IB Applications and Interpretation SL · 2D geometry · Section 4 of 5

Distance from a point to a line and perpendicular bisector

3 worked examples · By Álvaro López

Key idea

Two situations come up again and again in coordinate geometry: what is the shortest path from a point to a road, a pipe or a wall? And where are the points that are the same distance from two places? Both are solved with the same tool: a perpendicular line. In the IB exam they are the basis of Voronoi diagrams and location problems.

The shortest distance from a point to a line

A point $A$ and a line $r$ that does not pass through it. From $A$ you can reach $r$ by infinitely many straight paths, but only one is the shortest: the one that meets the line at a right angle.

[FIGURA mai-t4-a4-distancia-perpendicular: coordinate plane with axes and a soft grid. A downward-sloping line r (negative gradient) drawn in blue and labelled “r”. A point A in red, above and to the right of the line, labelled “A”. The perpendicular segment from A to the line in thick red, meeting r at a point D labelled “D”, with a small square at D marking the right angle. Two thin dashed grey segments from A to two other points on r (one on each side of D), labelled “longer than AD”. On the thick segment, label “d = AD, shortest distance”.]

Why is it the shortest? If you take another point $E$ on the line, triangle $A D E$ is right-angled at $D$ and $A E$ is its hypotenuse, which is always longer than the leg $A D$ .

You do not need a new formula to calculate it. You only need perpendicular lines (you already know that their gradients satisfy $m_{1} \cdot m_{2} = - 1$ ) and the distance between two points.

The perpendicular bisector of a segment

Now you have two points $A$ and $B$ and you are looking for the points in the plane that are the same distance from both.

The intuition: if $P$ is on the perpendicular bisector, triangles $P M A$ and $P M B$ are right-angled at $M$ , share the leg $P M$ and have the other leg equal ( $M A = M B$ ). By Pythagoras, their hypotenuses $P A$ and $P B$ are equal.

[FIGURA mai-t4-a4-mediatriz-equidistan: horizontal segment AB, with A on the left and B on the right, both points in blue and labelled. At the centre of the segment, the midpoint M labelled “M”, with a tick mark on each half to show that AM = MB. A vertical line through M, drawn in red, labelled “perpendicular bisector”, with a small right-angle square at M. On the line, two points P (above, at a medium distance) and Q (below), labelled. Thin dashed segments from P to A and B, and from Q to A and B; labels “PA = PB” near P and “QA = QB” near Q.]

To find its equation, follow the usual plan: one point and one gradient.

Notice that the perpendicular bisector does not pass through $A$ or $B$ : it passes through the midpoint. To check your equation, choose a point on the line and verify that it is the same distance from $A$ and from $B$ .

Worked examples

Example 1 Basic

The shortest route to a road

A campsite is at point $A (5, 4)$ on a map, with units in kilometres. A straight road has equation $2 x + y - 4 = 0$ .

a) Find the point on the road closest to the campsite.

b) Calculate the shortest distance from the campsite to the road.

Solution

a) The road has gradient $- 2$ (from $y = - 2 x + 4$ ). The perpendicular has gradient $\frac{1}{2}$ and passes through $A$ : $y - 4 = \frac{1}{2} (x - 5) \Rightarrow x - 2 y = - 3$ The closest point is the intersection of $2 x + y = 4$ with $x - 2 y = - 3$ . Using the GDC (or by substitution, $x = 2 y - 3$ ) you get $D (1, 2)$ .

b) The shortest distance is $A D$ : $d = (5 - 1)^{2} + (4 - 2)^{2} = 20 = 4.47 km$

Example 2 Medium

Two phone masts and the perpendicular bisector

Two phone masts are at points $A (- 2, 5)$ and $B (4, 1)$ on a plane, with units in kilometres. A third station must be the same distance from both.

a) Find the midpoint of segment $A B$ .

b) Find the equation of the perpendicular bisector of $A B$ in the form $y = m x + n$ .

c) The station is placed at the point on the perpendicular bisector whose $x$ -coordinate is $3$ . Find its $y$ -coordinate.

