{"id":3125,"date":"2012-11-29T09:00:33","date_gmt":"2012-11-29T17:00:33","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=3125"},"modified":"2024-06-03T16:48:02","modified_gmt":"2024-06-03T23:48:02","slug":"gmat-math-midpoints-and-parallel-vs-perpendicular-lines","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/","title":{"rendered":"GMAT Math: Midpoints and Parallel vs. Perpendicular lines"},"content":{"rendered":"<p>First, a few practice questions.<\/p>\n<p>1) Line A has the equation 3x + y = 7.\u00a0 Which of the following lines is perpendicular to Line A?<\/p>\n<ul>\n\t(A) y = 3x + 4<br \/>\n\t(B) y = \u20133x \u2013 6<br \/>\n\t(C) y = (1\/3)x \u2013 1<br \/>\n\t(D) y = (\u20131\/3)x + 2<br \/>\n\t(E) y = (\u20137\/3)x \u2013 5\n<\/ul>\n<p>2) Line P has a positive slope and a positive y-intercept.\u00a0 Line Q also has a positive slope and a positive y-intercept.\u00a0 The two slopes and the two y-intercepts are four different numbers, none equal.\u00a0 Lines P &amp; Q have a single intersection point &#8212; what is the full set of possible quadrants in which the intersection point could be?<\/p>\n<ul>\n\t(A) I only<br \/>\n\t(B) I &amp; III only<br \/>\n\t(C) I &amp; II only<br \/>\n\t(D) I &amp; II &amp; III only<br \/>\n\t(E) all four quadrants possible\n<\/ul>\n<p>3) The <strong>median<\/strong> of a triangle is the line from any vertex to the midpoint of the opposite side.\u00a0 Triangle ABC has vertices A = (0, 5), B = (\u20131, \u20131), and C = (5, 2).\u00a0 What is the slope of the median from A to the midpoint of BC?<\/p>\n<ul>\n\t(A) \u20133\/4<br \/>\n\t(B) >\u20134\/3<br \/>\n\t(C) \u20135<br \/>\n\t(D) \u20135\/2<br \/>\n\t(E) \u20139\/4\n<\/ul>\n<p>&nbsp;<\/p>\n<h2>Slope of horizontal and vertical lines.<\/h2>\n<p>In the <a href=\"https:\/\/magoosh.com\/gmat\/gmat-math-lines-slope-in-the-x-y-plane\/\">previous post<\/a>, I discuss the idea of slope, but I didn&#8217;t talk about the special cases of horizontal and vertical lines.<\/p>\n<p>A <strong>horizontal line<\/strong> is all run and no rise: therefore, its slope, its rise\/run, is <strong>zero<\/strong>.\u00a0 By contrast, a <strong>vertical line<\/strong> is all rise and no run &#8212; when we calculate its rise over run, we get a mathematical error, because we can&#8217;t divide by zero.\u00a0 This means, vertical lines have <strong>undefined<\/strong> slope; very roughly, we can think of vertical lines as having &#8220;infinite&#8221; slope, although technically, that way of interpreting vertical lines does not bear rigorous analysis of <a href=\"http:\/\/en.wikipedia.org\/wiki\/Transfinite_number\">the idea of infinity<\/a> &#8212;- which, of course, is well beyond anything on the GMAT.<\/p>\n<p>Horizontal lines have a slope equal to zero.\u00a0 Vertical lines have an undefined slope: the value of their slope does not make any mathematical sense.\u00a0 I would strongly recommend against saying that either has &#8220;no slope&#8221;, because that is a term ripe for confusion: does &#8220;no slope&#8221; mean a slope with a value of zero or does it mean not being able to compute the value of the slope?\u00a0 I recommend that you absolutely banish the term &#8220;no slope&#8221; from your vocabulary.<\/p>\n<p>&nbsp;<\/p>\n<h2>Parallel and perpendicular lines<\/h2>\n<p><strong>Parallel lines<\/strong> in the x-y plane have <strong>equal slope<\/strong>.\u00a0 If the think about what the slope means geometrically, you will see that this has to be true.\u00a0 The GMAT expects you to know this.<\/p>\n<p>The case with perpendicular lines is a little trickier.\u00a0 Let&#8217;s think about this.\u00a0 Consider any slanted line with a positive slope (i.e. moving up to the right) &#8212; then the line perpendicular to it will have to have a negative slope (i.e. moving down to the right, moving up to the left).\u00a0 Similarly, if the original line has a negative slope, the perpendicular would have to have a positive slope.\u00a0 Thus, the slope of a perpendicular line is always has a sign opposite of the sign of the original line.\u00a0 That&#8217;s part one of the idea.<\/p>\n<p>Now, let&#8217;s think about if we take a &#8220;rise over run triangle&#8221; and rotate it 90\u00b0.