{"id":3344,"date":"2024-09-03T17:20:29","date_gmt":"2024-09-03T17:20:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=3344"},"modified":"2025-08-13T15:26:12","modified_gmt":"2025-08-13T15:26:12","slug":"equations-of-lines-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/equations-of-lines-learn-it-6\/","title":{"raw":"Equations of Lines: Learn It 6","rendered":"Equations of Lines: Learn It 6"},"content":{"raw":"<h2>Parallel and Perpendicular Lines Cont.<\/h2>\r\n<h3>Writing Equations of Parallel Lines<\/h3>\r\nSuppose we are given the following equation:\r\n<p style=\"text-align: center;\">[latex]y=3x+1[\/latex]<\/p>\r\nWe know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to [latex]y=3x+1[\/latex]. So all of the following lines will be parallel to the given line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+6\\hfill &amp; \\\\ y=3x+1\\hfill &amp; \\\\ y=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nSuppose then we want to write the equation of a line that is parallel to [latex]y=3x+6[\/latex]\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with point-slope form of a line and then rewrite it in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill &amp; \\\\ y - 7=3\\left(x - 1\\right)\\hfill &amp; \\\\ y - 7=3x - 3\\hfill &amp; \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]y=3x+4[\/latex] is parallel to [latex]y=3x+1[\/latex] and passes through the point (1, 7).\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the equation of a LINE, write the equation of a line parallel to the given line that passes through A given point<\/strong>\r\n<ol>\r\n \t<li>Find the slope of the line.<\/li>\r\n \t<li>Substitute the given values into either point-slope form or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find a line parallel to the graph of [latex]y=3x+6[\/latex] that passes through the point [latex](3, 0)[\/latex].[reveal-answer q=\"672987\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"672987\"]The slope of the given line is [latex]3[\/latex]. If we choose to use slope-intercept form, we can substitute [latex]m\u00a0= 3[\/latex], [latex]x\u00a0= 3[\/latex], and [latex]y\u00a0= 0[\/latex] into slope-intercept form to find the [latex]y[\/latex]<em>-<\/em>intercept.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+b\\hfill &amp; \\\\ \\text{}0=3\\left(3\\right)+b\\hfill &amp; \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\r\nThe line parallel to\u00a0[latex]y=3x+6[\/latex]\u00a0that passes through [latex](3, 0)[\/latex] is [latex]y=3x - 9[\/latex].\r\n\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nWe can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/> Coordinate plane with two lines and labeled slopes[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]293763[\/ohm_question]<\/section>\r\n<h3>Writing Equations of Perpendicular Lines<\/h3>\r\nWe can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following line:\r\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\r\nThe slope of the line is 2 and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to [latex]y=2x+4[\/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=-\\frac{1}{2}x+4\\hfill &amp; \\\\ y=-\\frac{1}{2}x+2\\hfill &amp; \\\\ y=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nAs before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to [latex]y=2x+4[\/latex]\u00a0and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into slope-intercept form and solving for <em>b<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=mx+b\\hfill &amp; \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill &amp; \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill &amp; \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]y=-\\frac{1}{2}x+2[\/latex].\r\n\r\nSo [latex]y=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]y=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?\r\n\r\n<hr \/>\r\n\r\nNo. For two perpendicular linear functions, the product of their slopes is \u20131. As you will learn later, a vertical line is not a function so the definition is not contradicted.\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the equation of a LINE, write the equation of a line Perpendicular to the given line that passes through A given point<\/strong>\r\n<ol>\r\n \t<li>Find the slope of the given line.<\/li>\r\n \t<li>Determine the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Find the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point [latex](3, 0)[\/latex].[reveal-answer q=\"566588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"566588\"]The original line has slope [latex]m\u00a0= 3[\/latex], so the slope of the perpendicular line will be its negative reciprocal or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}y=-\\frac{1}{3}x+b\\hfill &amp; \\\\ \\text{}0=-\\frac{1}{3}\\left(3\\right)+b\\hfill &amp; \\\\ \\text{}1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{array}[\/latex]<\/p>\r\nThe line perpendicular to [latex]y=3x+3[\/latex]\u00a0that passes through [latex](3, 0)[\/latex] is [latex]y=-\\frac{1}{3}x+1[\/latex].\r\n\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nA graph of the two lines is shown below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/> Coordinate plane with intersecting lines[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18933[\/ohm2_question]<\/section>","rendered":"<h2>Parallel and Perpendicular Lines Cont.