Solution

a) $M = (\frac{- 2 + 4}{2}, \frac{5 + 1}{2}) = (1, 3)$ .

b) The gradient of $A B$ is $\frac{1 - 5}{4 - ( - 2 )} = - \frac{4}{6} = - \frac{2}{3}$ , so the gradient of the perpendicular bisector is $\frac{3}{2}$ . Using the point $M$ : $y - 3 = \frac{3}{2} (x - 1) \Rightarrow y = \frac{3}{2} x + \frac{3}{2}$

c) For $x = 3$ : $y = \frac{3}{2} \cdot 3 + \frac{3}{2} = 6$ . The point is $P (3, 6)$ .

Check: $P A^{2} = (3 + 2)^{2} + (6 - 5)^{2} = 26$ and $P B^{2} = (3 - 4)^{2} + (6 - 1)^{2} = 26$ . They are the same distance from both.

Example 3 Hard

A fire station equidistant from two villages

Two villages are at $A (0, 6)$ and $B (8, 2)$ (units in kilometres). A straight road has equation $x + y - 11 = 0$ . A fire station is to be built on the road and the same distance from both villages.

a) Find the equation of the perpendicular bisector of segment $A B$ in the form $y = m x + n$ .

b) Find the coordinates of the point on the road where the fire station must go.

c) Calculate the distance from the fire station to each village.

d) Village $A$ is to be connected to the road by the shortest route. Calculate its length.

Solution

[FIGURA mai-t4-a4-bomberos: coordinate plane with axes x and y graduated from 0 to 10. Points A(0,6) and B(8,2) in blue with their labels and coordinates. Segment AB in grey. The perpendicular bisector y = 2x − 4 in red, labelled. The road x + y − 11 = 0 in black, labelled “road”. The two lines meet at the point P(5,6), marked in green and labelled “station P(5, 6)”. Dashed segments PA and PB. In addition, the foot D(2.5; 8.5) of the perpendicular from A to the road, with segment AD in thick line and a small right-angle square at D, labelled “shortest route”.]

a) The midpoint is $M = (4, 4)$ . The gradient of $A B$ is $\frac{2 - 6}{8 - 0} = - \frac{1}{2}$ , and that of the perpendicular bisector is $2$ : $y - 4 = 2 (x - 4) \Rightarrow y = 2 x - 4$

b) The station must be on the perpendicular bisector and on the road. Solve the system $y = 2 x - 4$ , $x + y = 11$ with the GDC (or by substitution: $x + 2 x - 4 = 11 \Rightarrow x = 5$ ). This gives $P (5, 6)$ .

c) $P A = (5 - 0)^{2} + (6 - 6)^{2} = 5$ km. Check: $P B = (5 - 8)^{2} + (6 - 2)^{2} = 25 = 5$ km. They coincide, as they should.

d) The road has gradient $- 1$ , so the perpendicular through $A$ has gradient $1$ : $y = x + 6$ . Its intersection with $x + y = 11$ is $x + x + 6 = 11$ , that is $D (2.5; 8.5)$ . Then $d = (2.5 - 0)^{2} + (8.5 - 6)^{2} = 12.5 = 3.54 km$

In the exam: marks are awarded separately for the midpoint, the perpendicular gradient, the equation and the system. Write down which line you intersect with which in each part, and leave the square root with the numbers before the decimal. Checking that $P A = P B$ is a good way to spot an error.

Common mistakes

Summary

The distance from a point $A$ to a line $r$ is the length of the perpendicular segment $A D$ (with $D$ on $r$ ); it is the shortest path.
Method: gradient of $r$ , perpendicular gradient, line through $A$ , intersection point $D$ (with the GDC) and distance $A D$ .
The perpendicular bisector of $A B$ is the line perpendicular to $A B$ through its midpoint $M$ .
All points on the perpendicular bisector are the same distance from $A$ as from $B$ ( $P A = P B$ ), and only those points are.
Steps: midpoint, gradient of $A B$ , perpendicular gradient, point-gradient equation.
To locate a point equidistant from $A$ and $B$ on another line, intersect the perpendicular bisector with that line.
Check your answers: $m \cdot m^{'} = - 1$ and $P A = P B$ .

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