GmatPrep2?/ (Slope)

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by mals24 » Wed Sep 03, 2008 4:41 pm
The graph of this line will look like this
Image

From the question we know that the area of the triangle formed is 12

Area of the triangle = 1/2 (base*height)

The base is given as 4

equ for area: 12 = 1/2(4*H)
Solving the equ you will get height as 6 since the line intersects the y axis in the negative side the y intercept will be -6

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Re: GmatPrep2?/ (Slope)

by parallel_chase » Wed Sep 03, 2008 5:08 pm
dferm wrote:Please Help...
Always remember the below rule, though it marginally helps you in this question but it will definitely help you on your GMAT.

for ax+by+c = 0

x intercept = -c/a

y intercept = -c/b

slope = -a/b

x intercept is 4, slope is positive

slope =-a/b

since x intercept is positive the y intercept has to be negative. Eliminate all the positive options

if y intercept is negative and slope is positive

the line is passing through negative y axis or the fourth quadrant.

area = 1/2 * x intercept * y intercept

12 = 1/2 * 4 * y

y intercept = -6, since it has to be negative for slope to be positive.


Hope this helps.