{"id":16518,"date":"2019-10-03T17:08:42","date_gmt":"2019-10-03T17:08:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-perpendicualr-lines\/"},"modified":"2024-04-30T23:15:51","modified_gmt":"2024-04-30T23:15:51","slug":"read-or-watch-perpendicualr-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-or-watch-perpendicualr-lines\/","title":{"raw":"Equations of Parallel and Perpendicular Lines","rendered":"Equations of Parallel and Perpendicular Lines"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write equations of parallel and perpendicular lines<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"Write the equations of parallel and perpendicular lines\">Write the equations of parallel and perpendicular lines<\/h2>\r\nThe relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.\r\n\r\nLet\u2019s start with an example involving parallel lines.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]x\u2013y=5[\/latex] and goes through the point [latex](\u22122,1)[\/latex].\r\n\r\n[reveal-answer q=\"763534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"763534\"]\r\n\r\nRewrite the line you want to be parallel to into the\u00a0[latex]y=mx+b[\/latex] form, if needed.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u2212y=\u2212x+5\\\\y=x\u20135\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=1[\/latex] and [latex]b=\u22125[\/latex].\r\n\r\nSince [latex]m=1[\/latex], the slope is [latex]1[\/latex].\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is [latex]1[\/latex].\r\n\r\nUse the method for writing an equation from the slope and a point on the line. Substitute 1 for <i>m<\/i>, and the point [latex](\u22122,1)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=mx+b\\\\1=1(\u22122)+b\\end{array}[\/latex]<\/p>\r\nSolve for b.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=\u22122+b\\\\3=b\\end{array}[\/latex]<\/p>\r\nWrite the equation using the new slope for <i>m<\/i> and the <i>b<\/i> you just found.\r\n<h4>Answer<\/h4>\r\n[latex]y=x+3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Determine the Equation of a Line Parallel to a Line in General Form\">Determine the Equation of a Line Parallel to Another Line Through a Given Point<\/span><\/h2>\r\nhttps:\/\/youtu.be\/TQKz2XHI09E\r\n<h2>Determine the Equation of a Line Perpendicular to Another Line Through a Given Point<\/h2>\r\nWhen you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other.\u00a0To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal. \u00a0In other words, flip it and change the sign.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that contains the point [latex](1,5)[\/latex] and is perpendicular to the line [latex]y=2x\u2013 6[\/latex].\r\n\r\n[reveal-answer q=\"604282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604282\"]\r\n\r\nIdentify the slope of the line you want to be perpendicular to.\r\n\r\nThe given line is written in [latex]y=mx+b[\/latex] form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope is [latex]2[\/latex].\r\n\r\nTo find the slope of a perpendicular line, find the reciprocal, [latex] \\displaystyle \\frac{1}{2}[\/latex], then the opposite, [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nThe slope of the perpendicular line is [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nUse the method for writing an equation from the slope and a point on the line. Substitute [latex] \\displaystyle -\\frac{1}{2}[\/latex] for m, and the point [latex](1,5)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}y=mx+b\\\\5=-\\frac{1}{2}(1)+b\\end{array}[\/latex]<\/p>\r\nSolve for b.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,5=-\\frac{1}{2}+b\\\\\\frac{11}{2}=b\\end{array}[\/latex]<\/p>\r\nWrite the equation using the new slope for m and the b you just found.\r\n<h4>Answer<\/h4>\r\n[latex]y=-\\frac{1}{2}x+\\frac{11}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form<\/h2>\r\nhttps:\/\/youtu.be\/QtvtzKjtowA\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]y=4[\/latex] through the point [latex](0,10)[\/latex].\r\n\r\n[reveal-answer q=\"426450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426450\"]\r\n\r\nRewrite the line into [latex]y=mx+b[\/latex]\u00a0form, if needed.\r\n\r\nYou may notice without doing this that [latex]y=4[\/latex]\u00a0is a horizontal line 4 units above the <i>x<\/i>-axis. Because it is horizontal, you know its slope is zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=4\\\\y=0x+4\\end{array}[\/latex]<\/p>\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=4[\/latex].\r\n\r\nSince [latex]m=0[\/latex], the slope is [latex]0[\/latex]. This is a horizontal line.\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is also [latex]0[\/latex].\r\n\r\nSince the parallel line will be a horizontal line, its form is\r\n<p style=\"text-align: center;\">[latex]y=\\text{a constant}[\/latex]<\/p>\r\nSince we want this new line to pass through the point [latex](0,10)[\/latex], we will need to write the equation of the new line as:\r\n<p style=\"text-align: center;\">[latex]y=10[\/latex]<\/p>\r\nThis line is parallel to [latex]y=4[\/latex]\u00a0and passes through [latex](0,10)[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=10[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of a line that is perpendicular to the line [latex]y=-3[\/latex] through the point [latex](-2,5)[\/latex].\r\n\r\n[reveal-answer q=\"426550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426550\"]\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=-3[\/latex].