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
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GMAT Prep Geometry question
Source: Beat The GMAT — Problem Solving |
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jayhawk2001
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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.
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.
