Q.Find the perpendicular distance of the point $P(3, -5)$ from the line $3x - 4y - 26 = 0$.

The line is already in general form, so apply $d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$.

The perpendicular distance of $P(x_1, y_1)$ from the line $ax + by + c = 0$ is

The line $3x - 4y - 26 = 0$ is already in general form, so $a = 3$, $b = -4$, $c = -26$; and $(x_1, y_1) = (3, -5)$. Substitute:

$d = \dfrac{\lvert 3(3) + (-4)(-5) + (-26) \rvert}{\sqrt{3^2 + (-4)^2}} = \dfrac{\lvert 9 + 20 - 26 \rvert}{\sqrt{9 + 16}} = \dfrac{\lvert 3 \rvert}{\sqrt{25}} = \dfrac{3}{5}$ …

Sign errors in the term $b y_1 = (-4)(-5)$ — this equals $+20$, not $-20$. Also, using the line in a non-general form (say $y = \tfrac34 x - \tfrac{26}{4}$) a …