Straight lines and gradients

The idea

A line through $(x_1, y_1)$ with gradient $m$ is $y - y_1 = m(x - x_1)$. Parallel lines share a gradient; perpendicular gradients multiply to $-1$, so the normal gradient is $-\frac{1}{m}$.

1. Find the gradient: $m = \frac{y_2 - y_1}{x_2 - x_1}$, or rearrange to $y = mx + c$.
2. Use $y - y_1 = m(x - x_1)$ with a point on the line.
3. Give the form asked for, such as $ax + by + c = 0$ with integers.

Watch out Rearrange $3x + 4y = 12$ before reading the gradient: it is $-\frac{3}{4}$, not $3$.