Midpoint of an Interval
Midpoint of an Interval in NSW Year 10 Core maths finds the point halfway between two points (NESA outcome MA5‑LIN‑C‑01). The midpoint's coordinates are the averages of the endpoints' coordinates, so the midpoint of \((-4, 5)\) and \((8, 1)\) is \((2, 3)\).
You will find midpoints from coordinates and from a diagram, including ones with fractions and surds, and work backwards from a midpoint and one endpoint to the other endpoint. You will also find the centre of a circle from a diameter, place a water tank halfway between two farms, use the fact that a parallelogram's diagonals bisect each other, and combine midpoints with lengths. It is part of Year 10 Core Straight Line Graphs.
Every question with a fully worked solution.
- Midpoint of an Interval - Video - How To Find The Midpoint Using The Midpoint Formula Watch
Theory
The midpoint of an interval is the point exactly halfway between its endpoints. This Year 10 maths page shows how to find it by averaging the coordinates, how to work backwards to a missing endpoint or an unknown, and how midpoints locate the centre of a circle and points on maps.
The midpoint \(M\) of an interval \(AB\) is the point on \(AB\) that is the same distance from \(A\) as from \(B\).
Its \(x\)-coordinate is the average (mean) of the two \(x\)-values, and its \(y\)-coordinate is the average of the two \(y\)-values. Add, then halve.
Because \(M\) is halfway, the step from \(A\) to \(M\) (across and up or down) is the same as the step from \(M\) to \(B\). This gives a quick way to find a missing endpoint.
For a horizontal or vertical interval, only one coordinate changes, so only that one needs averaging.
The midpoint of the interval joining \((x_1, y_1)\) and \((x_2, y_2)\):
Working backwards, if \(M(m_x, m_y)\) and one endpoint \((x_1, y_1)\) are known:
How to find a midpoint
- Add the two \(x\)-values and divide by \(2\).
- Add the two \(y\)-values and divide by \(2\).
- Write the answer as a point \((x, y)\). Fractions or decimals are fine, such as \(\left(\dfrac{5}{2}, -1\right)\).
- Check on a sketch that \(M\) lies between the endpoints.
Other questions you will meet
- Reading from a grid: read both endpoints carefully, then average.
- Missing endpoint: take the same step again from \(M\), or solve \(\dfrac{x_1 + a}{2} = m_x\) (Example 2).
- Unknowns in the coordinates: make one equation for \(x\) and one for \(y\). If the same unknown appears in both, solve one and check the other.
- Points on the axes: a point on the \(x\)-axis is \((b, 0)\); a point on the \(y\)-axis is \((0, a)\).
- Circles: the centre is the midpoint of a diameter; if the centre and one end are known, find the other end with the step method.
- Parallelograms: the diagonals bisect each other, so they share a midpoint. Use the midpoint of one diagonal to find a missing vertex.
- Triangles: join a vertex to the midpoint of the opposite side, or join two midpoints, then use the length of an interval.
- Maps and tracks: \(\dfrac{1}{4}\) of the way along is the midpoint of the start and the midpoint. Change grid units to real distances with the scale.
- Surds: add like surds before halving, for example \(\sqrt{3} + \sqrt{12} = 3\sqrt{3}\).
Add the \(x\)-values and halve; add the \(y\)-values and halve. Adding a negative means subtracting.
| \(x\) | \(=\) | \(\dfrac{-5 + 2}{2} = \dfrac{-3}{2} = -1.5\) |
| \(y\) | \(=\) | \(\dfrac{3 + (-6)}{2} = \dfrac{-3}{2} = -1.5\) |
The midpoint is \(\left(\mathbf{-\dfrac{3}{2}}, \mathbf{-\dfrac{3}{2}}\right)\), or \((\mathbf{-1.5}, \mathbf{-1.5})\).
Check on the sketch: \(M\) sits on \(AB\), halfway along.
Method 1 (steps). From \(A\) to \(M\): across \(2 - (-3) = 5\), and up \(-1 - 4 = -5\) (that is, down \(5\)). Take the same step again from \(M\).
| \(x\) | \(=\) | \(2 + 5 = 7\) |
| \(y\) | \(=\) | \(-1 - 5 = -6\) |
Method 2 (equations). Let \(B = (a, b)\) and use the midpoint formula.
| \(\dfrac{-3 + a}{2}\) | \(=\) | \(2 \ \Rightarrow\ -3 + a = 4 \ \Rightarrow\ a = 7\) |
| \(\dfrac{4 + b}{2}\) | \(=\) | \(-1 \ \Rightarrow\ 4 + b = -2 \ \Rightarrow\ b = -6\) |
\(B\) is \((\mathbf{7}, \mathbf{-6})\). Check: \(\dfrac{-3 + 7}{2} = 2\) and \(\dfrac{4 + (-6)}{2} = -1\).
Set the average of the \(x\)-values equal to \(4\), and the average of the \(y\)-values equal to \(3\). Solve each equation.
| \(\dfrac{a + 9}{2}\) | \(=\) | \(4\) |
| \(a + 9\) | \(=\) | \(8\) |
| \(a\) | \(=\) | \(\mathbf{-1}\) |
| \(\dfrac{-2 + b}{2}\) | \(=\) | \(3\) |
| \(-2 + b\) | \(=\) | \(6\) |
| \(b\) | \(=\) | \(\mathbf{8}\) |
Multiply both sides by \(2\) first, then solve. Check: \(A(-1, -2)\) and \(B(9, 8)\) have midpoint \(\left(\dfrac{8}{2}, \dfrac{6}{2}\right) = (4, 3)\).
(i) the centre \(C\) of the circle
(ii) the radius, exactly and correct to two decimal places.
(i) The centre of a circle is the midpoint of any diameter.
| \(C\) | \(=\) | \(\left(\dfrac{-4 + 6}{2}, \dfrac{-3 + 1}{2}\right) = (\mathbf{1}, \mathbf{-1})\) |
(ii) The radius is the distance from the centre to an end of the diameter. Use the length of an interval.
| \(CQ\) | \(=\) | \(\sqrt{(6 - 1)^{2} + (1 - (-1))^{2}}\) |
| \(\) | \(=\) | \(\sqrt{25 + 4} = \mathbf{\sqrt{29}}\) |
| \(\) | \(=\) | \(5.385\ldots \approx \mathbf{5.39} \text{ units}\) |
Check: \(CP = \sqrt{(1 - (-4))^{2} + (-1 - (-3))^{2}} = \sqrt{29}\) too, as it must be.
Common pitfalls
Frequently asked questions
What is the midpoint formula?
The midpoint of \((x_1, y_1)\) and \((x_2, y_2)\) is \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\). Average the \(x\)-values and average the \(y\)-values.
How do you find an endpoint if you know the midpoint?
Work out the step from the known endpoint to the midpoint, then take the same step again from the midpoint. Or use \(x_2 = 2m_x - x_1\) and \(y_2 = 2m_y - y_1\).
Can a midpoint have fractions?
Yes. If the two \(x\)-values add to an odd number, the midpoint's \(x\)-coordinate is a half, such as \(\dfrac{5}{2}\) or \(2.5\). That is still the exact midpoint.
How do you find the centre of a circle from a diameter?
The centre is the midpoint of the diameter. Average the \(x\)-values and the \(y\)-values of the two ends of the diameter.
What is the difference between midpoint and distance?
The midpoint is a point, found by adding and halving the coordinates. The distance is a length, found with Pythagoras' theorem using the differences of the coordinates.