Right Bisectors and Distance to a Line

This page puts your three tools together: midpoint, slope and length. You’ll find the right bisector of a segment, the shortest distance from a point to a line, and the point that’s the same distance from all three corners of a triangle. These are real multi-step problems, so the key skill is planning the steps before you start.

Key ideas

The right bisector of a segment

The right bisector (or perpendicular bisector) of a segment $AB$ is the line that

• passes through the midpoint of $AB$, and
• is perpendicular to $AB$.

To find its equation:

1. Find the midpoint $M$ of $AB$.
2. Find the slope of $AB$.
3. Take the negative reciprocal to get the perpendicular slope (see parallel and perpendicular lines).
4. Write the equation of the line with that slope through $M$.

Special cases: if $AB$ is horizontal, its right bisector is a vertical line $x = \text{(midpoint's } x)$. If $AB$ is vertical, the right bisector is a horizontal line $y = \text{(midpoint's } y)$.

Why right bisectors matter

Every point on the right bisector of $AB$ is the same distance from $A$ as from $B$. That makes right bisectors useful whenever something must be “equally far” from two places.

Shortest distance from a point to a line

The distance from a point $P$ to a line means the shortest distance, and that’s always measured along the line through $P$ that is perpendicular to the given line. The plan:

1. Find the slope of the given line.
2. Write the equation of the line through $P$ with the perpendicular slope.
3. Solve the two equations as a linear system (by substitution or elimination) to find $F$, the point where they meet.
4. Use the length formula to find $PF$.

Centres of a triangle

Three special lines in a triangle each meet at a single point:

Lines	Meet at the	Special because
Right bisectors of the sides	circumcentre	it’s the same distance from all three vertices (centre of the circle through them)
Medians (vertex to midpoint of opposite side)	centroid	it’s the balance point of the triangle
Altitudes (from a vertex, perpendicular to the opposite side)	orthocentre	it’s where the three heights meet

To find any of these, find the equations of two of the lines and solve them as a system. The third line passes through the same point, which makes a good check.

Worked examples

Example 1: The right bisector of a segment

Find the equation of the right bisector of the segment joining $A(-2, 1)$ and $B(4, 5)$.

Solution.

Midpoint: $M = \left( \dfrac{-2 + 4}{2}, \dfrac{1 + 5}{2} \right) = (1, 3)$

Slope of $AB$: $m_{AB} = \dfrac{5 - 1}{4 - (-2)} = \dfrac{4}{6} = \dfrac{2}{3}$

Perpendicular slope: $-\dfrac{3}{2}$

Equation: substitute $M(1, 3)$ into $y = -\dfrac{3}{2}x + b$:

$$3 = -\frac{3}{2}(1) + b \quad\Rightarrow\quad b = 3 + \frac{3}{2} = \frac{9}{2}$$

The right bisector is $y = -\dfrac{3}{2}x + \dfrac{9}{2}$. Multiplying by $2$ and rearranging gives the standard form $3x + 2y - 9 = 0$.

Check: $(3, 0)$ is on this line, since $3(3) + 2(0) - 9 = 0$. Its distance to $A$ is $\sqrt{(3 + 2)^2 + (0 - 1)^2} = \sqrt{26}$, and to $B$ is $\sqrt{(3 - 4)^2 + (0 - 5)^2} = \sqrt{26}$. Equally far, as promised. ✓

Example 2: A horizontal segment

Find the right bisector of the segment joining $C(-3, 2)$ and $D(5, 2)$.

Solution. Both points have $y = 2$, so $CD$ is horizontal. Its midpoint is $\left( \dfrac{-3 + 5}{2}, 2 \right) = (1, 2)$.

A line perpendicular to a horizontal line is vertical, so the right bisector is $x = 1$.

Example 3: Distance from a point to a line

Find the shortest distance from $P(7, 0)$ to the line $y = 2x + 1$.

Solution.

Step 1. The line has slope $2$.

Step 2. The perpendicular slope is $-\dfrac{1}{2}$. Through $P(7, 0)$:

$$0 = -\frac{1}{2}(7) + b \quad\Rightarrow\quad b = \frac{7}{2}$$

so the perpendicular line is $y = -\dfrac{1}{2}x + \dfrac{7}{2}$.

