Let $3 x^{2} + 8 x y - 3 y^{2} = 0 r e p r e s e n t t h e l i n e s L_{1}, L_{2} a n d 3 x^{2} + 8 x y - 3 y^{2} + 2 x - 4 y - 1 = 0$ represent the lines $L_{3}, L_{4}$ . Let L be the line joining the points of intersection of $L_{1}, L_{3} a n d L_{2}, L_{4} .$ Then the area (in square units) of the triangle formed by L with the coordinate axes is

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$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 3 x^{2} + 8 x y - 3 y^{2} = 0 - - - (1) \\ a n d 3 x^{2} + 8 x y - 3 y^{2} + 2 x - 4 y + 1 = 0 - - - (2) \\ E q u a t i o n o f \bar{A C} i s (2) - (1) \\ \Rightarrow 2 x - 4 y + 1 = 0 \\ A r e a o f Δ^{l e} f o r m e d b y t h e l i n e \\ 2 x - 4 y + 1 = 0 w i t h c o - o r d i n a t e a x i s i s \frac{c^{2}}{2 | a b |} \\ \Rightarrow \frac{1}{2 | 2 \times - 4 |} = \frac{1}{16} \end{array}$