Find the equation of the line through (0, -3) and perpendicular to the line joining the points (-3, 2) and (9, 1).

The slope (m₁) of the line joining (-3, 2) and (9, 1) i.e. two points is $\dfrac{y$2 - y1}{x2 - x1} so,

$\text{m}_1 = \dfrac{1 - 2}{9 - (-3)} \\[1em] = -\dfrac{1}{12}.$

Let the slope of the line perpendicular to the above line be m₂.

Then, m₁ × m₂ = -1.

$\Rightarrow -\dfrac{1}{12} \times m$2 = -1 \\[1em] \Rightarrow m2 = 12.

So, the equation of the line passing through (0, -3) and slope 12 can be given by point-slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - (-3) = 12(x - 0) \\[1em] \Rightarrow y + 3 = 12x \\[1em] \Rightarrow 12x - y - 3 = 0.

Hence, the equation of the required line is 12x - y - 3 = 0.