Step 3. Solve the system by substitution (set the two expressions for $y$ equal):

$$\begin{aligned} 2x + 1 &= -\frac{1}{2}x + \frac{7}{2} \\ 4x + 2 &= -x + 7 && \text{multiply by } 2 \\ 5x &= 5 \\ x &= 1 \end{aligned}$$

Then $y = 2(1) + 1 = 3$, so the lines meet at $F(1, 3)$.

Step 4. Find $PF$:

$$PF = \sqrt{(1 - 7)^2 + (3 - 0)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71$$

The shortest distance is $\sqrt{45}$ (or $3\sqrt{5}$) units, about $6.71$ units.

The shortest path from $P$ to the line meets it at a right angle, at $F(1, 3)$.

Check: $F(1, 3)$ is on both lines: $2(1) + 1 = 3$ ✓ and $-\dfrac{1}{2}(1) + \dfrac{7}{2} = 3$ ✓. You can also check with graphing technology by plotting both lines and the point.

Example 4: The circumcentre (a cell tower)

Three towns are at $A(-2, 5)$, $B(4, 5)$ and $C(6, 1)$ on a map grid measured in kilometres. A cell tower is to be placed the same distance from all three towns. Where should it go, and how far is it from each town?

Solution. A point the same distance from $A$ and $B$ is on the right bisector of $AB$. A point the same distance from $B$ and $C$ is on the right bisector of $BC$. So the tower goes where those two lines meet: the circumcentre.

Right bisector of $AB$. Both points have $y = 5$, so $AB$ is horizontal. Its midpoint is $(1, 5)$, so the right bisector is $x = 1$.

Right bisector of $BC$. Midpoint: $\left( \dfrac{4 + 6}{2}, \dfrac{5 + 1}{2} \right) = (5, 3)$. Slope of $BC$: $\dfrac{1 - 5}{6 - 4} = \dfrac{-4}{2} = -2$. Perpendicular slope: $\dfrac{1}{2}$. Through $(5, 3)$:

$$3 = \frac{1}{2}(5) + b \quad\Rightarrow\quad b = \frac{1}{2}$$

so this right bisector is $y = \dfrac{1}{2}x + \dfrac{1}{2}$.

Intersect. Substitute $x = 1$: $y = \dfrac{1}{2} + \dfrac{1}{2} = 1$. The tower goes at $(1, 1)$.

Distances:

$$\begin{aligned} \text{to } A &: \sqrt{(-2 - 1)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5 \\ \text{to } B &: \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5 \\ \text{to } C &: \sqrt{(6 - 1)^2 + (1 - 1)^2} = \sqrt{25} = 5 \end{aligned}$$

The tower should go at $(1, 1)$, and it’s $5$ km from each town.

The right bisectors meet at the circumcentre $(1, 1)$, the centre of the circle through $A$, $B$ and $C$.

Common mistakes

Using the segment’s slope instead of the perpendicular slope. The right bisector is perpendicular to the segment. If $AB$ has slope $\dfrac{2}{3}$, the bisector has slope $-\dfrac{3}{2}$: flip the fraction and change the sign.

Putting the bisector through an endpoint. The right bisector goes through the midpoint of the segment. Using $A$ or $B$ gives a perpendicular line, but not the bisector.

Measuring distance to the wrong point on the line. The distance from a point to a line is the perpendicular distance. Picking any convenient point on the line, like its $y$-intercept, gives a longer distance. Always find the foot of the perpendicular $F$ first.

Stopping after finding the intersection point. In a distance problem, $F$ is not the answer. The answer is the length $PF$.

Mixing up the triangle centres. Right bisectors give the circumcentre, medians give the centroid, and altitudes give the orthocentre. Medians go to midpoints; altitudes are perpendicular to the opposite side; right bisectors do both but don’t have to pass through a vertex.

Arithmetic slips in the system. Fractions make substitution error-prone. Clear the fractions first (multiply every term by the denominator), and check that your intersection point satisfies both equations.

Practice

1. (Warm-up) Find the equation of the right bisector of each segment.

• (a) $(-3, -1)$ and $(5, -1)$
• (b) $(2, -4)$ and $(2, 6)$

Solution

(a) The segment is horizontal with midpoint $(1, -1)$, so the right bisector is the vertical line $x = 1$.

(b) The segment is vertical with midpoint $(2, 1)$, so the right bisector is the horizontal line $y = 1$.

2. (Warm-up) For the segment joining $(1, 2)$ and $(5, 10)$, find the midpoint and the slope of the right bisector.

