# Thread: Tough Coordinate Geometry problem! HELP!

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## Tough Coordinate Geometry problem! HELP!

Ive been doing this for thirty minutes, and Im sure its easy but even the explanation Manhattan GMAT gives is HARD! SOmeone please explain in an easier way? I really appreciate it!!!!

The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?

A) 169/3

B) 84.5

C) 75
D) 169
E) 225/4

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SORRY IF IT LOOKS CONFUSING.....THE ANSWER CHOICES AR ROOT 3 NOT THE POWER OF ROOT 3...duh

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Ans . A

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ok, obviously the first step is to find the length of PQ. There are many ways to do this but the easiest is to plot a simple graph. Mark the two points and join the points. Make a right angle triangle with PQ as the hypoteneuse. You'll realise that one side of the triangle is of length 5 and another of length 12. So, the hypoteneuse is 13. THerefore, PQ = 13.

So, the height of an equilateral triangle is 13. We need to find the area. The height of an equilateral triangle is ALWAYS [sqrt(3)/2]*a where a is the side. Therefore, a=26/sqrt(3). Now, we know the base and height. Therefore area = 169/sqrt(3), which can be re-written as 169*sqrt(3)/3.

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That's pretty easy if you know:
The height of an equilateral triangle is ALWAYS [sqrt(3)/2]*a

Thanks for the explanation

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PQ = 13.
we know height f the eq triangle = (sqrt3/2)a where a is the side of the triangle
solve for a = 26/sqrt3
we also know area = sqrt3*a^2/4

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