\u00a0\u00a0 Let&#8217;s ignore \u00b1 signs for a moment, and just think about absolute values.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3126\" alt=\"\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1.png\" width=\"612\" height=\"527\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1.png 612w, https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1-300x258.png 300w\" sizes=\"(max-width: 612px) 100vw, 612px\" \/><\/a><\/p>\n<p>So, we have two perpendicular lines, the original line AB and the perpendicular line AE, intersecting at point A.\u00a0 For each, we have constructed &#8220;rise over run&#8221; triangle for the slope.\u00a0 The slope of line AB = (original rise)\/(original run) = BC\/AC, and the slope of line AE = (new rise)\/(new run) = AD\/DE.\u00a0 But notice that these two triangles are congruent!\u00a0 Triangle ADE is just Triangle ACB rotated 90\u00b0 around point A.\u00a0 Because the triangle are congruent, the corresponding lengths are equal: AC = AD and BC = DE.\u00a0 In other words, the rise of one is the run of the other!\u00a0 When we go from the original line to a line perpendicular to it, the rise and the run switch places in the slope fraction.\u00a0 When the numerator and denominator of a fraction switch places, that&#8217;s called a <strong>reciprocal<\/strong>.\u00a0 For example, 2\/5 and 5\/2 are reciprocals; 3\/11 and 11\/3 are reciprocals; 7 and 1\/7 are reciprocals.<\/p>\n<p>Now, put these two facts together &#8212; when line #1 has a slope, and line #2 is perpendicular to it, the slope of line #2 must have both the opposite sign from and must be a reciprocal of the original slop of line #1.\u00a0 More elegantly, <strong>perpendicular slopes are opposite reciprocals<\/strong>.\u00a0 The GMAT expects you to know this.<\/p>\n<p>If one line has a slope = +4\/7, then the perpendicular line has a slope = \u20137\/4.\u00a0 If one line has a slope of \u20133, the perpendicular line has a slope = +1\/3.\u00a0 If you know the slope of one line, you flip that fraction over and change its sign to the opposite to make the slope of the perpendicular line.\u00a0 If this is a new idea to you, or one which learned but forget a long time ago, then I strongly recommend you actually get some graph paper and actually graph sets of perpendicular lines and find their slopes, so you have a visual understanding of this idea.<\/p>\n<p>&nbsp;<\/p>\n<h2>Midpoints<\/h2>\n<p>Occasionally, the GMAT will ask to you to find the coordinates of the midpoint of a segment.\u00a0\u00a0 You will be given the two endpoints of the segment &#8212; for example, (4, 1) and (10, 15).\u00a0 You find the coordinates of the midpoint by averaging the values they give you.\u00a0 The x-coordinate of the midpoint is the average of the x-coordinates of the end points; here, this average is (4 + 10)\/2 = 7.\u00a0\u00a0 Similarly, the y-coordinate of the midpoint is the average of the y-coordinates of the end points; here, this average is (1 + 15)\/2 = 8.\u00a0 Therefore, the coordinates of the midpoint are (7, 8).\u00a0 If you can take an average &#8212;- twice &#8212;- then you can find the coordinates of a midpoint.<\/p>\n<p>&nbsp;<\/p>\n<h2>Final Thoughts<\/h2>\n<p>Any question about any points or lines in the x-y plane is a visual question.\u00a0\u00a0 If you are not given a diagram, always sketch a diagram.\u00a0 You open up a whole new level of understanding when you add a <a href=\"https:\/\/magoosh.com\/gmat\/a-geometric-and-visual-approach-to-gmat-math\/\">visual approach<\/a>.<\/p>\n<p>If the two practice questions at the top gave you some difficulty when you first looked at them, now that you have read this article, take another look at them before you jump into the explanations below.<\/p>\n<p>&nbsp;<\/p>\n<h2>Practice problem explanations<\/h2>\n<p>1)\u00a0 What&#8217;s tricky about this problem: we have to begin by solving the given equation for y, so that we know its slope.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img2.