Solution

Midpoint: $\left( \dfrac{1 + 5}{2}, \dfrac{2 + 10}{2} \right) = (3, 6)$.

Slope of the segment: $\dfrac{10 - 2}{5 - 1} = \dfrac{8}{4} = 2$, so the right bisector has slope $-\dfrac{1}{2}$.

3. (Core) Find the equation of the right bisector of $PQ$, with $P(-1, 3)$ and $Q(5, -1)$. Then check that the point $(4, 4)$ is on it, and that it’s the same distance from $P$ and $Q$.

Solution

Midpoint: $\left( \dfrac{-1 + 5}{2}, \dfrac{3 + (-1)}{2} \right) = (2, 1)$.

Slope of $PQ$: $\dfrac{-1 - 3}{5 - (-1)} = \dfrac{-4}{6} = -\dfrac{2}{3}$, so the perpendicular slope is $\dfrac{3}{2}$.

Through $(2, 1)$: $1 = \dfrac{3}{2}(2) + b$, so $b = -2$.

The right bisector is $y = \dfrac{3}{2}x - 2$ (standard form $3x - 2y - 4 = 0$).

Check $(4, 4)$: $\dfrac{3}{2}(4) - 2 = 4$. ✓ Distances: to $P$, $\sqrt{(4 + 1)^2 + (4 - 3)^2} = \sqrt{26}$; to $Q$, $\sqrt{(4 - 5)^2 + (4 + 1)^2} = \sqrt{26}$. Equal. ✓

4. (Core) Find the shortest distance from $P(-3, 4)$ to the line $y = x - 1$.

Solution

The line has slope $1$, so the perpendicular slope is $-1$. Through $P(-3, 4)$: $4 = -1(-3) + b$, so $b = 1$, giving $y = -x + 1$.

Solve:

$$x - 1 = -x + 1 \quad\Rightarrow\quad 2x = 2 \quad\Rightarrow\quad x = 1$$

Then $y = 1 - 1 = 0$, so $F = (1, 0)$.

$$PF = \sqrt{(1 - (-3))^2 + (0 - 4)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66$$

The distance is $\sqrt{32}$ (or $4\sqrt{2}$) units, about $5.66$ units.

5. (Core) Find the shortest distance from the origin to the line $3x + 4y = 25$.

Solution

Rearrange: $4y = -3x + 25$, so $y = -\dfrac{3}{4}x + \dfrac{25}{4}$. The slope is $-\dfrac{3}{4}$, so the perpendicular slope is $\dfrac{4}{3}$. Through the origin, the perpendicular line is $y = \dfrac{4}{3}x$.

Substitute into $3x + 4y = 25$:

$$\begin{aligned} 3x + 4\left(\frac{4}{3}x\right) &= 25 \\ 9x + 16x &= 75 && \text{multiply by } 3 \\ 25x &= 75 \\ x &= 3 \end{aligned}$$

Then $y = \dfrac{4}{3}(3) = 4$, so $F = (3, 4)$.

$$\text{distance} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$

The distance is $5$ units. (So the line just touches the circle $x^2 + y^2 = 25$ at $(3, 4)$.)

6. (Core) On a farm map with a grid where $1$ unit $= 100$ m, a straight road follows the line $x + 2y = 6$. A farmhouse is at $H(5, 8)$. A water line will be laid along the shortest path from the house to the road. Where does it meet the road, and how long is it, to the nearest metre?

Solution

Rearrange the road: $y = -\dfrac{1}{2}x + 3$, slope $-\dfrac{1}{2}$. The perpendicular slope is $2$. Through $H(5, 8)$: $8 = 2(5) + b$, so $b = -2$, giving $y = 2x - 2$.

Substitute into $x + 2y = 6$:

$$x + 2(2x - 2) = 6 \quad\Rightarrow\quad 5x - 4 = 6 \quad\Rightarrow\quad x = 2$$

Then $y = 2(2) - 2 = 2$. The water line meets the road at $(2, 2)$. Check: $2 + 2(2) = 6$. ✓

$$\text{length} = \sqrt{(5 - 2)^2 + (8 - 2)^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.708 \text{ units}$$

Each unit is $100$ m, so the water line is about $670.8$ m, or $671$ m to the nearest metre.

7. (Core) Find the circumcentre of the triangle with vertices $D(-1, 0)$, $E(3, 2)$ and $F(5, -2)$, and show that it’s the same distance from all three vertices.

