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 Post subject: GMAT Coordinate Geometry
PostPosted: Fri Apr 30, 2010 1:30 pm 
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Joined: Tue Apr 13, 2010 8:48 am
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If the graphs of the functions y = x and y = x² + 3x + 1 intersect at the point (w, u), what is the equation of the line that contains the point (w, u) and that is perpendicular to the line y = (1/2)x – 2?

A. y = -2x - 3
B. y = -2x + 3
C. y = 2x + 3
D. y = 2x - 3
E. y = 3x - 2

(A) The first step is to determine the point (w, u). To accomplish this we set the two equations equal to each other in order to eliminate the y parameter (if they are both equal to y, they are both equal to each other):

x² + 3x + 1 = x.

Now we move all the terms to the left hand side: x² + 2x + 1 = 0.
Notice that the left hand side of this equation can be factored to (x + 1)² and equation rewritten as:

(x + 1)² = 0.

It should be clear that the only value of x that will satisfy the equation is x = -1. To determine the y coordinate we simply substitute x = -1 into either of the two given equations to get a value for y (you can convince yourself that it doesn't matter which one we plug into by testing x = -1 in both). The easiest route is to substitute x = -1 into the first equation (y = x) to get y = -1. The point (w, u) therefore, is the point (-1, -1).

To determine the answer there is one more set of operations we must compute. Recall that the product of the slopes of perpendicular lines is equal to -1. That means that the equation of the line we are asked to find must have a slope of -2 (since line y = (1/2)x – 2 has a slope of 1/2. Using the well known point-slope equation for lines we set up the ratio (y+1)/(x+1) = -2/1 and cross multiply before solving for y: y+1 = -2x -2, which implies: y = -2x - 3, or (A), the answer to our question.


---------------------------------

could use a bunch of graphics for the explanation built into the page.
Thanks.


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 Post subject: Re: Math: x,y - plane geometry question.
PostPosted: Fri Apr 30, 2010 2:08 pm 
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Joined: Fri Apr 09, 2010 2:11 pm
Posts: 459
This question should be solved analytically. You may lose quite some time graphing those equations.

However, if it is not clear what to do, graphing may help. So here it is:

Image

Here we can see given graphs in black and the required line in red.
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