Find the equation of the pair of straight lines passing through the point $\left(1,3\right)$ and perpendicular to the lines $2x - 3y + 1 = 0$ and $5x + y - 3 = 0$
The given lines are
$2x - 3y + 1 = 0$ $\;$ i.e. $y = \dfrac{2}{3}x + \dfrac{1}{3}$ $\;\;\; \cdots \; (1)$
and $5x + y - 3 = 0$ $\;$ i.e. $y = - 5 x + 3$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ Slope of line given by equation $(1)$ is $= m_1 = \dfrac{2}{3}$
and slope of line given by equation $(2)$ is $= m_2 = -5$
Let the slopes of the required pair of lines be $m_3$ and $m_4$.
Since the required lines are perpendicular to the given lines, their slopes are
$m_3 = \dfrac{-1}{m_1} = \dfrac{-3}{2}$ $\;$ and $\;$ $m_4 = \dfrac{-1}{m_2} = \dfrac{1}{5}$
Now, the required lines pass through the point $\left(1,3\right)$.
$\therefore$ $\;$ The equations of the individual lines are
$\left(y - 3\right) = \dfrac{-3}{2} \left(x - 1\right)$ $\;$ and $\;$ $\left(y - 3\right) = \dfrac{1}{5} \left(x - 1\right)$
i.e. $2y - 6 = - 3x + 3$ $\;$ and $\;$ $5y - 15 = x - 1$
i.e. $3x + 2 y - 9 = 0$ $\;$ and $\;$ $x - 5y + 14 = 0$
$\therefore$ $\;$ The combined equation of the required lines is
$\left(3x + 2y - 9\right) \left(x - 5y + 14\right) = 0$
i.e. $3x^2 - 15xy + 42x + 2 xy - 10 y^2 + 28 y - 9x + 45 y - 126 = 0$
i.e. $3x^2 - 13xy - 10y^2 + 33x + 73y - 126 = 0$ $\;\;\; \cdots \; (3)$
Equation $(3)$ gives the required equation of the pair of straight lines.