Question

A (3, 1) B (6, 5) and C (x, y) are three points such that angle ACB is a right angle and the area of triangle ABC = 7 sq. unit then the number of such point C is=

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

AB = $\sqrt{(6–3)^{2} + (5–1)^{2}}$Ž AB = 5

AB = diameter Ž 2r = 5 Ž r = 5/2 = 2.5

Area of triangle ACB

$\frac{1}{2}$×5 × h = 7 Ž h = $\frac{14}{5}$ = 2.8 Ž r = 2.5 ; r < h

which is not possible.

Hence no such point C possible.