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

Factor the given pair into two lines through the origin, replace each slope by its negative reciprocal, then recombine into a joint equation.

The given pair is $5x^2+2xy-3y^2=0$. Treating this as a quadratic in $x$:

$5x^2+2xy-3y^2=0 \implies x = \frac{-2y \pm \sqrt{4y^2+60y^2}}{10} = \frac{-2y\pm 8y}{10}$

So $x = \dfrac{6y}{10} = \dfrac{3y}{5}$ or $x=-y$, giving the two lines:

$5x-3y=0 \quad (\text{slope } m_1 = 5/3), \qquad x+y=0 \quad (\text{slope } m_2=-1)$

A line through the origin perpendicular to a line of slope $m$ has slope $-1/m$. So the required perpendicular lines have slopes:

$-\frac{1}{m_1} = -\frac{3}{5}, \qquad -\frac{1}{m_2} = 1$

These are the lines:

$y = -\frac35 x \implies 3x+5y=0, \qquad y=x \implies x-y=0$

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