Solution

Right bisector of $DE$. Midpoint $(1, 1)$. Slope of $DE$: $\dfrac{2 - 0}{3 - (-1)} = \dfrac{1}{2}$, so the perpendicular slope is $-2$. Through $(1, 1)$: $1 = -2 + b$, so $b = 3$, giving $y = -2x + 3$.

Right bisector of $EF$. Midpoint $(4, 0)$. Slope of $EF$: $\dfrac{-2 - 2}{5 - 3} = -2$, so the perpendicular slope is $\dfrac{1}{2}$. Through $(4, 0)$: $0 = 2 + b$, so $b = -2$, giving $y = \dfrac{1}{2}x - 2$.

Intersect:

$$\begin{aligned} -2x + 3 &= \frac{1}{2}x - 2 \\ -4x + 6 &= x - 4 && \text{multiply by } 2 \\ -5x &= -10 \\ x &= 2 \end{aligned}$$

Then $y = -2(2) + 3 = -1$. The circumcentre is $(2, -1)$.

Distances: to $D$, $\sqrt{(-1 - 2)^2 + (0 + 1)^2} = \sqrt{10}$; to $E$, $\sqrt{(3 - 2)^2 + (2 + 1)^2} = \sqrt{10}$; to $F$, $\sqrt{(5 - 2)^2 + (-2 + 1)^2} = \sqrt{10}$. All equal (about $3.16$). ✓

8. (Challenge) Triangle $ABC$ has vertices $A(-4, -2)$, $B(2, 6)$ and $C(8, -4)$. Find the equations of two medians and their point of intersection (the centroid). Then show that the third median passes through the same point.

Solution

Median from $A$. Midpoint of $BC$: $(5, 1)$. Slope from $A(-4, -2)$: $\dfrac{1 - (-2)}{5 - (-4)} = \dfrac{3}{9} = \dfrac{1}{3}$. Through $A$: $-2 = \dfrac{1}{3}(-4) + b$, so $b = -\dfrac{2}{3}$, giving $y = \dfrac{1}{3}x - \dfrac{2}{3}$.

Median from $B$. Midpoint of $AC$: $(2, -3)$. Both $B(2, 6)$ and $(2, -3)$ have $x = 2$, so this median is the vertical line $x = 2$.

Intersect: at $x = 2$, $y = \dfrac{2}{3} - \dfrac{2}{3} = 0$. The centroid is $(2, 0)$.

Median from $C$. Midpoint of $AB$: $(-1, 2)$. Slope from $C(8, -4)$: $\dfrac{2 - (-4)}{-1 - 8} = \dfrac{6}{-9} = -\dfrac{2}{3}$. Through $C$: $-4 = -\dfrac{2}{3}(8) + b$, so $b = -4 + \dfrac{16}{3} = \dfrac{4}{3}$, giving $y = -\dfrac{2}{3}x + \dfrac{4}{3}$.

At $x = 2$: $y = -\dfrac{4}{3} + \dfrac{4}{3} = 0$. So the third median also passes through $(2, 0)$. ✓

(Notice that $(2, 0)$ is the average of the three vertices: $\left(\dfrac{-4 + 2 + 8}{3}, \dfrac{-2 + 6 - 4}{3}\right) = (2, 0)$.)

9. (Challenge) Find the orthocentre of the triangle with vertices $A(-4, 0)$, $B(4, 0)$ and $C(2, 6)$. (An altitude goes from a vertex to the opposite side, perpendicular to that side.)

Solution

Altitude from $C$. The opposite side $AB$ is horizontal (on the $x$-axis), so the altitude is vertical through $C$: $x = 2$.

Altitude from $A$. Slope of $BC$: $\dfrac{6 - 0}{2 - 4} = \dfrac{6}{-2} = -3$, so the altitude has slope $\dfrac{1}{3}$. Through $A(-4, 0)$: $0 = -\dfrac{4}{3} + b$, so $b = \dfrac{4}{3}$, giving $y = \dfrac{1}{3}x + \dfrac{4}{3}$.

Intersect: at $x = 2$, $y = \dfrac{2}{3} + \dfrac{4}{3} = 2$. The orthocentre is $(2, 2)$.

Check with the third altitude (from $B$). Slope of $AC$: $\dfrac{6 - 0}{2 - (-4)} = 1$, so the altitude has slope $-1$. Through $B(4, 0)$: $y = -x + 4$. At $x = 2$, $y = 2$. ✓