Midpoint of a Line Segment

The midpoint of a line segment is the point exactly halfway between its endpoints. Once you can find midpoints on a grid, you can find the centre of a circle from a diameter, split a trip in half on a map, and build special lines in triangles called medians. It’s the first of three tools (midpoint, length, and slope) that this whole unit runs on.

Key ideas

Halfway means “average”

Start with a horizontal segment from $(2, 3)$ to $(8, 3)$. Halfway between $x = 2$ and $x = 8$ is $x = 5$, which is the average of $2$ and $8$:

$$\frac{2 + 8}{2} = 5$$

So the midpoint is $(5, 3)$. A vertical segment works the same way with the $y$-coordinates.

For a slanted segment, the run and the rise both split in half at the midpoint. Going halfway along the segment means going halfway across and halfway up. So you average the $x$-coordinates and average the $y$-coordinates separately.

The midpoint formula

The midpoint $M$ of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is

$$M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right)$$

In words: add the $x$‘s and halve, add the $y$‘s and halve. It doesn’t matter which endpoint you call $A$, because addition works in either order.

Finding a missing endpoint

Sometimes you know the midpoint $M$ and one endpoint $A$, and you need the other endpoint $B$. Two ways to think about it:

• Step method. Find the step from $A$ to $M$ (how far across, how far up). Take the same step again from $M$ to land on $B$.
• Equation method. Put the unknown endpoint $(x, y)$ into the midpoint formula and solve each coordinate. This works out to $x = 2(\text{midpoint } x) - x_1$, and the same for $y$.

Medians of a triangle

A median of a triangle is the segment from a vertex to the midpoint of the opposite side. Every triangle has three medians, one from each vertex. To find the equation of a median:

1. Find the midpoint of the side opposite the vertex.
2. Find the slope from the vertex to that midpoint.
3. Use the slope and a point to write the equation, as on equations of lines.

Worked examples

Example 1: Finding a midpoint

Find the midpoint of the segment joining $A(-3, 5)$ and $B(7, -1)$.

Solution. Average each coordinate:

$$M = \left( \frac{-3 + 7}{2},\ \frac{5 + (-1)}{2} \right) = \left( \frac{4}{2},\ \frac{4}{2} \right) = (2, 2)$$

Check: from $A$ to $M$ is $5$ right and $3$ down. From $M$ to $B$ is also $5$ right and $3$ down. Equal steps, so $M$ is halfway. ✓

Example 2: Finding a missing endpoint

The point $M(4, -1)$ is the midpoint of $PQ$, and $P$ is $(1, 3)$. Find $Q$.

Solution (step method). From $P(1, 3)$ to $M(4, -1)$, $x$ goes up by $3$ and $y$ goes down by $4$. Take the same step again from $M$:

$$Q = (4 + 3,\ -1 - 4) = (7, -5)$$

Solution (equation method). Let $Q = (x, y)$. Then

$$\frac{1 + x}{2} = 4 \quad\Rightarrow\quad 1 + x = 8 \quad\Rightarrow\quad x = 7$$ $$\frac{3 + y}{2} = -1 \quad\Rightarrow\quad 3 + y = -2 \quad\Rightarrow\quad y = -5$$

Both methods give $Q(7, -5)$.

Check: the midpoint of $(1, 3)$ and $(7, -5)$ is $\left(\dfrac{8}{2}, \dfrac{-2}{2}\right) = (4, -1)$. ✓

Example 3: Midpoints of the sides of a triangle

A triangle has vertices $A(-2, 4)$, $B(6, 2)$ and $C(0, -4)$. Find the midpoint of each side.

Solution.

$$\begin{aligned} \text{Midpoint of } AB &= \left( \frac{-2 + 6}{2},\ \frac{4 + 2}{2} \right) = (2, 3) \\ \text{Midpoint of } BC &= \left( \frac{6 + 0}{2},\ \frac{2 + (-4)}{2} \right) = (3, -1) \\ \text{Midpoint of } AC &= \left( \frac{-2 + 0}{2},\ \frac{4 + (-4)}{2} \right) = (-1, 0) \end{aligned}$$

Example 4: The equation of a median

For the triangle in Example 3, find the equation of the median from $A$.

Solution. The median from $A$ goes to the midpoint of the opposite side, $BC$. From Example 3, that’s $M(3, -1)$.

Slope from $A(-2, 4)$ to $M(3, -1)$:

$$m = \frac{-1 - 4}{3 - (-2)} = \frac{-5}{5} = -1$$

Substitute $m = -1$ and the point $(3, -1)$ into $y = mx + b$:

$$-1 = -1(3) + b \quad\Rightarrow\quad -1 = -3 + b \quad\Rightarrow\quad b = 2$$

The median from $A$ is $y = -x + 2$.

Check with the other point: at $A(-2, 4)$, $-(-2) + 2 = 4$. ✓

The median from $A$ ends at $M(3, -1)$, the midpoint of the opposite side $BC$.

Common mistakes

Subtracting instead of adding. The midpoint formula adds the coordinates: $\dfrac{x_1 + x_2}{2}$. Subtracting is for slope and length. If your midpoint isn’t between the two endpoints on a quick sketch, check the signs.

Mixing up $x$‘s and $y$‘s. Average the two $x$-coordinates together and the two $y$-coordinates together. Never add an $x$ to a $y$. Writing the coordinates in a little table (one row for $x$, one for $y$) helps.

Losing a negative sign. With $(-3, 5)$ and $(7, -1)$, the $y$-sum is $5 + (-1) = 4$, not $6$. Put negative numbers in brackets when you substitute.

