If a pair of perpendicular lines through the origin together with the straight line $2 x+3 y=6$ form an isosceles triangle, then the area of that triangle (in sq units) is

• A

  $\frac{6}{\sqrt{13}}$
• B

  $\frac{6}{13}$
• Correct answer
  C

  $\frac{36}{13}$
• D

  $\frac{27}{13}$

Solution

Since, $\triangle A O B$ is a right angled triangle, so $$ \angle A=\angle B=\frac{\pi}{4} $$
$$ \begin{aligned} So, \quad & A B=2 A P=2 O P \\ Now, \quad & O P=\frac{|6|}{\sqrt{4+9}}=\frac{6}{\sqrt{13}} \end{aligned} $$ So, Area of $\triangle A O B=\frac{1}{2} \times A B \times O P=O P^2=\frac{36}{13}$

Hence, the correct answer is option: 'C'.