Find the equation of the line passing through $\left(-6, 10\right)$ and perpendicular to the line $7x + 8y = 1$
Equation of given line $\;$ $7x + 8y = 1$
i.e. $\;$ $y = \dfrac{-7}{8} x + \dfrac{1}{8}$
$\therefore \;$ Slope of the given line $= m = \dfrac{-7}{8}$
Since the required line is perpendicular to the given line,
slope of the required line $= \dfrac{-1}{m} = \dfrac{8}{7}$
The required line passes through the point $\left(x_1, y_1\right) = \left(-6, 10\right)$
$\therefore \;$ The equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - 10 = \dfrac{8}{7} \left(x + 6\right)$
i.e. $\;$ $7y - 70 = 8x + 48$
i.e. $\;$ $8x - 7y + 118 = 0$