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3127\" alt=\"\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img2.png\" width=\"288\" height=\"27\" \/><\/a><\/p>\n<p>The slope of the original line is m = \u20133, and the <strong><em>negative reciprocal<\/em><\/strong> of that is +1\/3, so the perpendicular line must have a slope of +1\/3.\u00a0 Among the answer choices, the only line with a slope of +1\/3 is <strong>(C)<\/strong>.\u00a0 Just so you have a visual for this, here&#8217;s a diagram of these line (whenever possible, always verify questions about the x-y plane visually, with at least a quick sketch!)<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/perpendicular-diagram.jpg\"><img decoding=\"async\" alt=\"perpendicular diagram\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/perpendicular-diagram.jpg\" width=\"385\" height=\"426\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p>2) If you forget quadrants, take a look at <a href=\"https:\/\/magoosh.com\/gmat\/quadrants-on-the-gmat-the-cartesian-plane\/\">this post<\/a>.\u00a0\u00a0 This is a tricky one, because you have to be careful to think about all cases.\u00a0 The first important point to realize is: any line with both a positive slope and a positive y-intercept goes through quadrants I &amp; II &amp; III, but never quadrant IV.\u00a0 Neither of these lines is ever in Quadrant IV, so they can&#8217;t intersect there.\u00a0 As it turns out, for different values of the slope, they can intersect in any of the other three quadrants.\u00a0 Here are visual examples.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img4.png\"><img decoding=\"async\" class=\"alignnone  wp-image-3129\" alt=\"\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img4.png\" width=\"640\" height=\"188\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img4.png 800w, https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img4-300x88.png 300w\" sizes=\"(max-width: 640px) 100vw, 640px\" \/><\/a><\/p>\n<p>Therefore, the answer = <strong>(D)<\/strong><\/p>\n<p>3) For your visual understanding, here&#8217;s a diagram of the situation.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img5.png\"><img decoding=\"async\" class=\"alignnone  wp-image-3130\" alt=\"\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img5.png\" width=\"515\" height=\"500\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img5.png 2938w, https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img5-300x291.png 300w, https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img5-1024x995.png 1024w\" sizes=\"(max-width: 515px) 100vw, 515px\" \/><\/a><\/p>\n<p>First to find the midpoint of B &amp; C &#8212; average the x-coordinates: (\u20131+ 5)\/2 = 2; and average the y-coordinates: (\u20131 + 2)\/2 = 1\/2.\u00a0 Thus, the midpoint has coordinates (2, 1\/2).\u00a0 We want the slope from A = (0, 5) to (2, 1\/2).\u00a0\u00a0 The rise is the change in the y-coordinates: 1\/2 \u2013 5 = -9\/2.\u00a0 The run is the change in the x-coordinates: 2 \u2013 0 = 2.\u00a0 Slope = rise\/run = [\u20139\/2]\/2 = \u20139\/4.\u00a0\u00a0 Answer = <strong>(E)<\/strong><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>First, a few practice questions. 1) Line A has the equation 3x + y = 7.\u00a0 Which of the following lines is perpendicular to Line A? (A) y = 3x + 4 (B) y = \u20133x \u2013 6 (C) y = (1\/3)x \u2013 1 (D) y = (\u20131\/3)x + 2 (E) y = (\u20137\/3)x \u2013 [&hellip;]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[124],"tags":[],"ppma_author":[13209],"class_list":["post-3125","post","type-post","status-publish","format-standard","hentry","category-geometry"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>GMAT Math: Midpoints and Parallel vs. Perpendicular Lines | Magoosh Study Resources<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"GMAT Math: Midpoints and Parallel vs. Perpendicular lines\" \/>\n<meta property=\"og:description\" content=\"First, a few practice questions. 1) Line A has the equation 3x + y = 7.