Midpoints
The midpoint of an interval is halfway between its ends, found by averaging the coordinates: \(M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). This NSW Year 9 Advanced maths subtopic uses the midpoint formula (NESA outcome MA5‑LIN‑C‑01).
You will find midpoints from coordinates and diagrams, check whether a point is a midpoint, and find a missing end point or coordinate from a known midpoint. You will show that the diagonals of a parallelogram bisect each other and use this to find a fourth vertex, find a circle's radius from a diameter, and show that the line joining two midpoints is half the third side. It is part of Year 9 Advanced Coordinate Geometry and Graphs.
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 end points. Its coordinates are the averages of the end points' coordinates. This Year 9 Advanced maths page covers the midpoint formula, working backwards to a missing end point, circles, medians and the fourth vertex of a parallelogram.
The midpoint \(M\) of an interval \(AB\) is the point on \(AB\) that is the same distance from \(A\) as from \(B\). It cuts the interval into two equal halves; we say it bisects the interval.
To find it, take the average (mean) of the two \(x\)-coordinates and the average of the two \(y\)-coordinates. Halfway across and halfway up is halfway along.
For a horizontal or vertical interval only one coordinate changes. The midpoint of \((-3,6)\) and \((5,6)\) is \((1,6)\).
Working backwards, if you know \(M\) and one end, the other end is the same step again: \(B = 2M - A\) for each coordinate.
Midpoints solve geometry problems: the centre of a circle is the midpoint of a diameter, a median joins a vertex to the midpoint of the opposite side, and the diagonals of a parallelogram share a midpoint.
The midpoint of \((x_1,y_1)\) and \((x_2,y_2)\) is:
Missing end point \(B\), given \(A\) and \(M\):
How to find a midpoint
- Write the two end points, and decide which is \((x_1,y_1)\) and which is \((x_2,y_2)\). Either order works.
- Average the \(x\)-coordinates: add them and halve.
- Average the \(y\)-coordinates the same way.
- Check on a sketch: \(M\) must lie between the end points.
Problems built on midpoints
- Is this the midpoint? Work out the true midpoint and compare both coordinates. One coordinate wrong means no.
- Missing end point or missing coordinate: set each average equal to the known midpoint coordinate and solve (Example 2). Only the coordinate that is unknown needs an equation.
- Points on the axes: a point on the \(x\)-axis is \((a,0)\) and a point on the \(y\)-axis is \((0,b)\). Use the midpoint to find \(a\) and \(b\).
- Circle: the centre is the midpoint of a diameter; the radius is the distance from the centre to an end (Example 3).
- Median or other lengths: find the midpoint first, then use the length of an interval. Give exact answers in simplest surd form when asked.
- Equal parts: to cut \(AE\) into four equal parts, find the midpoint \(C\) of \(AE\), then the midpoints of \(AC\) and \(CE\).
- Parallelogram: use equal midpoints of the diagonals (Example 4). If the order of the vertices is not given, \(D\) could be opposite \(A\), \(B\) or \(C\), giving three possible points.
- Midpoints of two sides of a triangle: the interval joining them is parallel to the third side and half as long.
Average the \(x\)-coordinates, then average the \(y\)-coordinates. Add carefully with the negatives.
| \(x_M\) | \(=\) | \(\dfrac{-5+2}{2} = \dfrac{-3}{2} = -1.5\) |
| \(y_M\) | \(=\) | \(\dfrac{2+(-7)}{2} = \dfrac{-5}{2} = -2.5\) |
So \(\boldsymbol{M(-1.5,\,-2.5)}\). A midpoint need not be on a grid point. Fractions \(\left(-\dfrac{3}{2},-\dfrac{5}{2}\right)\) are equally correct.
Let \(B=(p,q)\). Each coordinate of \(M\) is an average, so write two equations.
| \(\dfrac{-4+p}{2}\) | \(=\) | \(2\) |
| \(-4+p\) | \(=\) | \(4\) |
| \(p\) | \(=\) | \(8\) |
| \(\dfrac{3+q}{2}\) | \(=\) | \(-1\) |
| \(3+q\) | \(=\) | \(-2\) |
| \(q\) | \(=\) | \(-5\) |
So \(\boldsymbol{B(8,-5)}\).
Quick check by steps: \(A\) to \(M\) is \(6\) right and \(4\) down. The same step again from \(M\) lands on \((8,-5)\).
The centre of a circle is the midpoint of any diameter. The radius is the distance from the centre to either end.
| \(x_C\) | \(=\) | \(\dfrac{-5+3}{2} = -1\) |
| \(y_C\) | \(=\) | \(\dfrac{-1+5}{2} = 2\) |
So \(\boldsymbol{C(-1,2)}\). From \(C\) to \(Q\): run \(4\), rise \(3\).
| \(r^2\) | \(=\) | \(4^2+3^2 = 25\) |
| \(r\) | \(=\) | \(\mathbf{5}\text{ units}\) |
Check: the diameter \(PQ\) has run \(8\) and rise \(6\), so it is \(10\) units long, twice the radius.
The diagonals \(AC\) and \(BD\) bisect each other, so they have the same midpoint.
| \(\text{mid } AC\) | \(=\) | \(\left(\dfrac{-3+4}{2},\ \dfrac{0+2}{2}\right) = (0.5,\ 1)\) |
Let \(D=(p,q)\). The midpoint of \(BD\) must also be \((0.5,1)\).
| \(\dfrac{1+p}{2}\) | \(=\) | \(0.5\) |
| \(p\) | \(=\) | \(0\) |
| \(\dfrac{-2+q}{2}\) | \(=\) | \(1\) |
| \(q\) | \(=\) | \(4\) |
So \(\boldsymbol{D(0,4)}\).
Check with steps: \(B\) to \(A\) is \(4\) left and \(2\) up, and \(C\) to \(D\) is also \(4\) left and \(2\) up.
Common pitfalls
Frequently asked questions
How do you find the midpoint of two points?
Add the two \(x\)-coordinates and divide by \(2\), then do the same with the \(y\)-coordinates. The two averages are the coordinates of the midpoint.
What is the midpoint formula?
The midpoint of \((x_1,y_1)\) and \((x_2,y_2)\) is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
How do you find an end point if you know the midpoint?
Double the midpoint coordinate and subtract the known end point: \(x_B = 2x_M - x_A\), and the same for \(y\). Or take the step from the known end to the midpoint and repeat it.
Can a midpoint have fractions or decimals?
Yes. If the two coordinates add to an odd number, the average is a half: the midpoint of \(1\) and \(4\) is \(2.5\).
How do you find the centre of a circle from a diameter?
The centre is the midpoint of the diameter. Average the coordinates of its two ends. The radius is then the distance from the centre to either end.