3.5 Perpendicular bisectorsIB Maths: Applications and Interpretation SL: Revision notes

For points $A(x_1,y_1)$ and $B(x_2,y_2)$: $\text{midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\qquad m=\frac{y_2-y_1}{x_2-x_1},\qquad AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$ Example: $A(2,1)$ and $B(8,5)$ give midpoint $(5,3)$, gradient $\frac{4}{6}=\frac23$ and $AB=\sqrt{36+16}=7.21$.

Two lines are perpendicular when their gradients satisfy $m_1m_2=-1$. The perpendicular gradient is the negative reciprocal: $m_{\perp}=-\frac{1}{m}.$ Example: a line of gradient $\frac23$ has perpendicular gradient $-\frac32$. A horizontal line (gradient 0) is perpendicular to a vertical line (undefined gradient).

The perpendicular bisector of $[AB]$ cuts the segment at its midpoint at right angles. Every point on it is equidistant from $A$ and $B$.

1. Find the midpoint $M$ of $[AB]$.
2. Find the gradient of $AB$ and take the negative reciprocal.
3. Use $y-y_1=m(x-x_1)$ with $M$ and the perpendicular gradient. Example: $P(-3,7)$, $Q(5,3)$. Midpoint $(1,5)$; gradient of $PQ=-\frac12$, so perpendicular gradient 2. Then $y-5=2(x-1)$, i.e. $y=2x+3$. Check: $(2,7)$ is on the line and $\sqrt{25}=5$ from both points.

Sometimes you are given the equation of the line containing the segment and its midpoint, not the end points. The gradient comes from the equation and the point from the midpoint. Example: $[CD]$ lies on $3x-4y=10$ with midpoint $M(6,2)$. Rearranged, $y=\frac34x-\frac52$ has gradient $\frac34$. The perpendicular gradient is $-\frac43$, so $y-2=-\frac43(x-6)$, which gives $4x+3y=30$. If one end point is known, use the midpoint to find the other: from $C(2,-1)$ and $M(6,2)$, $D=(2\times6-2,\ 2\times2-(-1))=(10,5)$.

Give your equation in the form the question asks for:

• gradient–intercept: $y=mx+c$
• point–gradient: $y-y_1=m(x-x_1)$
• general: $ax+by=d$, where $a$, $b$ and $d$ are usually integers. Example: $y-3=-\frac32(x-5)$ becomes $2y-6=-3x+15$, so $3x+2y=21$. A vertical line is $x=k$ and a horizontal line is $y=k$. If $A$ and $B$ have the same $x$-coordinate, the perpendicular bisector is horizontal, $y=\frac{y_1+y_2}{2}$.

The perpendicular bisector is the set of points equidistant from $A$ and $B$, so it models a route, boundary or meeting place that is fair to both.

• A point on the bisector with a given $x$ or $y$: substitute into the equation.
• A point equidistant from three locations lies where two perpendicular bisectors cross: solve the two equations simultaneously (or use your GDC), then find the distance with $\sqrt{(\Delta x)^2+(\Delta y)^2}$. Example: schools $A(1,2)$, $B(9,8)$, $C(9,-2)$. The bisector of $[AB]$ is $4x+3y=35$ and that of $[BC]$ is $y=3$, meeting at $(6.5,3)$, which is 5.59 km from each school.