\u00a0 Which of the following lines is perpendicular to Line A? (A) y = 3x + 4 (B) y = \u20133x \u2013 6 (C) y = (1\/3)x \u2013 1 (D) y = (\u20131\/3)x + 2 (E) y = (\u20137\/3)x \u2013 [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\" \/>\n<meta property=\"og:site_name\" content=\"Magoosh Blog \u2014 GMAT\u00ae Exam\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/MagooshGMAT\/\" \/>\n<meta property=\"article:published_time\" content=\"2012-11-29T17:00:33+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-06-03T23:48:02+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1.png\" \/>\n<meta name=\"author\" content=\"Mike M\u1d9cGarry\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@MagooshGMAT\" \/>\n<meta name=\"twitter:site\" content=\"@MagooshGMAT\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Mike M\u1d9cGarry\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\"},\"author\":{\"name\":\"Mike M\u1d9cGarry\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b\"},\"headline\":\"GMAT Math: Midpoints and Parallel vs. Perpendicular lines\",\"datePublished\":\"2012-11-29T17:00:33+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\"},\"wordCount\":1307,\"commentCount\":16,\"publisher\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/#organization\"},\"articleSection\":[\"GMAT Geometry\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\",\"url\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\",\"name\":\"GMAT Math: Midpoints and Parallel vs. Perpendicular Lines | Magoosh Study Resources\",\"isPartOf\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/#website\"},\"datePublished\":\"2012-11-29T17:00:33+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/magoosh.com\/gmat\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"GMAT Math: Midpoints and Parallel vs. Perpendicular lines\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#website\",\"url\":\"https:\/\/magoosh.com\/gmat\/\",\"name\":\"Magoosh Blog \u2014 GMAT\u00ae Exam\",\"description\":\"Everything you need to know about the GMAT\",\"publisher\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/magoosh.com\/gmat\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#organization\",\"name\":\"Magoosh\",\"url\":\"https:\/\/magoosh.com\/gmat\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png\",\"contentUrl\":\"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png\",\"width\":265,\"height\":60,\"caption\":\"Magoosh\"},\"image\":{\"@id\":\"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/MagooshGMAT\/\",\"https:\/\/twitter.com\/MagooshGMAT\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b\",\"name\":\"Mike M\u1d9cGarry\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/image\/15a1e36ef1c2c3940179212433de141a\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g\",\"caption\":\"Mike M\u1d9cGarry\"},\"description\":\"Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's Youtube video explanations.\",\"sameAs\":[\"https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured\"],\"award\":[\"Magna cum laude from Harvard\"],\"knowsAbout\":[\"GMAT\"],\"knowsLanguage\":[\"English\"],\"jobTitle\":\"Content Creator\",\"worksFor\":\"Magoosh\",\"url\":\"https:\/\/magoosh.com\/gmat\/author\/mikemcgarry\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"GMAT Math: Midpoints and Parallel vs. Perpendicular Lines | Magoosh Study Resources","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/","og_locale":"en_US","og_type":"article","og_title":"GMAT Math: Midpoints and Parallel vs. Perpendicular lines","og_description":"First, a few practice questions. 1) Line A has the equation 3x + y = 7.\u00a0 Which of the following lines is perpendicular to Line A? (A) y = 3x + 4 (B) y = \u20133x \u2013 6 (C) y = (1\/3)x \u2013 1 (D) y = (\u20131\/3)x + 2 (E) y = (\u20137\/3)x \u2013 [&hellip;]","og_url":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/","og_site_name":"Magoosh Blog \u2014 GMAT\u00ae Exam","article_publisher":"https:\/\/www.facebook.com\/MagooshGMAT\/","article_published_time":"2012-11-29T17:00:33+00:00","article_modified_time":"2024-06-03T23:48:02+00:00","og_image":[{"url":"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las2_img1.png"}],"author":"Mike M\u1d9cGarry","twitter_card":"summary_large_image","twitter_creator":"@MagooshGMAT","twitter_site":"@MagooshGMAT","twitter_misc":{"Written by":"Mike M\u1d9cGarry","Est. reading time":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#article","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/"},"author":{"name":"Mike