Treating the midpoint as the missing endpoint. If $M$ is the midpoint and you know $A$, the other endpoint is past $M$, not between $A$ and $M$. Averaging $A$ and $M$ gives a quarter point, not the endpoint. Use the step method and check that $M$ really is the midpoint of your answer.

Drawing the median to the wrong side. The median from a vertex goes to the midpoint of the side opposite that vertex: from $A$ to the midpoint of $BC$, never to the midpoint of $AB$ or $AC$.

Practice

1. (Warm-up) Find the midpoint of each segment.

• (a) $(2, 8)$ and $(6, 4)$
• (b) $(-5, 3)$ and $(1, -7)$
• (c) $(0, 0)$ and $(9, -4)$

Solution

(a) $\left(\dfrac{2 + 6}{2}, \dfrac{8 + 4}{2}\right) = (4, 6)$

(b) $\left(\dfrac{-5 + 1}{2}, \dfrac{3 + (-7)}{2}\right) = \left(\dfrac{-4}{2}, \dfrac{-4}{2}\right) = (-2, -2)$

(c) $\left(\dfrac{0 + 9}{2}, \dfrac{0 + (-4)}{2}\right) = (4.5, -2)$. A midpoint doesn’t have to land on a grid point.

2. (Warm-up) Find the midpoint of the segment joining $(-1, 4)$ and $(4, -3)$. Give the coordinates as fractions.

Solution

$$M = \left( \frac{-1 + 4}{2},\ \frac{4 + (-3)}{2} \right) = \left( \frac{3}{2},\ \frac{1}{2} \right)$$

3. (Core) $M(-2, 5)$ is the midpoint of $AB$, and $A$ is $(3, 8)$. Find $B$.

Solution

From $A(3, 8)$ to $M(-2, 5)$, $x$ goes down by $5$ and $y$ goes down by $3$. Take the same step again from $M$:

$$B = (-2 - 5,\ 5 - 3) = (-7, 2)$$

Check: $\left(\dfrac{3 + (-7)}{2}, \dfrac{8 + 2}{2}\right) = (-2, 5)$. ✓

4. (Core) A diameter of a circle has endpoints $(-4, -1)$ and $(6, 5)$. Find the centre of the circle.

Solution

The centre of a circle is the midpoint of any diameter:

$$\left( \frac{-4 + 6}{2},\ \frac{-1 + 5}{2} \right) = (1, 2)$$

5. (Core) A triangle has vertices $P(1, 5)$, $Q(7, 1)$ and $R(-3, -1)$. Find the midpoint of each side.

Solution

$$\begin{aligned} \text{Midpoint of } PQ &= \left( \frac{1 + 7}{2},\ \frac{5 + 1}{2} \right) = (4, 3) \\ \text{Midpoint of } QR &= \left( \frac{7 + (-3)}{2},\ \frac{1 + (-1)}{2} \right) = (2, 0) \\ \text{Midpoint of } PR &= \left( \frac{1 + (-3)}{2},\ \frac{5 + (-1)}{2} \right) = (-1, 2) \end{aligned}$$

6. (Core) For the triangle in question 5, find the equation of the median from $Q$.

Solution

The median from $Q$ goes to the midpoint of $PR$, which is $(-1, 2)$.

$$m = \frac{2 - 1}{-1 - 7} = \frac{1}{-8} = -\frac{1}{8}$$

Substitute $(7, 1)$ into $y = -\dfrac{1}{8}x + b$:

$$1 = -\frac{7}{8} + b \quad\Rightarrow\quad b = \frac{15}{8}$$

The median is $y = -\dfrac{1}{8}x + \dfrac{15}{8}$, or $x + 8y - 15 = 0$ in standard form.

Check with $(-1, 2)$: $-1 + 8(2) - 15 = 0$. ✓

7. (Core) On a map with a grid in kilometres, Lakeview is at $(2, 9)$ and Brookdale is at $(14, 3)$. A rest stop is being built halfway between them along the straight road joining the towns. Where should it go?

Solution

$$\left( \frac{2 + 14}{2},\ \frac{9 + 3}{2} \right) = (8, 6)$$

The rest stop goes at $(8, 6)$ on the map grid.

8. (Challenge) The points $A(-2, 1)$, $B(4, 3)$ and $C(6, -1)$ are three vertices of parallelogram $ABCD$. The diagonals of a parallelogram bisect each other (they have the same midpoint). Use this to find $D$.

Solution

The diagonals are $AC$ and $BD$. The midpoint of $AC$ is

$$\left( \frac{-2 + 6}{2},\ \frac{1 + (-1)}{2} \right) = (2, 0)$$

So $(2, 0)$ must also be the midpoint of $BD$. From $B(4, 3)$ to $(2, 0)$ is $2$ left and $3$ down. Take the same step again:

$$D = (2 - 2,\ 0 - 3) = (0, -3)$$

Check: the midpoint of $B(4, 3)$ and $D(0, -3)$ is $(2, 0)$. ✓

9. (Challenge) Find the three points that divide the segment from $A(-6, 2)$ to $B(10, -6)$ into four equal parts.

Solution

The middle point is the midpoint of $AB$:

$$M = \left( \frac{-6 + 10}{2},\ \frac{2 + (-6)}{2} \right) = (2, -2)$$

Then split each half in half again:

$$\begin{aligned} \text{Midpoint of } AM &= \left( \frac{-6 + 2}{2},\ \frac{2 + (-2)}{2} \right) = (-2, 0) \\ \text{Midpoint of } MB &= \left( \frac{2 + 10}{2},\ \frac{-2 + (-6)}{2} \right) = (6, -4) \end{aligned}$$

The points are $(-2, 0)$, $(2, -2)$ and $(6, -4)$. Check: each step is $4$ right and $2$ down. ✓