GMAT Prep Geometry question

[gviren]

Please help

Thanks

[jayhawk2001]

slope of line connecting origin and 1, -sqrt-3 = -1/sqrt-3

Slope of perpendicular line = sqrt-3

Eqn of perpendicular line is y = sqrt-3*x. (s,t) is a point on this. So,
t = sqrt-3*s

Distance between s,t and origin = s^2 + t^2 = 4*s^2

This should be the same distance between 1, -sqrt-3 i.e. 4

so, 4*s^2 = 4. s = 1

[gmatfan]

I do it a little different, not using the slope, but just 30, 60, 90 triangles.

With the point given, you can make a triangle on the negative side of the x axis that has sides root 3 and 1. That makes the radius 2, and you have a classic 30, 60, 90 triangle.

Since the leg of the triangle opposite the 60 degree angle is the root 3 leg, I know that there are 30 degrees remaining of the 90 degree angle shown for the positive side of the x axis.

The side opposite the 30 degree angle is the smallest leg, so 1 is the distance for s.