_gmt":"2024-06-03T18:49:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1677"},"modified":"2024-12-20T17:51:34","modified_gmt":"2024-12-20T17:51:34","slug":"module-7-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/module-7-background-youll-need-2\/","title":{"raw":"Linear Functions: Background You'll Need 2","rendered":"Linear Functions: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Find the equation of the line.<\/li>\r\n<\/ul>\r\n<\/section>\r\n\r\n<h2>The Equation of a Line<\/h2>\r\nWhen data is collected, a linear model can be created from two data points. Let's see how to find an equation of a line when two points are given by following the steps below.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Find the equation of a line given two points:<\/strong>\r\n\r\n<ol>\r\n \t<li>Find the slope using the given points.<\/li>\r\n \t<li>Choose one point and label its coordinates <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">[latex](x_1, y_1)[\/latex].<\/span><\/li>\r\n \t<li>Plug [latex]m[\/latex], [latex]x_1[\/latex], and [latex]y_1[\/latex] into the point-slope form, [latex]y - y_1 = m(x - x_1)[\/latex].<\/li>\r\n \t<li>Rewrite the equation in slope-intercept form, [latex]y = mx+b[\/latex].<\/li>\r\n<\/ol>\r\n\r\n<\/section><section class=\"textbox example\">\r\nFind the equation of a line containing the points [latex](-4, -3)[\/latex] and [latex](1, -5)[\/latex].\r\n<ol>\r\n \t<li>Calculate the slope.\r\n<center>[latex]\\begin{align*} m &amp;= \\frac{y_2 - y_1}{x_2 - x_1} \\\\ &amp;= \\frac{-5 - (-3)}{1 - (-4)} \\\\ &amp;= \\frac{-5 + 3}{1 + 4} \\\\ &amp;= \\frac{-2}{5} \\end{align*}[\/latex]<\/center><\/li>\r\n \t<li>Let's use [latex](-4, -3)[\/latex] as <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">[latex](x_1, y_1)[\/latex].<\/span><\/li>\r\n \t<li>Plug the values above into the point-slope form of the equation [latex]y - y_1 = m(x - x_1)[\/latex].\r\n<center>[latex]\\begin{align*} y - (-3) &amp;= -\\frac{2}{5}(x - (-4)) \\\\ y + 3 &amp;= -\\frac{2}{5}(x + 4) \\end{align*}[\/latex]<\/center><\/li>\r\n \t<li>Simplify to the slope-intercept form, [latex]y = mx+b[\/latex].\r\n<center>[latex]\\begin{align*} y + 3 &amp;= -\\frac{2}{5}x - \\frac{2}{5} \\cdot 4 \\\\ y + 3 &amp;= -\\frac{2}{5}x - \\frac{8}{5} \\\\ y &amp;= -\\frac{2}{5}x - \\frac{8}{5} - 3 \\\\ y &amp;= -\\frac{2}{5}x - \\frac{8}{5} - \\frac{15}{5} \\\\ y &amp;= -\\frac{2}{5}x - \\frac{23}{5} \\end{align*}[\/latex]<\/center><\/li>\r\n<\/ol>\r\n<strong>So, the equation of the line is: <\/strong>\r\n<center>[latex]y = -\\dfrac{2}{5}x - \\dfrac{23}{5}[\/latex]<\/center>\r\n\r\n[reveal-answer q=\"204995\"]Standard Form[\/reveal-answer]\r\n[hidden-answer a=\"204995\"]We can also rewrite the equation of the line in the standard form [latex]Ax+By = C[\/latex].\r\n\r\n<center>[latex]\\begin{align*} &amp; \\text{Start with the slope-intercept form:} &amp; y &amp;= -\\frac{2}{5}x - \\frac{23}{5} \\\\ &amp; \\text{Eliminate the fractions by multiplying through by 5:} &amp; 5y &amp;= -2x - 23 \\\\ &amp; \\text{Rearrange the terms to get } Ax + By = C: &amp; 2x + 5y &amp;= -23 \\end{align*}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]23244[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]23245[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]23246[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find the equation of the line.<\/li>\n<\/ul>\n<\/section>\n<h2>The Equation of a Line<\/h2>\n<p>When data is collected, a linear model can be created from two data points. Let&#8217;s see how to find an equation of a line when two points are given by following the steps below.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Find the equation of a line given two points:<\/strong><\/p>\n<ol>\n<li>Find the slope using the given points.<\/li>\n<li>Choose one point and label its coordinates <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">[latex](x_1, y_1)[\/latex].<\/span><\/li>\n<li>Plug [latex]m[\/latex], [latex]x_1[\/latex], and [latex]y_1[\/latex] into the point-slope form, [latex]y - y_1 = m(x - x_1)[\/latex].<\/li>\n<li>Rewrite the equation in slope-intercept form, [latex]y = mx+b[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\nFind the equation of a line containing the points [latex](-4, -3)[\/latex] and [latex](1, -5)[\/latex].<\/p>\n<ol>\n<li>Calculate the slope.\n<div style=\"text-align: center;\">[latex]\\begin{align*} m &= \\frac{y_2 - y_1}{x_2 - x_1} \\\\ &= \\frac{-5 - (-3)}{1 - (-4)} \\\\ &= \\frac{-5 + 3}{1 + 4} \\\\ &= \\frac{-2}{5} \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Let&#8217;s use [latex](-4, -3)[\/latex] as <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">[latex](x_1, y_1)[\/latex].<\/span><\/li>\n<li>Plug the values above into the point-slope form of the equation [latex]y - y_1 = m(x - x_1)[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align*} y - (-3) &= -\\frac{2}{5}(x - (-4)) \\\\ y + 3 &= -\\frac{2}{5}(x + 4) \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Simplify to the slope-intercept form, [latex]y = mx+b[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align*} y + 3 &= -\\frac{2}{5}x - \\frac{2}{5} \\cdot 4 \\\\ y + 3 &= -\\frac{2}{5}x - \\frac{8}{5} \\\\ y &= -\\frac{2}{5}x - \\frac{8}{5} - 3 \\\\ y &= -\\frac{2}{5}x - \\frac{8}{5} - \\frac{15}{5} \\\\ y &= -\\frac{2}{5}x - \\frac{23}{5} \\end{align*}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p><strong>So, the equation of the line is: <\/strong><\/p>\n<div style=\"text-align: center;\">[latex]y = -\\dfrac{2}{5}x - \\dfrac{23}{5}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q204995\">Standard Form<\/button><\/p>\n<div id=\"q204995\" class=\"hidden-answer\" style=\"display: none\">We can also rewrite the equation of the line in the standard form [latex]Ax+By = C[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} & \\text{Start with the slope-intercept form:} & y &= -\\frac{2}{5}x - \\frac{23}{5} \\\\ & \\text{Eliminate the fractions by multiplying through by 5:} & 5y &= -2x - 23 \\\\ & \\text{Rearrange the terms to get } Ax + By = C: & 2x + 5y &= -23 \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm23244\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=23244&theme=lumen&iframe_resize_id=ohm23244&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm23245\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=23245&theme=lumen&iframe_resize_id=ohm23245&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm23246\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=23246&theme=lumen&iframe_resize_id=ohm23246&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":164,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1677"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1677\/revisions"}],"predecessor-version":[{"id":6811,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1677\/revisions\/6811"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/164"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1677\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1677"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1677"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1677"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1677"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}