{"id":117358,"date":"2022-03-08T10:15:57","date_gmt":"2022-03-08T16:15:57","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=117358"},"modified":"2026-03-28T11:50:52","modified_gmt":"2026-03-28T16:50:52","slug":"equations-of-perpendicular-lines","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/equations-of-perpendicular-lines\/","title":{"rendered":"Equations of Perpendicular Lines"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_zy5RL99NpRQ\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_zy5RL99NpRQ\" data-source-videoID=\"zy5RL99NpRQ\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Equations of Perpendicular Lines Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Equations of Perpendicular Lines\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_zy5RL99NpRQ:hover {cursor:pointer;} img#videoThumbnailImage_zy5RL99NpRQ {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/2326-thumbnail-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_zy5RL99NpRQ\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_zy5RL99NpRQ\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Equations of Perpendicular Lines\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_zy5RL99NpRQ\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_zy5RL99NpRQ\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_zy5RL99NpRQ\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction rO6_Function() {\n  var x = document.getElementById(\"rO6\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"rO6_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"rO6\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Practice_Questions\" class=\"smooth-scroll\">Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hello! Today we\u2019re going to take a look at how to find the equation of a line that passes through a given point and is perpendicular to another line.<\/p>\n<h2><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h2>\n<p>\nLet\u2019s start with an example.<\/p>\n<p>Find the equation of a line that passes through the point \\((1,7)\\) and is perpendicular to the line \\(y=4x-3\\).<\/p>\n<p>Remember, perpendicular lines have slopes that are negative reciprocals of one another. So, to determine the slope of the line we are looking for, we must first determine the slope of the line we are given. The line \\(y=4x-3\\) is in slope-intercept form \\((y=mx+b)\\), so we know that our slope \\((m)\\) is 4.<\/p>\n<div class=\"examplesentence\">\\(y=4x-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we need to find the negative reciprocal of 4 to determine the slope of our line. The negative reciprocal of 4 is \\(-\\frac{1}{4}\\), so the slope of our line is \\(-\\frac{1}{4}\\).<\/p>\n<div class=\"examplesentence\">\\(m=-\\frac{1}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we know the slope of the line and that the line passes through the point \\((1,7)\\), we can plug these values into either our general slope-intercept equation or our general point-slope equation. For this example, we\u2019ll use the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/slope-intercept-and-point-slope-forms\/\">slope-intercept equation<\/a>.<\/p>\n<div class=\"examplesentence\">\\(y=mx+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe know our value for \\(m\\) and values for \\(x\\) and \\(y\\) that make our equation true (because the line passes through them), so let\u2019s plug in these three values and solve for \\(b\\).<\/p>\n<div class=\"examplesentence\">\\(7=-\\frac{1}{4}(1)+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nFirst, multiply.<\/p>\n<div class=\"examplesentence\">\\(7=-\\frac{1}{4}+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we need to convert 7 into a fraction that can be added with \\(\\frac{1}{4}\\). So to do that, we need to remember that a whole number is always a fraction over 1. So \\(\\frac{7}{1}\\) is the same as 7. Now, to convert \\(\\frac{7}{1}\\) into a fraction with a denominator of 4, we&#8217;ll have to multiply both the denominator and the numerator by 4. That will give us the fraction \\(\\frac{28}{4}\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{28}{4}=-\\frac{1}{4}+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, to solve for \\(b\\), we&#8217;re going to add \\(\\frac{1}{4}\\) to both sides of the equation.<\/p>\n<div class=\"examplesentence\">\\(\\frac{29}{4}=b\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we know what \\(b\\) is, we can use our values for \\(m\\) and \\(b\\) to create the equation for our line.<\/p>\n<div class=\"examplesentence\">\\(y=-\\frac{1}{4}x+\\frac{29}{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nDon\u2019t be scared by these fractions! As you continue on to higher math, you will start using fractions more and more frequently, especially improper fractions. They\u2019re actually easier to work with than decimals, so you don\u2019t need to convert these values to decimals, leave them as fractions.<\/p>\n<h2><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h2>\n<p>\nLet\u2019s try another example!<\/p>\n<p>Find the equation of a line that passes through the point \\((-2,6)\\) and is perpendicular to the line \\(y=-2x+3\\).<\/p>\n<p>First, find the slope of the line we are given. Again, this is in slope-intercept form, so our slope is \\(-2\\).<\/p>\n<div class=\"examplesentence\">\\(y=-2x+3\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, use this value to find the slope of the line we are looking for. Remember, the slopes of perpendicular lines are negative reciprocals of one another, so the slope of the line we are looking for is:<\/p>\n<div class=\"examplesentence\">\\(m=\\frac{1}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor this example, we are going to find our equation by plugging in our slope and point into the point-slope equation.