Two lines are perpendicular exactly when $m_1\times m_2=-1$, so $m_2=-\frac{1}{m_1}$ — the negative reciprocal: flip the fraction, then change the sign. Both steps are required — e.g. if $m_1=-\frac23$, then $m_2=+\frac32$, not $-\frac32$.

This "stacks" three skills: midpoint, gradient, and perpendicular rule.

1. Midpoint of $A(x_A,y_A)$ and $B(x_B,y_B)$: $M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right)$.
2. Gradient of $AB$: $m_1=\frac{y_B-y_A}{x_B-x_A}$.
3. Perpendicular gradient: $m_2=-\frac{1}{m_1}$.
4. Equation: $y-y_M=m_2(x-x_M)$.

Worked example: Perpendicular bisector of $A(2,-1)$ and $B(8,7)$.

Midpoint: $(5,3)$. Gradient of $AB$: $\frac{7-(-1)}{8-2}=\frac86=\frac43$. Perpendicular gradient: $-\frac34$.

A tangent to a circle is always perpendicular to the radius at the point of contact — so find the radius's gradient, then take the negative reciprocal.

Worked example: Tangent to $x^2+y^2=25$ at $(3,4)$ (centre $(0,0)$).

For a circle centred at $(a,b)$ rather than the origin, the method is identical — just use $m_{\text{radius}}=\frac{y_1-b}{x_1-a}$.

For a curve $y=f(x)$ there's no centre or radius — the gradient at any point comes from differentiation, $\frac{dy}{dx}$. The power rule: if $y=ax^n$, then $\frac{dy}{dx}=anx^{n-1}$ (differentiate term-by-term for sums). Rewrite roots/fractions as powers first, e.g. $y=\frac{6}{x^2}=6x^{-2} \implies \frac{dy}{dx}=-12x^{-3}$.

Method: (1) find $y_1$ by substituting $x_1$ into $y=f(x)$; (2) find $\frac{dy}{dx}$; (3) substitute $x_1$ into $\frac{dy}{dx}$ to get the gradient $m_T$; (4) use $y-y_1=m_T(x-x_1)$.

Worked example: Tangent to $y=x^3-5x$ at $x=3$.

$y_1=27-15=12$. $\frac{dy}{dx}=3x^2-5$, so $m_T=3(9)-5=22$.

If you're given a circle equation ($x^2+y^2=r^2$ or similar) → it's a Coordinate Geometry problem: find $m_{\text{radius}}$, then flip.

If you're given a function ($y=$ some expression in $x$) → it's a Calculus problem: differentiate, then substitute.

The normal is perpendicular to the tangent at the point of contact, so $m_N=-\frac{1}{m_T}$.

Shortcut for circles: since both the radius and the normal are perpendicular to the tangent at the same point, they must be the same line — so the normal's gradient is simply $m_{\text{radius}}$, with no need to find the tangent at all.

Worked example: Normal to $x^2+y^2=100$ at $(6,-8)$: $m_N=\frac{-8-0}{6-0}=-\frac43$. Using the centre $(0,0)$: $y=-\frac43x$.

For a curve, there's no shortcut — find $m_T=\frac{dy}{dx}$ at the point, then $m_N=-\frac{1}{m_T}$, then apply the point-gradient formula as usual.

Find the equation of the normal to $y=x^2-6x+5$ at $(7,12)$.

$\frac{dy}{dx}=2x-6$, so at $x=7$: $m_T=14-6=8$.