\r\n\r\nA perpendicular line will have a slope that is the negative reciprocal of the slope of\u00a0[latex]y=-3[\/latex], but\u00a0what does that mean in this case?\r\n\r\nThe reciprocal of [latex]0[\/latex] is [latex]\\frac{1}{0}[\/latex], but we know that dividing by [latex]0[\/latex] is undefined.\r\n\r\nThis means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.\r\n\r\nThe form of a vertical line is [latex]x=\\text{a constant}[\/latex], where every x-value on the line is equal to some constant. \u00a0Since we are looking for a line that goes through the point [latex](-2,5)[\/latex], all of the [latex]x[\/latex]-values on this line must be [latex]-2[\/latex].\r\n\r\nThe equation of a line passing through [latex](-2,5)[\/latex] that is perpendicular to the horizontal line\u00a0[latex]y=-3[\/latex] is therefore,\r\n<p style=\"text-align: center;\">[latex]x=-2[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=-2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]1738[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line<\/span><\/h2>\r\nhttps:\/\/youtu.be\/Qpn3f3wMeIs\r\n<h2 class=\"yt watch-title-container\"><\/h2>\r\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form\"><\/span>Summary<\/h2>\r\nWhen lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a [latex]90\u00b0[\/latex] angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[\/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write equations of parallel and perpendicular lines<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"Write the equations of parallel and perpendicular lines\">Write the equations of parallel and perpendicular lines<\/h2>\n<p>The relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.<\/p>\n<p>Let\u2019s start with an example involving parallel lines.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is parallel to the line [latex]x\u2013y=5[\/latex] and goes through the point [latex](\u22122,1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q763534\">Show Solution<\/span><\/p>\n<div id=\"q763534\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the line you want to be parallel to into the\u00a0[latex]y=mx+b[\/latex] form, if needed.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u2212y=\u2212x+5\\\\y=x\u20135\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Identify the slope of the given line.<\/p>\n<p>In the equation above, [latex]m=1[\/latex] and [latex]b=\u22125[\/latex].<\/p>\n<p>Since [latex]m=1[\/latex], the slope is [latex]1[\/latex].<\/p>\n<p>To find the slope of a parallel line, use the same slope.<\/p>\n<p>The slope of the parallel line is [latex]1[\/latex].<\/p>\n<p>Use the method for writing an equation from the slope and a point on the line. Substitute 1 for <i>m<\/i>, and the point [latex](\u22122,1)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=mx+b\\\\1=1(\u22122)+b\\end{array}[\/latex]<\/p>\n<p>Solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=\u22122+b\\\\3=b\\end{array}[\/latex]<\/p>\n<p>Write the equation using the new slope for <i>m<\/i> and the <i>b<\/i> you just found.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=x+3[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Determine the Equation of a Line Parallel to a Line in General Form\">Determine the Equation of a Line Parallel to Another Line Through a Given Point<\/span><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine the Equation of a Line Parallel to a Line in General Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TQKz2XHI09E?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Determine the Equation of a Line Perpendicular to Another Line Through a Given Point<\/h2>\n<p>When you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other.\u00a0To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal. \u00a0In other words, flip it and change the sign.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that contains the point [latex](1,5)[\/latex] and is perpendicular to the line [latex]y=2x\u2013 6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604282\">Show Solution<\/span><\/p>\n<div id=\"q604282\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify the slope of the line you want to be perpendicular to.<\/p>\n<p>The given line is written in [latex]y=mx+b[\/latex] form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope is [latex]2[\/latex].<\/p>\n<p>To find the slope of a perpendicular line, find the reciprocal, [latex]\\displaystyle \\frac{1}{2}[\/latex], then the opposite, [latex]\\displaystyle -\\frac{1}{2}[\/latex].<\/p>\n<p>The slope of the perpendicular line is [latex]\\displaystyle -\\frac{1}{2}[\/latex].<\/p>\n<p>Use the method for writing an equation from the slope and a point on the line. Substitute [latex]\\displaystyle -\\frac{1}{2}[\/latex] for m, and the point [latex](1,5)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}y=mx+b\\\\5=-\\frac{1}{2}(1)+b\\end{array}[\/latex]<\/p>\n<p>Solve for b.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,5=-\\frac{1}{2}+b\\\\\\frac{11}{2}=b\\end{array}[\/latex]<\/p>\n<p>Write the equation using the new slope for m and the b you just found.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=-\\frac{1}{2}x+\\frac{11}{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form<\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QtvtzKjtowA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is parallel to the line [latex]y=4[\/latex] through the point [latex](0,10)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426450\">Show Solution<\/span><\/p>\n<div id=\"q426450\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rewrite the line into [latex]y=mx+b[\/latex]\u00a0form, if needed.