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Drawn from origin are two mutually perpendicular lines forming an isosceles triangle together with the straight line $2 x + y = a .$ Then the area of this triangle is

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$\frac{a^{2}}{3}$

$\frac{a^{2}}{2}$

$\frac{a^{2}}{5}$

$\frac{a^{2}}{7}$

answer is C.

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Detailed Solution

Let the two perpendiculars through the intersect $2 x + y = a$ A and B so that the triangle OAB is isosceles.

$O M =$ length of perpendicular from

O to AB, $O M = \frac{a}{\sqrt{5}}$ . Also $A M = M B = O M$

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$\Rightarrow A B = \frac{2 a}{\sqrt{5}} .$ Area of $Δ O A B = \frac{1}{2} A B . O M$

$= \frac{1}{2} . \frac{2 a}{\sqrt{5}} . \frac{a}{\sqrt{5}} = \frac{a^{2}}{5}$

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