M\u1d9cGarry","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b"},"headline":"GMAT Math: Midpoints and Parallel vs. Perpendicular lines","datePublished":"2012-11-29T17:00:33+00:00","mainEntityOfPage":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/"},"wordCount":1307,"commentCount":16,"publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"articleSection":["GMAT Geometry"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/","url":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/","name":"GMAT Math: Midpoints and Parallel vs. Perpendicular Lines | Magoosh Study Resources","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/#website"},"datePublished":"2012-11-29T17:00:33+00:00","breadcrumb":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/magoosh.com\/gmat\/gmat-math-midpoints-and-parallel-vs-perpendicular-lines\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/magoosh.com\/gmat\/"},{"@type":"ListItem","position":2,"name":"GMAT Math: Midpoints and Parallel vs. Perpendicular lines"}]},{"@type":"WebSite","@id":"https:\/\/magoosh.com\/gmat\/#website","url":"https:\/\/magoosh.com\/gmat\/","name":"Magoosh Blog \u2014 GMAT\u00ae Exam","description":"Everything you need to know about the GMAT","publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/magoosh.com\/gmat\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/magoosh.com\/gmat\/#organization","name":"Magoosh","url":"https:\/\/magoosh.com\/gmat\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/","url":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","contentUrl":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","width":265,"height":60,"caption":"Magoosh"},"image":{"@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/MagooshGMAT\/","https:\/\/twitter.com\/MagooshGMAT"]},{"@type":"Person","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b","name":"Mike M\u1d9cGarry","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/image\/15a1e36ef1c2c3940179212433de141a","url":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","caption":"Mike M\u1d9cGarry"},"description":"Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's Youtube video explanations.","sameAs":["https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured"],"award":["Magna cum laude from Harvard"],"knowsAbout":["GMAT"],"knowsLanguage":["English"],"jobTitle":"Content Creator","worksFor":"Magoosh","url":"https:\/\/magoosh.com\/gmat\/author\/mikemcgarry\/"}]}},"authors":[{"term_id":13209,"user_id":26,"is_guest":0,"slug":"mikemcgarry","display_name":"Mike M\u1d9cGarry","avatar_url":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","user_url":"","last_name":"M\u1d9cGarry","first_name":"Mike","description":"Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as \"member of the month\" for over two years at <a href=\"https:\/\/gmatclub.com\/blog\/2012\/09\/mike-mcgarrys-gmat-experience\/\" rel=\"noopener noreferrer\">GMAT Club<\/a>. Mike holds an A.B. in Physics (graduating <em>magna cum laude<\/em>) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's <a href=\"https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured\" rel=\"noopener noreferrer\">Youtube <\/a>video explanations and resources like <a href=\"https:\/\/magoosh.com\/gmat\/whats-a-good-gmat-score\/\" rel=\"noopener noreferrer\">What is a Good GMAT Score?<\/a> and the <a href=\"https:\/\/magoosh.com\/gmat\/gmat-diagnostic-test\/\" rel=\"noopener noreferrer\">GMAT Diagnostic Test<\/a>."}],"_links":{"self":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/3125","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/users\/26"}],"replies":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/comments?post=3125"}],"version-history":[{"count":0,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/3125\/revisions"}],"wp:attachment":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/media?parent=3125"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/categories?post=3125"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/tags?post=3125"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/ppma_author?post=3125"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}