Find the joint equation of the pair of lines passing through the origin, which are perpendicular to the lines represented by $5 x^{2} + 2 x y - 3 y^{2} = 0$ .

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Solution

Given homogeneous equation is $5 x^{2} + 2 x y - 3 y^{2} = 0$ which is factorisable as
$5 x^{2} + 5 x y - 3 x y - 3 y^{2} = 0$

$\Rightarrow 5 x (x + y) - 3 y (x + y) = 0$

$\Rightarrow (x + y) (5 x - 3 y) = 0$

$x + y = 0$ and $5 x - 3 y = 0$ are the two lines represented by the given equation.

$\Rightarrow$ Their slopes are $- 1$ and $\frac{5}{3}$ .

Required two lines are respectively perpendicular to these lines.

$∴$ Slopes of required lines are $1$ and $- \frac{3}{5}$ and the lines pass through origin.

$∴$ Their individual equations are $y = 1 \cdot x$ and $y = - \frac{3}{5} x$

$i . e ., x - y = 0, 3 x + 5 y = 0$

$∴$ Their joint equation is $(x - y) (3 x + 5 y) = 0$

$\Rightarrow 3 x^{2} - 3 x y + 5 x y - 5 y^{2} = 0$

$\Rightarrow 3 x^{2} + 2 x y - 5 y^{2} = 0$

Hence $3 x^{2} + 2 x y - 5 y^{2} = 0$ is the required joint equation.

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