Area of triangle formed by pair of lines $(x + 2 a)^{2} - 3 y^{2} = 0$ and $x = a$ is

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Solution

$(x + 2 a)^{2} - 3 y^{2} = 0$ and $x = a$ is

Finding the vertices of the triangle

$\Rightarrow (x + 2 a - 3 y) (x + 2 a + 3 y) = 0$ Two line equations

$x - 3 y = - 2 a$
$x + 3 y = - 2 a$
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$2 x = - 4 a$ $x = - 2 a$
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$- 2 a - 3 y = - 2 a \Rightarrow y = 0$

$1^{s t}$ point $(- 2 a, 0) \to (x_{1}, y_{1})$

$\Rightarrow x + 2 a - 3 y = 0$ and $x = a$

$3 a - 3 y = 0 \Rightarrow y = a$

$2^{n d}$ point $(a, a) \to (x_{2}, y_{2})$

$\Rightarrow x + 2 a + 3 y = 0$ and $x = a$

$3 a + 3 y = 0 \Rightarrow y = - a$

$3^{r d}$ point $(a, - a) \to (x_{3}, y_{3})$

Area of a triangle formed by three points

$= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} | [(- 2 a) (a - (- a)) + a (- a - 0) + a (0 - a)] |$

$= \frac{1}{2} | - 4 a^{2} - a^{2} - a^{2} | = \frac{1}{2} \times 6 a^{2}$

$= 3 a^{2}$

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