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

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$\frac{6}{\sqrt{13}}$

$\frac{36}{13}$

$\frac{6}{13}$

$\frac{27}{13}$

answer is C.

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Detailed Solution

$\begin{array}{l} O A = O B a n d ∠ A O B = 90^{0} \\ ∴ θ = 45^{0} \\ O D = ⊥ r d i s \tan c e f r o m (0, 0) t o 2 x + 3 y - 6 = 0 \\ = \frac{|- 6|}{\sqrt{2^{2} + 3^{2}}} = \frac{6}{\sqrt{4 + 9}} = \frac{6}{\sqrt{13}} \\ I n Δ^{l e} O A D \sin θ = \frac{O D}{O A} \Rightarrow \sin 45^{0} = \frac{O D}{O A} \\ \Rightarrow \frac{1}{\sqrt{2}} = \frac{O D}{O A} \Rightarrow O A = \sqrt{2} O D \Rightarrow \frac{6 \sqrt{2}}{\sqrt{13}} \end{array}$

$\begin{array}{l} A r e a o f Δ^{l e} O A B = \frac{1}{2} O A . O B \\ \Rightarrow \frac{1}{2} (\frac{6 \sqrt{2}}{\sqrt{13}}) (\frac{6 \sqrt{2}}{\sqrt{13}}) \Rightarrow \frac{18 (2)}{13} \Rightarrow \frac{36}{13} \end{array}$