<\/p>\n<div class=\"examplesentence\">\\(y-y_{1}=m(x-x_{1})\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, since this form is created to tell you a specific point on the line, we will plug in our point \\((-2,6)\\) for \\((x_{1},y_{1})\\) instead of \\((x,y)\\) like we did last time. Let\u2019s plug in what we know.<\/p>\n<div class=\"examplesentence\">\\(y-6=\\frac{1}{2}(x-(-2))\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, if we want our answer in point-slope form, we can stop right here. However, if we want to have it in slope-intercept form, we need to do a little bit of manipulating. First, change the subtracting a negative number to addition.<\/p>\n<div class=\"examplesentence\">\\(y-6=\\frac{1}{2}(x+2)\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, distribute the \\(\\frac{1}{2}\\) on the right side of the equation.<\/p>\n<div class=\"examplesentence\">\\(y-6=\\frac{1}{2}x+1\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, add 6 to both sides.<\/p>\n<div class=\"examplesentence\">\\(y-6+6=\\frac{1}{2}x+1+6\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\">\\(y=\\frac{1}{2}x+7\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd there you have it! Before we go, let\u2019s try one more problem together.<\/p>\n<p>Find the equation of a line that passes through the point \\((5,-7)\\) and is perpendicular to the line \\(y=-\\frac{1}{3}x-9\\).<\/p>\n<p>First, find the slope of the given line. This line is in slope-intercept form, so the slope is \\(-\\frac{1}{3}\\).<\/p>\n<div class=\"examplesentence\">\\(y=-\\frac{1}{3}x-9\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, find the opposite reciprocal of this number to find the slope of the perpendicular line.<\/p>\n<div class=\"examplesentence\">\\(m=3\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, plug in your slope and point into either the slope-intercept equation or the point-slope equation. We\u2019ll work through both ways for this example, starting with the slope-intercept equation.<\/p>\n<div class=\"examplesentence\">\\(y=mx+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nPlug in the values for \\(m\\), \\(x\\), and \\(y\\). Then, solve for \\(b\\).<\/p>\n<div class=\"examplesentence\">\\(-7=3(5)+b\\)<br \/>\n\\(-7=15+b\\)<br \/>\n\\(-22=b\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we know our value for \\(b\\), we can plug it into the slope-intercept equation, along with the value for \\(m\\).<\/p>\n<div class=\"examplesentence\">\\(y=3x-22\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we wanted to use the point-slope equation instead, we would get the same answer. Let\u2019s try it that way!<\/p>\n<div class=\"examplesentence\">\\(y-y_{1}=m(x-x_{1})\\)<\/div>\n<p>\n&nbsp;<br \/>\nPlug in the values for \\(m\\), \\(x_{1}\\), and \\(y_{1}\\).<\/p>\n<div class=\"examplesentence\">\\(y-(-7)=3(x-5)\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we simplify the subtracting a negative number part, we get:<\/p>\n<div class=\"examplesentence\">\\(y+7=3(x-5)\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we can distribute and rearrange to convert our equation to slope-intercept form.<\/p>\n<div class=\"examplesentence\">\\(y+7=3x-15\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubtract 7 from both sides of the equation.<\/p>\n<div class=\"examplesentence\">\\(y=7-7=3x-15-7\\)<br \/>\n\\(y=3x-22\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur answers match, just like they should! This proves that either way you want to solve the equation, using slope-intercept form or point-slope form, will get you the correct answer.<\/p>\n<p>I hope that this video on equations of perpendicular lines was helpful. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\">Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the equation of the line that contains the point \\((-10,6)\\) and is perpendicular to the line \\(y=5x+3\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(y=-5x-44\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(y=-5x+56\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\(y=-\\frac{1}{5}x+4\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(y=-\\frac{1}{5}x-4\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Two lines are perpendicular if their slopes are negative reciprocals of each other. The given line is in slope-intercept form \\(y=mx+b\\), where \\(m\\) is the slope and \\(b\\) is the \\(y\\)-intercept. For the given line, the slope is \\(m=5\\). The negative reciprocal of 5 is \\(-\\frac{1}{5}\\), so the slope of the perpendicular line is \\(m=-\\frac{1}{5}\\).<\/p>\n<p>Substitute the given point and the slope into the slope-intercept form of a line to find the equation for the line containing the point \\((-10,6)\\). Substitute the values into slope-intercept form, then solve for \\(b\\).<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(6=-\\frac{1}{5}(-10)+b\\)<br \/>\n\\(6=2+b\\)<br \/>\n\\(6-2=2+b-2\\)<br \/>\n\\(4=b\\)<\/p>\n<p>Plug \\(b=4\\) and \\(m=-\\frac{1}{5}\\) into the slope-intercept form of a line. The equation of the new line is:<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(y=-\\frac{1}{5}x+4\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the equation of the line that contains the point \\((-3,8)\\) and is perpendicular to the line \\(y=-4x+1\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\(y=\\frac{1}{4}x+\\frac{35}{4}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(y=\\frac{1}{4}x-5\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(y=4x+20\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(y=4x-35\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Two lines are perpendicular if their slopes are negative reciprocals of each other. The given line is in slope-intercept form \\(y=mx+b\\), where \\(m\\) is the slope and \\(b\\) is the \\(y\\)-intercept. For the given line, the slope is \\(m=-4\\). The negative reciprocal of \u20134 is \\(\\frac{1}{4}\\), so the slope of the perpendicular line is \\(m=\\frac{1}{4}\\).