Tough Coordinate Geometry problem! HELP!

[Cam54]

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

Answer: A

[Cam54]

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

[krishna_k1]

Ans . A

[raja501]

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.

[umgrad76]

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

Thanks for the explanation

[nikiforos]

My answer is A

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