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This page sets out the method for midpoint of a line segment in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Midpoint of a line segment topic page.
When you see 'find the midpoint of ' with two coordinates given.
When you are given the midpoint and one end point.
The midpoint of (2, 6) and (8, 2) is (5, 4).
The x-coordinate of the midpoint of (1, 8) and (7, 2) is 4.
Average the y-coordinates and leave x out of the answer.
The distance between (1, 2) and (4, 6) is 5, from a 3-4-5 triangle.
A midpoint is halfway, so divide the sum by 2.
Average x with x. Average y with y. Do not mix them.
Write the two little fractions before you calculate. For the ends (2, 6) and (8, 6), the midpoint is ((2+8)/2, (6+6)/2)=(5, 6). The y-coordinates were equal, so the midpoint stays on that horizontal line, which is what a sketch would show. Subtraction would have given the length of the run, 6, which is not a coordinate of the middle. Those two fractions are the method. A midpoint written with no working has nothing for a marker to credit if the average is mis-added.
Keep the x values in the x fraction. Averaging 2 with 6 because they are "the first numbers you see" mixes a coordinate from one point with a coordinate from the other in the wrong slot. Label the points A and B and copy x_A, x_B, y_A, y_B into the formula. The habit removes the mix-up.
Method: 1. Write the x average and the y average as two fractions before you simplify them. 2. Average (2, 6) and (8, 6) to get ((2+8)/2, (6+6)/2), which is (5, 6). 3. Equal y-coordinates keep the midpoint on that same horizontal line.
From (-3, 4) to (5, -2), the x average is (-3+5)/2=1. The y average is (4+(-2))/2=1. The midpoint is (1, 1). Adding a negative is subtracting its size. Treating -3+5 as -8, or 4+(-2) as 6, moves the midpoint out of the segment. A small sketch of the two ends shows immediately whether your middle point sits between them.
Halves are allowed. The midpoint of (1, 2) and (6, -3) has x=3.5 and y=-0.5. Rounding 3.5 to 4 shifts the point by half a unit, which is enough to lose the accuracy mark. Leave the decimal, or write it as a fraction 7/2.
Method: 1. From (-3, 4) to (5, -2), the x average is (-3+5)/2=1. 2. The y average is (4+(-2))/2=1. 3. Write the midpoint as the pair (1, 1).
If the midpoint M is (2, -1) and one end A is (-4, 3), the other end B(x, y) satisfies (-4+x)/2=2 and (3+y)/2=-1. Multiply both equations by 2. -4+x=4, so x=8. 3+y=-2, so y=-5. The other end is (8, -5).
This is not "halve the midpoint". Halving again finds a point halfway from the origin, which has nothing to do with B. You double the step from A to M. The vector from A to M is (2-(-4), -1-3)=(6, -4). The same step again lands on (2+6, -1+(-4))=(8, -5). Both methods must agree. If they do not, the sign of a negative coordinate is the usual cause.
The same average finds the centre of a rectangle from a pair of opposite corners. Opposite corners of a rectangle are the ends of a diagonal, and the diagonals share a midpoint. If one diagonal runs from (1, 2) to (7, 8), the centre is (4, 5). You do not need the other two corners to find it. A quick check is that the centre sits inside the rectangle on a sketch, between the corners in both directions. The other diagonal, once you have the remaining corners, must give that same centre again too.
Method: 1. If the midpoint is (2, -1) and one end is (-4, 3), set (-4+x)/2=2 and (3+y)/2=-1. 2. Multiply both of those equations by 2. 3. The y equation becomes 3+y=-2, so y=-5.
When the two coordinates in a pair add to an odd number, the midpoint has a half in it. The midpoint of (3, -4) and (6, 1) is ((3 + 6)/2, (-4 + 1)/2) = (4.5, -1.5). This is a normal answer. It means the midpoint sits halfway between grid lines, and rounding it to a whole number would lose the marks.
Take extra care with negatives. Add the two numbers first, using a number line if it helps, and only then halve the result. -4 + 1 = -3, and half of -3 is -1.5. A common slip is to treat -4 + 1 as -5 or 3. Write the sum on its own line. An exam question will say whether to give the answer as a decimal or as a fraction, and either is fine unless it asks for a particular form.
Method: 1. Write the x values and add them, then halve. 2. Write the y values and add them, then halve. 3. Keep decimals such as 4.5 and -1.5 exactly as they come. 4. Write the answer as a pair in brackets, x first.
Tip: Add first, then halve. Do not halve one number and add the other.
If a line AB is a diameter of a circle, the centre of the circle is the midpoint of AB. So a question that gives the end points of a diameter and asks for the centre is a midpoint question in disguise. The radius is half the length of the diameter. When the two points have the same y value, the length is just the difference between the x values.
For a sloping diameter, find the length of AB with Pythagoras: the horizontal distance and the vertical distance form a right-angled triangle. For A(-3, -2) and B(5, 4) the distances are 8 and 6, and √(64 + 36) = 10. The radius is 5. To test whether a point lies on the circle, find its distance from the centre. If that distance equals the radius, the point is on the circle.
Method: 1. Find the midpoint of the two ends of the diameter to get the centre. 2. Find the horizontal and vertical distances between the ends. 3. Use Pythagoras for the diameter and halve it for the radius. 4. To test a point, find its distance to the centre and compare with the radius.
The diagonals of a parallelogram cut each other exactly in half. That means they share the same midpoint. If you are given three corners of a parallelogram, find the midpoint of the diagonal that joins two of the known corners, then use it to find the missing corner. The same fact is true for a rectangle, a square and a rhombus.
Suppose A(1, 2), B(7, 3) and C(9, 8) are three corners of a parallelogram ABCD. The diagonal AC has midpoint (5, 5). The other diagonal is BD, with the same midpoint, so D = 2 × (5, 5) - (7, 3) = (3, 7). Check by looking at the vectors: the step from A to B must equal the step from D to C.
Method: 1. Identify the two diagonals from the order of the letters. 2. Find the midpoint of the diagonal whose end points you know. 3. Use that midpoint and the known end of the other diagonal to find the missing end. 4. Check that opposite sides have the same step in x and in y.
Tip: Diagonals of ABCD join A to C and B to D. Sides go A to B, B to C, C to D and D to A.
A harder question puts a letter inside a coordinate. If the midpoint of A(2k, 7) and B(6, 3k) has x-coordinate 5, then (2k + 6)/2 = 5. Solve this for k to get k = 2, then use k to find the other coordinate. Always treat the midpoint formula as an equation you can solve backwards.
Many questions link parts. After finding the midpoint you might be asked for the gradient of AB, and then for the equation of the perpendicular bisector. That line passes through the midpoint and has a gradient that is the negative reciprocal of the gradient of AB. Work through the parts in order and carry each answer forward, and write the final equation as y = mx + c.
Method: 1. Write the midpoint formula for each coordinate with the letter included. 2. Solve the equation for the letter, then substitute back. 3. For a perpendicular bisector, find the midpoint and the gradient of AB. 4. Use the negative reciprocal gradient through the midpoint and write y = mx + c.
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