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 Post subject: GRE Coordinate Geometry (Select One)Posted: Sun Oct 23, 2011 1:22 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
The vertices of a triangle are located at the points (-2, 1), (3, 1), (x, y) on the coordinate plane. What is the area of the triangle if |y – 1| = 1?
A. 1
B. 2.5
C. 5
D. 25
E. It cannot be determined.

The correct answer is (B). Imagine or sketch the two definite points first.

From |y – 1| = 1 we can determine that y is either 2 or 0. So the third vertex lies on the line y = 2 or on the line y = 0. In either case the height of the triangle is 1. The base of the triangle is 5 (the distance between the two given definite points), so we can calculate the area of the triangle. The area is (1/2) × height × base . We know that the height is 1 and the base is 5.
A = (1/2) × 1 × 5
A = 2.5

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We need to know that x and y can only be integers.

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 Post subject: Re: t.4, qt.30: x,y - geometry, data sufficiency, geomteryPosted: Sun Oct 23, 2011 1:53 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
We need to know that x and y can only be integers.
No, x can possesss any value.

For any x and y = 0 or y = 2 the length of a height to the base (between points (-2, 1) and (3, 1)) is the same.

Let's consider the case y = 2 (y = 0 has the same reasoning).
One side of the triangle (between points (-2, 1) and (3, 1)) is set. The point (x, y) lies somewhere on the line y = 2:

Any such triangle has the same base and the length of the height to the base:

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