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Here are a set of practice GMAT questions about the Cartesian plane.<\/p>\nHere are a set of practice GMAT questions about the Cartesian plane.<\/p>\n
1) What is the equation of the line that goes through (\u20132, 3) and (5, \u20134)?<\/p>\n
<\/p>\n
Slope is a measure of how steep a line is.\u00a0 There is very algebraic formula for the slope, and if you know that, that's great!\u00a0 If you don't know that formula, or used to know it and can't remember it, I will say:\u00a0fuhgeddaboudit<\/em><\/a>!\u00a0\u00a0\u00a0 Here's a much better way of thinking about slope.\u00a0 Slope is\u00a0rise over run<\/strong>.<\/p>\n To calculate rise and run, first have to put the two points in order.\u00a0 It actually doesn't matter which one we say is the first and which one, the second: all that matters is that we are consistent.<\/p>\n The\u00a0rise<\/strong>\u00a0is the vertical change --- the change in y-coordinate (second point minus first).\u00a0 The\u00a0run<\/strong>\u00a0is the horizontal change --- the change in the x-coordinate (again, second minus first).\u00a0 Once we have rise & run, divide them, rise divided by run, to find the slope.<\/p>\n For example, suppose our points are (\u20132, 4) and (5, 1).\u00a0 For the sake of argument, we'll say that's the order -- (\u20132, 4) is the \"first\" and (5, 1) is the \"second.\"\u00a0 The rise is the change in height, the change in y-coordinate: 1 \u2013 4 = \u20133 (notice, we had to do second minus first, which gave us a negative here!)\u00a0\u00a0 The run is the horizontal change, the change in x-coordinate: 5 \u2013 (\u20132) = 5 + 2 = 7 (remember: subtracting a negative is the same as adding a positive!).\u00a0 Now, rise\/run = \u20133\/7 ---- that's the slope.\u00a0 Slope is definitely something you need to understand for the GMAT Quantitative section.<\/p>\n Whenever you find a slope, I strongly suggest doing a rough sketch, just to verify that the sign of the slope (positive or negative) and the value of the slope are approximately correct.\u00a0 Here's a sketch of this particular calculation:<\/p>\n Your sketch, of course, does not need to be this precise.\u00a0 Even a rough sketch would verify that, yes, the slope should be negative.\u00a0\u00a0 Again, I highly recommend performing this visual check every time you calculate slope.<\/p>\n <\/p>\n Let's get a bit philosophical for a moment.\u00a0 The technical name of the x-y plane is the Cartesian plane, named after its inventor, Mr.\u00a0Rene Descartes<\/a>.\u00a0 Although you may have met this sometime in middle school math and may now take it for granted, it is actually a brilliant mathematical device.\u00a0 It allowed for the unification of two ancient branches of mathematics: algebra and geometry.\u00a0 In more practical terms, every equation (an algebraic object) corresponds to a picture (a geometric object).\u00a0 That is a very deep idea.<\/p>\n <\/p>\n A straight line is a very simple picture, and not surprisingly it has a very simple equation.\u00a0 There are a few different ways to write a line, but the most popular and easiest to understand is\u00a0y = mx + b<\/strong>.\u00a0 The\u00a0m<\/strong>\u00a0is the slope of the line.\u00a0 The\u00a0b<\/strong>\u00a0is the y-intercept: where the line crosses the y-axis.\u00a0 For any given line, m & b are constants: for a given line, both m & b equal a fixed number.\u00a0 By contrast,\u00a0x<\/strong>\u00a0&\u00a0y<\/strong>\u00a0(sometimes call the \"graphing variables\") do not equal just one thing.\u00a0 This is not the \"x\" of ordinary solve-for-x algebra.\u00a0 This is a very deep idea --- x & y don't equal any one pair of values; rather,\u00a0every single point (x, y) on the line, the entire continuous infinity of points that make up that line --- every single one of them satisfies the equation of the line<\/strong>.\u00a0 That is a powerful and often underappreciated mathematical idea.<\/p>\n <\/p>\n Sometimes the GMAT will give you the equation of a line already in y = mx + b form.