If a pair of perpendicular lines through the origin together with the straight line $2 x+3 y=6$ form an…

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
  1. $\frac{6}{\sqrt{13}}$
  2. $\frac{6}{13}$
  3. $\frac{36}{13}$
  4. $\frac{27}{13}$

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 $\triangle A O B=\frac{1}{2} \times A B \times O P=O P^2=\frac{36}{13}$

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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