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Area Enclosed

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by raajan_p » Wed Nov 26, 2008 10:49 am
If c > 0 and d > 0, then which of the following expressions is the area enclosed by the line with the equation x = 0, the line with the equation y = 0, and the line with the equation y = cx + d ?

A) d/2c
B) d^2/2c
C) d^2/c
D) c^2/2d
E) cd/2
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by jimmiejaz » Wed Nov 26, 2008 11:50 am
Is the ans B?
Am a bit sceptical about the answer though.
contained the image.

Image

as given in the question, c>0 and d>0
the area enclosed will be
since d is the y intercept. and x intercept is -d/c
Area = 1/2 * d*(-d/c)
= -d^2/2c
Hence IMO ans is B.
but i would love to see the explanation by an expert though.
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Re: Area Enclosed

by Stuart@KaplanGMAT » Wed Nov 26, 2008 12:15 pm
raajan_p wrote:If c > 0 and d > 0, then which of the following expressions is the area enclosed by the line with the equation x = 0, the line with the equation y = 0, and the line with the equation y = cx + d ?

A) d/2c
B) d^2/2c
C) d^2/c
D) c^2/2d
E) cd/2
If we're not sure about the algebra, we can attack this question by picking numbers instead.

Let's let our equation be y = 2x + 6
so, we've chosen:
c = 2
d = 6

If we draw out our triangle, we see that the 3 vertices are at (0,0), (0,6) and (-3,0), giving us a base of 3 and a height of 6.

area = 1/2(b)(h) = 1/2(3)(6) = 9

Now we plug our values for c and d into the choices:

(a) 6/4... wrong
(b) 36/4 = 9... right
(c) 36/2... wrong
(d) 4/12... wrong
(e) 12/2... wrong

Only (b) worked out, therefore (b) is correct!

If more than 1 choice worked (for example, if we had picked c=2 and d=4, both (b) and (e) work), then we'd have to try another set of numbers, only checking the choices we haven't been able to eliminate. As a general rule, the stranger the numbers you choose, the less likely it is that you'll get duplicates.

If you like algebra, jimmiejaz's solution is great.
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