\u00a0 Sometimes, the GMAT give you the line in another form (e.g. 3x + 7y = 22), and you will have to do a little algebraic re-arranging ---- essentially, solve for y ---- to bring the equation into y = mx + b form.\u00a0 Sometimes, though, as in problem #2 above, they give you two points and ask you to find the equation of a line.\u00a0 Here the procedure.\u00a0 First, find the slope (as demonstrated above).\u00a0\u00a0 Now, plug the slope in for \"m\" in the y = mx + b equation, and pick either point (it doesn't matter which one) and plug those coordinates in for x & y in this equation.\u00a0 This will produce an equation in which everything has a numerical value except for \"b\" --- that means, you can solve this equation for the value of b.\u00a0\u00a0 Once you know m & b, you know the equation of the line.<\/p>\n <\/p>\n After this introduction, go back and try those practice problems again before reading the solutions below.\u00a0 Here's another relevant practice question:<\/p>\n 4)\u00a0https:\/\/gmat.magoosh.com\/questions\/821<\/a><\/p>\n In the next post, I will discuss midpoints and the issue of parallel & perpendicular lines.<\/p>\n <\/p>\n 1) Here, we will follow the procedure we demonstrated in the last section.\u00a0 Call (\u20132, 3) the \"first\" point and (5, \u20134), the \"second.\"\u00a0 Rise = \u20134 \u2013 3 = \u20137.\u00a0 Run = 5 \u2013 (\u20132) = 7.\u00a0 Slope = rise\/run = \u20137\/7 = \u20131.\u00a0 Visual check:<\/p>\n Yes, it makes sense that the slope is negative.\u00a0 We have the slope, so plug m = \u20131 and (x, y) = (\u20132, 3) into y = mx + b:<\/p>\n 3 = (\u20131)*(\u20132) + b<\/p>\n 3 = 2 + b<\/p>\n 1 = b<\/p>\n So, plugging in m = \u20131 and b = 1, we get an equation y = \u2013x + 1.\u00a0 Answer =\u00a0A<\/strong><\/p>\n 2) Here, we already have the slope, so we just need to follow the second half of the \"finding the equation\" procedure.\u00a0 Plug (x, y) = (7, \u20131) into this equation:<\/p>\n [pmath]{-1} \u00a0= \u00a0{{5(7)}\/3} + b \u00a0= \u00a0{35\/3} + b[\/pmath]<\/p>\n [pmath]b \u00a0= \u00a0{-35\/3} - 1\u00a0= \u00a0{-35\/3} - {3\/3} \u00a0= \u00a0-{38\/3}[\/pmath]<\/p>\n Answer =\u00a0E<\/strong><\/p>\n 3) This is a considerably more difficult one, which will involve some algebra.\u00a0 Let's say that the \"first\" point is (\u20131, \u20134) and the \"second,\" (3, k).\u00a0 The rise = k + 4, which involves a variable.\u00a0 The run = 3 \u2013 (\u20131) = 4.\u00a0 There the slope is (k + 4)\/4, and we can set this equal to k and solve for k.<\/p>\n [pmath]{{k + 4}\/4} = k [\/pmath]<\/p>\n k + 4 = 4k<\/p>\n 4 = 3k<\/p>\n k = 4\/3<\/p>\n Answer =\u00a0C<\/strong><\/p>\n This post was written by Mike McGarry, GMAT expert at Magoosh<\/a>, and originally posted here<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":" Here are a set of practice GMAT questions about the Cartesian plane. 1) What is the equation of the line that goes through (\u20132, 3) and (5, \u20134)? y =…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,736],"tags":[],"class_list":["post-15639","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-quant-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/15639","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=15639"}],"version-history":[{"count":2,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/15639\/revisions"}],"predecessor-version":[{"id":15642,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/15639\/revisions\/15642"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=15639"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=15639"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=15639"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/magoosh.com\/gmat\/files\/2012\/11\/las_img1.png\")
<\/a><\/p>\nEquations in the x-y plane<\/h2>\n
Equations of lines<\/h2>\n
Finding the equation of a line<\/h2>\n
Practice<\/h2>\n
Explanation of practice questions<\/h2>\n
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