Midpoints
The midpoint of an interval is halfway between its ends, found by averaging the two \(x\)-coordinates and the two \(y\)-coordinates. This NSW Year 9 Core maths subtopic is Stage 5 Core content under NESA outcome MA5‑LIN‑C‑01.
You will find midpoints of horizontal, vertical and sloping intervals, including ones with negative coordinates and answers that land on a half, such as \(\left(-\frac{3}{2}, \frac{3}{2}\right)\). You will check whether a given point really is the midpoint, find midpoints of the sides and diagonals of triangles, squares and parallelograms, and work backwards from a midpoint and one end to the other end. You will also divide an interval into four equal parts, place a rest stop halfway between two towns and find the centre of a circle from the ends of a diameter. It is part of Year 9 Core Straight Line.
Every question with a fully worked solution.
- Midpoints - Video - Midpoint Formula Watch
Theory
The midpoint of an interval is the point exactly halfway between its two ends. Find it by taking the average of the \(x\)-coordinates and the average of the \(y\)-coordinates. This Year 9 maths page shows the midpoint formula, how to work backwards to a missing end point, and how midpoints are used in shapes and circles.
The midpoint \(M\) of an interval is the point halfway between its two end points. It cuts the interval into two equal parts.
The midpoint is an average. Its \(x\)-coordinate is the mean of the two \(x\)-coordinates, and its \(y\)-coordinate is the mean of the two \(y\)-coordinates: add the two numbers and halve.
For a horizontal or vertical interval, one coordinate stays the same, so only the other one needs averaging.
Midpoints often land on a half, such as \(\left(\dfrac{5}{2}, 3\right)\). That is a correct answer; do not round it.
The centre of a circle is the midpoint of any diameter.
The midpoint of the interval joining \((x_1, y_1)\) and \((x_2, y_2)\) is:
Working backwards, from one end \(A\) and the midpoint \(M\) to the other end \(B\):
How to find a midpoint
- Write down the two points. It does not matter which is first.
- Add the two \(x\)-coordinates and halve. Then add the two \(y\)-coordinates and halve.
- Keep any halves as fractions or decimals.
- Going backwards? Double the midpoint, then subtract the known end.
- Check on a sketch: the midpoint should sit halfway along the interval.
Average the \(x\)-coordinates, then average the \(y\)-coordinates.
| \(x_M\) | \(=\) | \(\dfrac{3 + 9}{2} = \dfrac{12}{2} = 6\) |
| \(y_M\) | \(=\) | \(\dfrac{2 + 10}{2} = \dfrac{12}{2} = 6\) |
The midpoint is \(\mathbf{(6, 6)}\).
| \(x_M\) | \(=\) | \(\dfrac{-5 + 2}{2} = \dfrac{-3}{2}\) |
| \(y_M\) | \(=\) | \(\dfrac{4 + (-1)}{2} = \dfrac{3}{2}\) |
Neither coordinate is whole, and that is fine. The midpoint is \(\mathbf{\left(-\dfrac{3}{2}, \dfrac{3}{2}\right)}\), or \((-1.5, 1.5)\).
Double the midpoint, then subtract the end you already know.
| \(x_B\) | \(=\) | \(2 \times 1 - 4 = -2\) |
| \(y_B\) | \(=\) | \(2 \times 2 - (-3) = 7\) |
So \(B = \mathbf{(-2, 7)}\). Check: the average of \(4\) and \(-2\) is \(1\), and the average of \(-3\) and \(7\) is \(2\).
Find the midpoint \(M\) of \(AB\), then the midpoint of \(A\) and \(M\). Half of a half is a quarter.
| \(M\) | \(=\) | \(\left(\dfrac{-1 + 7}{2}, \dfrac{2 + 10}{2}\right)\) |
| \(\) | \(=\) | \((3, 6)\) |
| \(\text{quarter point}\) | \(=\) | \(\left(\dfrac{-1 + 3}{2}, \dfrac{2 + 6}{2}\right)\) |
| \(\) | \(=\) | \(\mathbf{(1, 4)}\) |
The midpoint of \(M\) and \(B\) would be the point three quarters of the way along.
Common pitfalls
Frequently asked questions
How do you find the midpoint of two points?
Add the two \(x\)-coordinates and divide by \(2\), then add the two \(y\)-coordinates and divide by \(2\). The midpoint of \((3, 2)\) and \((9, 10)\) is \((6, 6)\).
What is the midpoint formula?
\(M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). It is the average of the two \(x\)-coordinates and the average of the two \(y\)-coordinates.
How do you find an endpoint when you know the midpoint?
Double each coordinate of the midpoint, then subtract the matching coordinate of the end you know. With \(A(4, -3)\) and midpoint \((1, 2)\), the other end is \((2 - 4, 4 + 3) = (-2, 7)\).
Can a midpoint have fractions?
Yes. If two coordinates add to an odd number, the midpoint lands halfway between grid lines, such as \(\dfrac{3}{2}\). Leave it as a fraction or a decimal.
How do you find the centre of a circle from a diameter?
The centre is the midpoint of the diameter. Average the \(x\)-coordinates and the \(y\)-coordinates of its two ends.