Name the three points.
Two ends and a middle. Only one of them is unknown.
| Point | x | y |
|---|---|---|
| Endpoint 1 | −2 | 1 |
| Midpoint | 4 | 3 |
| Endpoint 2 | x | y |
Keep x under x and y under y. The two columns never mix, so this is really two small problems.
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- The missing endpoint
The Missing Endpoint
When you are given the midpoint — the midpoint is the average of the ends.
The problem
A line segment has one endpoint (−2, 1) and a midpoint of (4, 3). Find the other endpoint (x, y).
A midpoint is an average, so reverse it: double the midpoint and subtract the known end. x = 2(4) − (−2) = 10, y = 2(3) − 1 = 5. B averages a second time instead of undoing the average.
Write the midpoint formula, then split it in two.
M = ( x1 + x2 2 , y1 + y2 2 )
A midpoint is just the average of the two ends.
The x part
4 = −2 + x2
The y part
3 = 1 + y2
Where the 4 and the 3 come from
The midpoint (4, 3) is not one number. Its x-value 4 belongs to the x equation and its y-value 3 belongs to the y equation.
Solve the x part.
· 2 both sides
4 = −2 + x2
the x equation
8 = −2 + x
the fraction is gone
10 = x
+ 2 both sides
Multiply by 2 first. Halving is what hid x, so undoing that is always the first move.
Solve the y part the same way.
· 2 both sides
3 = 1 + y2
the y equation
6 = 1 + y
the fraction is gone
5 = y
− 1 both sides
Same three moves, different letters. Nothing from the x part is reused here.
Put the two answers back into one point.
x and y were solved apart. The answer is a single pair.
x = 10
y = 5
The one-line shortcut
missing end = 2 · midpoint − known end
2(4) − (−2) = 10 · 2(3) − 1 = 5
The trap: averaging a second time
The formula looks like an average, so the hand wants to average the two points you were given: (−2 + 4) ÷ 2 = 1 and (1 + 3) ÷ 2 = 2, giving (1, 2). That is the midpoint of the known end and the midpoint — a point a quarter of the way along, not the far end. The midpoint is already the average; undo it by doubling, never average again.
When an endpoint sits on an axis.
M(−2, 4) is the midpoint of AB. A lies on the x-axis and B lies on the y-axis. Find A and B.
A = (a, 0), B = (0, b)
on the x-axis y = 0; on the y-axis x = 0
M = (a ÷ 2, b ÷ 2) = (−2, 4)
each zero drops out of its average
a = −4, b = 8
double each coordinate
A = (−4, 0), B = (0, 8)
check: (−4 + 0) ÷ 2 = −2, (0 + 8) ÷ 2 = 4 ✓
Do not swap the axes
"On the x-axis" fixes the y-coordinate at 0, not the x. The point on the x-axis takes its x from the midpoint's x (doubled); the point on the y-axis takes its y from the midpoint's y. Read which point the question asks for — the other one is always among the choices.
Variations: a fractional midpoint, unknowns in both points, extending past an end.
Decimals in the midpoint. M(−1.5, 2.5) is the midpoint of AB and A is (4, −3). Find B.
x = 2(−1.5) − 4 = −3 − 4 = −7
double the midpoint, take away the known end
y = 2(2.5) − (−3) = 5 + 3 = 8
minus a negative adds
B = (−7, 8)
check: (4 − 7) ÷ 2 = −1.5, (−3 + 8) ÷ 2 = 2.5 ✓
What is different: the halves in the midpoint vanish when we double, so the far end usually comes out whole. Do not round the midpoint first.
An unknown in each point. The midpoint of (k, 3) and (5, m) is (1, −2). Find k and m.
(k + 5) ÷ 2 = 1 → k + 5 = 2
the x equation, × 2
k = −3
− 5 both sides
(3 + m) ÷ 2 = −2 → 3 + m = −4
the y equation, × 2
m = −7
− 3 both sides
What is different: the unknowns sit in different points, but each still lives in its own one-line equation — x with x, y with y.
Extending a segment. A is (−3, 2) and B is (1, 4). AB is extended past B to C so that BC = AB. Find C.
B is the midpoint of AC
the given point is now the middle
C = 2B − A = (2 − (−3), 8 − 2)
2 · midpoint − known end
C = (5, 6)
check: (−3 + 5) ÷ 2 = 1, (2 + 6) ÷ 2 = 4 ✓
What is different: no midpoint is named. Decide which point is in the middle before using the rule.
Find the middle point first
Averaging A and B gives (−1, 3), the midpoint of AB — a point between them, not beyond B. Here B is the middle, so B is the one we double.
Hidden midpoints: diameters, diagonals, quarter points.
Many questions never say "midpoint". They describe a shape whose centre is a midpoint, and then the same doubling rule finds the missing point.
| The question says | The midpoint hiding in it |
|---|---|
| centre of a circle, one end of a diameter | the centre is the midpoint of the diameter |
| three vertices of a parallelogram | the diagonals share one midpoint |
| points split a segment into four equal parts | the middle one is the midpoint; the others are midpoints of the halves |
Worked: PQRS is a parallelogram with P(2, 1), Q(6, 2) and R(7, 6). Find S.
diagonals: PR and QS
opposite corners — P with R, Q with S
midpoint of PR = (9 ÷ 2, 7 ÷ 2) = (4.5, 3.5)
it is also the midpoint of QS
S = 2(4.5, 3.5) − (6, 2) = (9 − 6, 7 − 2)
2 · midpoint − known end
S = (3, 5)
check: P→Q is 4 right, 1 up; S→R is 4 right, 1 up ✓
Same move for a circle: centre (−1, 2) and one end of a diameter (3, −1) give the other end 2(−1, 2) − (3, −1) = (−5, 5).
Find the end, then a length. M(2, 3) is the midpoint of AB and A is (−1, −1). Find the length of AB and the distance of B from the origin.
B = (2(2) − (−1), 2(3) − (−1)) = (5, 7)
2 · midpoint − known end
AM = √(2 − (−1))² + (3 − (−1))² = √9 + 16 = 5
distance = √across² + up²
AB = 2 × AM = 10
M is halfway, so the whole is double the half
OB = √5² + 7² = √74
from (0, 0), the legs are just B's coordinates
Pair the opposite corners
In PQRS the diagonals join P to R and Q to S. Averaging P and Q (neighbours) builds a different shape. Follow the letters round the figure before you pair.
On the exam
Expect the midpoint to be disguised — an axis point, a circle's centre, a parallelogram's diagonals — or to be one step of a longer question (find B, then its distance from something). The one line 2 · midpoint − known end does all of them; check by averaging your answer with the known end.
Check: is (4, 3) really halfway?
midpoint = ( −2 + 102 , 1 + 52 ) = (4, 3) ✓
Now you try
Practice problem
Find endpoint A
The midpoint of segment AB is (1, 0). Endpoint B is (5, −4). Find endpoint A.
x = 2(1) − 5 = −3 and y = 2(0) − (−4) = 4. D gets the y sign wrong — subtracting −4 adds 4. Check: the midpoint of (−3, 4) and (5, −4) really is (1, 0). ✓
Hint: here the missing end is the first point, and one midpoint value is 0. The formula does not change — the signs do.