A pair of straight lines drawn through the origin forms an isosceles triangle right angled at the origin with the line $2 x + 3 y = 6 .$ The area (in sq. units) of the triangle so formed is

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$\frac{36}{13}$

$\frac{32}{13}$

$\frac{28}{9}$

$\frac{26}{9}$

answer is A.

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Detailed Solution

$y = m x$

It makes an angle of $\pm 45^{\circ} with 2 x + 3 y = 6 (lope - \frac{2}{3})$

$∴ \tan (\pm 45^{\circ}) = \frac{m - (- 2 / 3)}{1 + m (- 2 / 3)} = \pm 1$

or $3 m + 2 = \pm (3 - 2 m) ∴ m = \frac{1}{5}, - 5$

Hence the sides are $x - 5 y = 0, 5 x + y = 0$ and $2 x + 3 y = 6$

If $p = A D$ be perpendicular from $A t o B C$ then

$p = \frac{6}{\sqrt{13}} = A D = D B = D C ∴ B C = 2 p$

In right angled triangle the mid-point of hypotenuse is equidistant from the vertices.

$∴ Area = \frac{1}{2} p \cdot 2 p = p^{2} = \frac{36}{13}$

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