<\/h2>\n<h3>Writing Equations of Parallel Lines<\/h3>\n<p>Suppose we are given the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]y=3x+1[\/latex]<\/p>\n<p>We know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to [latex]y=3x+1[\/latex]. So all of the following lines will be parallel to the given line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+6\\hfill & \\\\ y=3x+1\\hfill & \\\\ y=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Suppose then we want to write the equation of a line that is parallel to [latex]y=3x+6[\/latex]\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with point-slope form of a line and then rewrite it in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\\\ y - 7=3\\left(x - 1\\right)\\hfill & \\\\ y - 7=3x - 3\\hfill & \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]y=3x+4[\/latex] is parallel to [latex]y=3x+1[\/latex] and passes through the point (1, 7).<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the equation of a LINE, write the equation of a line parallel to the given line that passes through A given point<\/strong><\/p>\n<ol>\n<li>Find the slope of the line.<\/li>\n<li>Substitute the given values into either point-slope form or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find a line parallel to the graph of [latex]y=3x+6[\/latex] that passes through the point [latex](3, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q672987\">Show Solution<\/button><\/p>\n<div id=\"q672987\" class=\"hidden-answer\" style=\"display: none\">The slope of the given line is [latex]3[\/latex]. If we choose to use slope-intercept form, we can substitute [latex]m\u00a0= 3[\/latex], [latex]x\u00a0= 3[\/latex], and [latex]y\u00a0= 0[\/latex] into slope-intercept form to find the [latex]y[\/latex]<em>&#8211;<\/em>intercept.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=3x+b\\hfill & \\\\ \\text{}0=3\\left(3\\right)+b\\hfill & \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\n<p>The line parallel to\u00a0[latex]y=3x+6[\/latex]\u00a0that passes through [latex](3, 0)[\/latex] is [latex]y=3x - 9[\/latex].<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>We can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.<\/p>\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/><figcaption class=\"wp-caption-text\">Coordinate plane with two lines and labeled slopes<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm293763\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=293763&theme=lumen&iframe_resize_id=ohm293763&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Writing Equations of Perpendicular Lines<\/h3>\n<p>We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following line:<\/p>\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\n<p>The slope of the line is 2 and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to [latex]y=2x+4[\/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y=-\\frac{1}{2}x+4\\hfill & \\\\ y=-\\frac{1}{2}x+2\\hfill & \\\\ y=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to [latex]y=2x+4[\/latex]\u00a0and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into slope-intercept form and solving for <em>b<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllll}y=mx+b\\hfill & \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill & \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill & \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]y=-\\frac{1}{2}x+2[\/latex].<\/p>\n<p>So [latex]y=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]y=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/p>\n<hr \/>\n<p>No. For two perpendicular linear functions, the product of their slopes is \u20131. As you will learn later, a vertical line is not a function so the definition is not contradicted.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the equation of a LINE, write the equation of a line Perpendicular to the given line that passes through A given point<\/strong><\/p>\n<ol>\n<li>Find the slope of the given line.<\/li>\n<li>Determine the negative reciprocal of the slope.<\/li>\n<li>Substitute the slope and point into either point-slope form or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">Find the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point [latex](3, 0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q566588\">Show Solution<\/button><\/p>\n<div id=\"q566588\" class=\"hidden-answer\" style=\"display: none\">The original line has slope [latex]m\u00a0= 3[\/latex], so the slope of the perpendicular line will be its negative reciprocal or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}y=-\\frac{1}{3}x+b\\hfill & \\\\ \\text{}0=-\\frac{1}{3}\\left(3\\right)+b\\hfill & \\\\ \\text{}1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{array}[\/latex]<\/p>\n<p>The line perpendicular to [latex]y=3x+3[\/latex]\u00a0that passes through [latex](3, 0)[\/latex] is [latex]y=-\\frac{1}{3}x+1[\/latex].<\/p>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>A graph of the two lines is shown below.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/><figcaption class=\"wp-caption-text\">Coordinate plane with intersecting lines<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18933\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18933&theme=lumen&iframe_resize_id=ohm18933&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":75,"module-header":"learn_it","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\/3344"}],"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\/15"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3344\/revisions"}],"predecessor-version":[{"id":7597,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3344\/revisions\/7597"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/75"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3344\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=3344"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=3344"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=3344"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=3344"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}