<\/p>\n<p>You may notice without doing this that [latex]y=4[\/latex]\u00a0is a horizontal line 4 units above the <i>x<\/i>-axis. Because it is horizontal, you know its slope is zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=4\\\\y=0x+4\\end{array}[\/latex]<\/p>\n<p>Identify the slope of the given line.<\/p>\n<p>In the equation above, [latex]m=0[\/latex] and [latex]b=4[\/latex].<\/p>\n<p>Since [latex]m=0[\/latex], the slope is [latex]0[\/latex]. This is a horizontal line.<\/p>\n<p>To find the slope of a parallel line, use the same slope.<\/p>\n<p>The slope of the parallel line is also [latex]0[\/latex].<\/p>\n<p>Since the parallel line will be a horizontal line, its form is<\/p>\n<p style=\"text-align: center;\">[latex]y=\\text{a constant}[\/latex]<\/p>\n<p>Since we want this new line to pass through the point [latex](0,10)[\/latex], we will need to write the equation of the new line as:<\/p>\n<p style=\"text-align: center;\">[latex]y=10[\/latex]<\/p>\n<p>This line is parallel to [latex]y=4[\/latex]\u00a0and passes through [latex](0,10)[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=10[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of a line that is perpendicular to the line [latex]y=-3[\/latex] through the point [latex](-2,5)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426550\">Show Solution<\/span><\/p>\n<div id=\"q426550\" class=\"hidden-answer\" style=\"display: none\">\n<p>In the equation above, [latex]m=0[\/latex] and [latex]b=-3[\/latex].<\/p>\n<p>A perpendicular line will have a slope that is the negative reciprocal of the slope of\u00a0[latex]y=-3[\/latex], but\u00a0what does that mean in this case?<\/p>\n<p>The reciprocal of [latex]0[\/latex] is [latex]\\frac{1}{0}[\/latex], but we know that dividing by [latex]0[\/latex] is undefined.<\/p>\n<p>This means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.<\/p>\n<p>The form of a vertical line is [latex]x=\\text{a constant}[\/latex], where every x-value on the line is equal to some constant. \u00a0Since we are looking for a line that goes through the point [latex](-2,5)[\/latex], all of the [latex]x[\/latex]-values on this line must be [latex]-2[\/latex].<\/p>\n<p>The equation of a line passing through [latex](-2,5)[\/latex] that is perpendicular to the horizontal line\u00a0[latex]y=-3[\/latex] is therefore,<\/p>\n<p style=\"text-align: center;\">[latex]x=-2[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=-2[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1738\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1738&theme=oea&iframe_resize_id=ohm1738&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\">Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line<\/span><\/h2>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Qpn3f3wMeIs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 class=\"yt watch-title-container\"><\/h2>\n<h2 class=\"yt watch-title-container\"><span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form\"><\/span>Summary<\/h2>\n<p>When lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a [latex]90\u00b0[\/latex] angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[\/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.<\/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-16518\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Determine the Equation of a Line Parallel to a Line in General Form. <strong>Authored by<\/strong>: Mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/TQKz2XHI09E\">https:\/\/youtu.be\/TQKz2XHI09E<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form. <strong>Authored by<\/strong>: Mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QtvtzKjtowA\">https:\/\/youtu.be\/QtvtzKjtowA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line. <strong>Authored by<\/strong>: Mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Qpn3f3wMeIs\">https:\/\/youtu.be\/Qpn3f3wMeIs<\/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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Determine the Equation of a Line Parallel to a Line in General Form\",\"author\":\"Mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/TQKz2XHI09E\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form\",\"author\":\"Mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/QtvtzKjtowA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line\",\"author\":\"Mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Qpn3f3wMeIs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"33769246025745d88f2fea56ee77e766, 3a202cc33fe4496da42ba211f9614e3e","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16518","chapter","type-chapter","status-publish","hentry"],"part":8524,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16518","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":8,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16518\/revisions"}],"predecessor-version":[{"id":19847,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16518\/revisions\/19847"}],"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\/16518\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16518"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16518"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16518"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}