Midpoint and Length of a Line Segment
Theory
The midpoint of an interval is the point halfway between its ends: average the \(x\)-coordinates and average the \(y\)-coordinates. The length of an interval comes from Pythagoras' theorem, with the run and the rise as the two shorter sides. This Year 9 maths page covers both formulas, negative coordinates, halves, surds and rounding, finding a missing end point, points a quarter of the way along, and lengths in shapes and on scaled maps.
An interval (or line segment) is the part of a line between two end points.
The midpoint \(M\) is the point halfway between the two ends. It 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. Midpoints often land on a half, such as \(\left(\dfrac{5}{2}, 3\right)\); that is a correct answer, so do not round it.
The length of an interval is the straight-line distance between its ends. Go across from one point and up (or down) to the other. The distance across is the run, the difference of the \(x\)-coordinates; the distance up is the rise, the difference of the \(y\)-coordinates.
The run and the rise are the two shorter sides of a right-angled triangle, and the interval is the hypotenuse, so Pythagoras' theorem gives its length. A horizontal interval's length is just the run; a vertical interval's length is just the rise.
A length is never negative, and it does not matter which point you start from.
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 \(B\):
For the points \((x_1, y_1)\) and \((x_2, y_2)\):
How to find a midpoint
- 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.
How to find the length of an interval
- Find the run (\(x_2 - x_1\)) and the rise (\(y_2 - y_1\)). Take care when subtracting a negative.
- Square the run and the rise separately, then add them to get \(d^2\).
- Take the square root to get \(d\). Leave it as a surd, or round at the end if the question asks.
- Check on a sketch: \(d\) must be longer than both the run and the rise, and the midpoint must sit halfway along.
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 \(\mathbf{(6, 6)}\).
| \(\text{run}\) | \(=\) | \(1 - (-4) = 5\) |
| \(\text{rise}\) | \(=\) | \(-3 - 2 = -5\) |
| \(d^2\) | \(=\) | \(5^2 + (-5)^2\) |
| \(\) | \(=\) | \(25 + 25 = 50\) |
| \(d\) | \(=\) | \(\sqrt{50} = 7.071\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.1} \text{ units}\) |
The negative rise does no harm, because squaring makes it positive. Leave \(\sqrt{50}\) as it is if the question does not ask for decimals.
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 \(2\).
Work out the square of each side. There is no need to take square roots to compare them.
| \(AB^2\) | \(=\) | \(6^2 + 2^2 = 40\) |
| \(AC^2\) | \(=\) | \(2^2 + 6^2 = 40\) |
| \(BC^2\) | \(=\) | \((2 - 6)^2 + (6 - 2)^2\) |
| \(\) | \(=\) | \(16 + 16 = 32\) |
\(AB^2 = AC^2\), so \(AB = AC\) and the triangle is isosceles. Since \(32 + 40 \ne 40\), it is not right-angled.
Common pitfalls
Frequently asked questions
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.
What is the distance formula?
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). It is Pythagoras' theorem with the run and the rise as the shorter sides.
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.
Can a distance be negative?
No. A distance is a length, so it is always positive or zero. A negative run or rise becomes positive when it is squared.