Question

A right-angled isosceles triangle right-angled at origin has

2x + 3y = 6 as its base. Area of the triangle, is-

• A. $\sqrt { 13 }$
• B. 36/13
• C. 6
• D. 6 or 13

Answer 1

Answer: (B)

36/13

Answer 2

Let OAB be the right-angled isosceles triangle.

Let ŠAOX = q, Then

ŠBOY = q (see fig.).

Let OA = OB = r. Then, we have

A ŗ (r cos q, r sin q) and B ŗ (–r sin q, r cos q)

Since A and B both lie on the given line 2x + 3y = 6, therefore we have

2 cos q + 3 sin q = $\frac { 6 } { r }$ … (1)

and –2 sin q + 3 cos q = … (2)

squaring and adding equations (1) and (2), we have

2² + 3² =

given r² = $\frac { 36 } { 13 } \cdot 2$

Hence, area of D OAB, is = 36/13.