01 · Midpoint
Find the midpoint of (−7, 2) and (5, 8).
Hint
Average x and y separately.
Worked solution
M = ((−7 + 5)/2, (2 + 8)/2)
M = (−1, 5)
Understand · explore · practise
Midpoints and perpendicular bisectors
Find a midpoint, recover a missing endpoint and construct perpendicular bisectors. Understand why every point on a bisector is equally far from the two endpoints.
Before you startCoordinates, distances and perpendicular line equations
01 / Midpoints
Average the coordinates separately.
The midpoint is halfway along the segment joining two endpoints. Average the two x-coordinates and, separately, the two y-coordinates.
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
For A(−3, −1) and B(5, 3), M = (1, 1). The change from A to M is (4, 2), exactly half the change from A to B.
Do not average an x-coordinate with a y-coordinate. Negative and fractional values work in exactly the same way.
Keep exact coordinatesWorked example
A = (−5, 4), B = (2, −1)
Average each pair of like coordinates.
M = ((−5 + 2)/2, (4 − 1)/2)
M = (−3/2, 3/2)
The midpoint need not have integer coordinates.
A = (√3 − 2, 1), B = (√3 + 4, 7)
M = (√3 + 1, 4)
Combine exact surds before dividing by 2.
02 / Missing endpoints
Double the midpoint, then subtract the known endpoint.
If M is the midpoint of AB, then xB = 2xM − xA and yB = 2yM − yA. This is a reflection of A in M.
A = (−2, 5), M = (3, 1)
B = (2 × 3 − (−2), 2 × 1 − 5)
B = (8, −3)
A circle’s centre is the midpoint of a diameter. You can use the same calculation to find the opposite endpoint. A chord that is not a diameter need not have the centre as its midpoint.
Coordinates containing letters
A = (2a, −b), M = (−a, 3b)
B = (−4a, 7b)
Check: the averages are (−a, 3b), as required.
03 / Bisectors
Use both the midpoint and the perpendicular direction.
The perpendicular bisector passes through a segment’s midpoint and meets the segment at 90°. These are two separate conditions.
Find the midpoint, calculate the segment’s gradient, take the negative reciprocal, then use point-gradient form. Treat horizontal and vertical segments directly.
Construct the whole lineWorked example
A = (−3, −1), B = (5, 3)
M = (1, 1)
First find the midpoint.
mAB = 4/8 = 1/2
m⊥ = −2
Now find the perpendicular direction.
y − 1 = −2(x − 1)
y = −2x + 3
This line passes through M and is perpendicular to AB.
04 / Equal distances
The bisector is the set of points equally far from A and B.
For distinct points A and B, a point P lies on their perpendicular bisector exactly when PA = PB. This gives an alternative algebraic method: equate the squared distances.
(x + 3)² + (y + 1)²
= (x − 5)² + (y − 3)²
16x + 8y − 24 = 0
y = −2x + 3
The x² and y² terms cancel, leaving a line. Move P in the model: the two distances change together. Squared distances suffice because distances are non-negative.
For any chord of a circle, its centre is equally far from the endpoints. It must therefore lie on the chord’s perpendicular bisector.
The same distance from A and BEqual axis scales
A(−3,−1), B(5,3) have midpoint M(1,1). P(0,3) lies on their perpendicular bisector; PA² = PB² = 25.
Watch equal distances along a bisector
Pause, replay or seek freely. The notes explain the same idea and stay in view.
05 / Special cases
A horizontal segment has a vertical bisector.
For A(−3, 1), B(5, 1), the midpoint is (1, 1) and the bisector is x = 1. For A(1, −3), B(1, 5), the midpoint is again (1, 1), but the bisector is y = 1.
If A and B coincide, every point is equally far from them. There is no unique segment direction and no unique perpendicular bisector.
Why a triangle’s midpoint line is parallel to its third side
Take A(0,0), B(2u,2v), D(2s,2t). The midpoints of AB and BD are M(u,v) and N(u+s,v+t). The displacement from M to N is (s,t), half the displacement (2s,2t) from A to D. Thus MN has the same direction and half the length of AD. This argument also handles vertical sides.
06 / Your turn
Check both the midpoint and the direction.
Keep coordinates exact and give line equations explicitly.
02 · Opposite end of a diameter
A circle has centre (−2, 3). One diameter endpoint is (4, −1). Find the other.
Hint
Double the centre and subtract the endpoint.
Worked solution
B = (−4 − 4, 6 − (−1))
B = (−8, 7)
03 · A sloping segment
Find the perpendicular bisector of A(−1, 2), B(5, 6).
Hint
M = (2,4), and mAB = 2/3.
Worked solution
y − 4 = −3(x − 2)/2
3x + 2y − 14 = 0
04 · Vertical endpoints
Find the perpendicular bisector of (−2, −5) and (−2, 7).
Hint
The midpoint is (−2,1).
Worked solution
The segment is vertical. Its bisector is the horizontal line y = 1.
05 · Unknown coordinates
A(1,p) and B(q,−2) are diameter endpoints of a circle with centre (4,3). Find p and q.
Hint
(1 + q)/2 = 4 and (p − 2)/2 = 3.
Worked solution
q = 7, p = 8
06 · Use the given bisector
A(−2,1), B(4,k) have perpendicular bisector y = −2x + c. Find k and c.
Hint
AB must have gradient 1/2. Then calculate its midpoint.
Worked solution
(k − 1)/6 = 1/2 ⇒ k = 4
M = (1, 5/2)
5/2 = −2 + c ⇒ c = 9/2
07 / Recap
A bisector combines a position and a direction.
- Average like coordinates for the midpoint.
- Use B = 2M − A to recover an endpoint.
- A perpendicular bisector passes through the midpoint at 90°.
- Every point on it is equally far from the two endpoints.
- Handle horizontal, vertical and coincident endpoints explicitly.
Section 1 of 7 · Midpoints