<\/p>\n<p>Substitute the given point and the slope into the slope-intercept form of a line to find the equation for the line containing the point \\((-3,8)\\). Substitute the values into the slope-intercept form, then solve for \\(b\\).<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(8=\\frac{1}{4}(-3)+b\\)<br \/>\n\\(8=-\\frac{3}{4}+b\\)<br \/>\n\\(8+\\frac{3}{4}=-\\frac{3}{4}+b+\\frac{3}{4}\\)<br \/>\n\\(\\frac{35}{4}=b\\)<\/p>\n<p>Plug \\(b=\\frac{35}{4}\\) and \\(m=\\frac{1}{4}\\) into the slope-intercept form of a line, the equation of the new line is:<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(y=\\frac{1}{4}x+\\frac{35}{4}\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the equation of the line that contains the point \\((2,-5)\\) and is perpendicular to the line \\(y=-\\frac{1}{2}x-6\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(y=\\frac{1}{2}x+\\frac{9}{2}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(y=\\frac{1}{2}x-6\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(y=\\frac{1}{2}x-6\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(y=2x-9\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Two lines are perpendicular if their slopes are negative reciprocals of each other. The given line is in slope-intercept form \\(y=mx+b\\), where \\(m\\) is the slope and \\(b\\) is the \\(y\\)-intercept. For the given line, the slope is \\(m=-\\frac{1}{2}\\). The negative reciprocal of \\(-\\frac{1}{2}\\) is 2, so the slope of the perpendicular line is \\(m=2\\).<\/p>\n<p>We can substitute the given point and the slope into the slope-intercept form of a line to find the equation for our line containing the point \\((2,-5)\\). Substitute the values into the slope-intercept form, then solve for \\(b\\).<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(-5=2(2)+b\\)<br \/>\n\\(-5=4+b\\)<br \/>\n\\(-5-4=4+b-4\\)<br \/>\n\\(-9=b\\)<\/p>\n<p>Plug \\(b=-9\\) and \\(m=2\\) into the slope-intercept form of a line, the equation of the line containing the point \\((2,-5)\\) is:<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=2x+(-9)\\)<br \/>\n\\(y=2x-9\\)<\/p>\n<p>Alternatively, we can use the point-slope formula to write the equation of the line, we will get the same result. The point-slope form of a line is:<br \/>\n<\/p>\n<p style=\"text-align: center;\">\\(y-y_1=m(x-x_1)\\)<\/p>\n<p>where \\(m\\) is the slope of the line and \\((x_1,y_1)\\) is a point on the line. Substitute the values into point-slope form and simplify.<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y-(-5)=2(x-2)\\)<br \/>\n\\(y+5=2x-2(2)\\)<br \/>\n\\(y+5=2x-4\\)<br \/>\n\\(y+5-5=2x-4-5\\)<br \/>\n\\(y=2x-9\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nThe length of one side of a rectangle lies on the line \\(y=3x-2\\). One of the vertices of the rectangle lies on this line at the point \\((6,16)\\). What is the equation of the line that the width of the rectangle lies on that also contains the point \\((6,16)\\)?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-4-1\">\\(y=-\\frac{1}{3}x+18\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(y=-\\frac{1}{3}x+\\frac{34}{3}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(y=-3x+24\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(y=-3x+54\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Since the length and width of a rectangle are perpendicular to each other, the slopes of the lines they lie on are negative reciprocals of each other. The given vertex of the rectangles is a point where two sides of the rectangle meet. We want to find the equation of the line for the width of the rectangle containing the point \\((6,16)\\). So, its slope is the negative reciprocal of the slope of the line \\(y=3x-2\\). The slope of \\(y=3x-2\\) is \\(m=3\\), so the slope of the perpendicular line is \\(m=-\\frac{1}{3}\\). <\/p>\n<p>We can use the slope-intercept form \\(y=mx+b\\), where \\(m\\) is the slope and \\(b\\) is the \\(y\\)-intercept to write the equation of the line that contains the width.<br \/>\nSubstitute the slope in the equation.<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(y=mx+b\\)<br \/>\n\\(y=-\\frac{1}{3}x+b\\)<\/p>\n<p>Now, substitute the point that lies on the width, then solve for \\(b\\).<br \/>\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(16=-\\frac{1}{3}(6)+b\\)<br \/>\n\\(16=-2+b\\)<br \/>\n\\(16+2=-2+b+2\\)<br \/>\n\\(18=b\\)<\/p>\n<p>Plug \\(b=18\\) and \\(m=-\\frac{1}{3}\\) into the slope-intercept form of a line, the equation of the line containing the width is:<br \/>\n<\/p>\n<p style=\"text-align: center;\">\\(y=-\\frac{1}{3}x\\ +18\\)<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":22,"featured_media":117493,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-117358","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_type-video","7":"content_type-practice-questions","8":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/117358","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=117358"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/117358\/revisions"}],"predecessor-version":[{"id":281147,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/117358\/revisions\/281147"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/117493"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=117358"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}