Question

A pair of perpendicular straight lines drawn through the origin form an isosceles triangle with line $2x+3y=6$ then area of the triangle so formed is

• A. $\frac{36}{13}$
• B. $\frac{12}{17}$
• C. $\frac{13}{5}$
• D. $\frac{17}{13}$

Answer 1

The correct option is A $\frac{36}{13}$

$ax+by+c=0d=\left|\frac{{ax}_1+{by}_1+c}{\sqrt{a^2+b^2}}\right|2x+3y=6OD=\frac{2\left(0\right)+3\left(0\right)-6}{\sqrt{2^2+3^2}}OD=\frac{6}{\sqrt{13}}$
In$\Delta OAB\sin 45-\frac{OD}{OA}OA=OD\csc 45OA=\frac{6}{\sqrt{13}}\times \sqrt{2}OA=OB=\frac{6\sqrt{2}}{\sqrt{13}}$
Area$=\frac{1}{2}\times OA\times OB=\frac{1}{2}\times \frac{6\sqrt{2}}{\sqrt{13}}\times \frac{6\sqrt{2}}{\sqrt{13}}=\frac{36}{13}$