Question

A, B, C are the three vertices of a triangle such that they are equidistant from the origin. A and B lie on the positive X and Y axes respectively and C lies on the line y + x = a. If AB = $\sqrt{2}$a, find the area of the triangle. (C lies in the first quadrant)

• A.
• B.
• C.
• D. 0

Answer 1

The correct option is D

0

A and B are equidistant from origin and AB is $\sqrt{2}$a.
If we assume the points A and B to be (x,0) and (0,x) respectively,
AB= $\sqrt{x^2+x^2}$ = $\sqrt{2}$x = $\sqrt{2}$a
x=a
So A(0,a) and B(a,0) are the vertices.
Observe that A,B both lie on the line x+y=a and since C also lies on the line y+x = a, they are collinear.
Hence area of the triangle formed by the three points is 0.