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by sanju09 » Wed May 02, 2012 3:12 am
The base of an isosceles triangle has endpoints at coordinates (1, 1) and (13, 1). The area of the triangle is 48. The third point of the triangle could be located at:
(A) (6, -6)
(B) (6, -11)
(C) (7, 9)
(D) (7, -11)
(E) (13, 13)



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by Anurag@Gurome » Wed May 02, 2012 4:31 am
sanju09 wrote:The base of an isosceles triangle has endpoints at coordinates (1, 1) and (13, 1). The area of the triangle is 48. The third point of the triangle could be located at:
(A) (6, -6)
(B) (6, -11)
(C) (7, 9)
(D) (7, -11)
(E) (13, 13)

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Length of the base = √[(13 - 1)² + (1 - 1)²] = √(12²) = 12
Area of the triangle = (1/2) * base * height
48 = (1/2) * 12 * height
48 = 6 * height
height = 8
Now the altitude to the base of an isosceles triangle bisects the base.
Midpoint of (1, 1) and (13, 1) = (7, 1)
Let the coordinates of the 3rd point = (a, b)
Then √[(a - 7)² + (b - 1)²] = 8
Only answer choice (7, 9) satisfies the above equation.

The correct answer is C.
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