Legs: $x_1-x_2$ and $y_1-y_2$, then Pythagoras. Reflex: Subtract coordinates in the same order in both brackets.

Average the x's, average the y's. Reflex: Add and halve — nothing gets subtracted.

Gradient $= \frac{\text{rise}}{\text{run}}$, and in the triangle standing on the $x$-axis, $\tan\theta = \frac{\text{rise}}{\text{run}}$ — the same fraction, so gradient $= \tan\theta$. Here θ is the acute angle the line makes with the positive $x$-axis. Reflex: Need θ? Use $\theta = \tan^{-1}(\text{gradient})$.

Similar triangles form, so horizontal and vertical pieces split in the same ratio $n:m$. Reflex: Solve the x-equation and y-equation separately, on their own.

Faster than $y=mx+c$ since you skip solving for $c$. Reflex: Have a gradient and ANY point? Go straight to $y - y_1 = m(x - x_1)$.

Parallel lines have the same gradient. To find where two lines intersect, solve their equations simultaneously.

Reflex: See "parallel to" → copy that line's gradient.

Perpendicular gradients multiply to $-1$. Flip the fraction, change the sign. Reflex: $\frac{1}{2} \to -2$, $-\frac{3}{5} \to \frac{5}{3}$ — flip & flip sign.

Midpoint gives the point; negative reciprocal of $m_{AB}$ gives the gradient. Then form the equation. Reflex: Sub in the MIDPOINT — never $A$ or $B$.

↘ products are added, ↗ products subtracted; the modulus of half the total is the area. Reflex: List vertices anti-clockwise, repeat the first point at the end.

Parallelogram: diagonals share the same midpoint. Rhombus: same, plus diagonals perpendicular. Kite: diagonals perpendicular, axis bisects the other diagonal.

Reflex: Bisect → equate midpoints; perpendicular → negative-reciprocal gradients.

Common exam pair: part (i) find the area by shoelace; part (ii) "hence" find the perpendicular distance from $C$ to $AB$. Same triangle, so rearrange the area formula: double the area, then divide by the length $AB$ (distance formula). Reflex: After an area, "hence find the distance" means $h = 2 \times \text{Area} \div \text{base}$.