Maths · 4 Coordinate geometry

4.3 Length and midpoint

Before you start

Plotting and reading Cartesian coordinates, including negative ones (4.1).
Pythagoras' theorem: in a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides (7.1).
Subtracting a negative number, e.g. $2 - (- 3) = 5$ .
The squares of whole numbers from 1 to 15, and their square roots.
Simplifying surds, e.g. $\sqrt{40} = 2 \sqrt{10}$ (1.17).

Syllabus objectives covered (2)

Calculate the length of a line segment.
Find the coordinates of the midpoint of a line segment.

Part 1 of 6

A length on a grid is a hypotenuse

A line segment is the straight piece of line between two points, called its endpoints. Its length is how far apart those two points are, measured along that straight line.
You already have the tool for this: Pythagoras' theorem. Every slanted segment on a grid is the hypotenuse of a right-angled triangle whose other two sides run along the grid lines.
Take $A (1, 2)$ and $B (7, 10)$ . Go across from $A$ until you are directly below $B$ , which is the point $(7, 2)$ . Then go straight up to $B$ .
graph diagram — not drawn yet
Coordinate axes from 0 to 11 on both axes with grid lines; points A(1, 2) and B(7, 10) joined by a solid segment; a dashed horizontal line from A to (7, 2) labelled 6, and a dashed vertical line from (7, 2) up to B labelled 8, with a right-angle mark at (7, 2).
The segment AB is the hypotenuse. The two dashed sides are the change in x and the change in y.
The horizontal side is the change in $x$ : $7 - 1 = 6$ . The vertical side is the change in $y$ : $10 - 2 = 8$ .
These two sides meet at a right angle, because one is horizontal and the other is vertical. So Pythagoras applies to them directly:
$A B^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$
So $A B = \sqrt{100} = 10$ . Nothing here is new: it is Pythagoras, with the two shorter sides read from the coordinates instead of from a diagram.

Part 2 of 6

The length formula, and why the signs do not matter

Write that method once with letters and you get the formula. For endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :
$length = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
This formula is not on the List of formulas printed on page 2 of Papers 2 and 4, so you have to carry it yourself. If you ever doubt it, sketch the triangle and use Pythagoras: that is all the formula is.
A difference can be negative. From $A (- 3, 4)$ to $B (2, - 8)$ the change in $y$ is $- 8 - 4 = - 12$ . That only says the segment goes downwards. A length is never negative.
Squaring removes the sign: ${(- 12)}^{2} = 144$ , the same as $12^{2}$ . This is also why it does not matter which point you call $(x_{1}, y_{1})$ . Swapping the points only changes the sign of each difference, and the squares stay the same.
The trap is subtracting a negative coordinate. The change in $x$ from $- 3$ to $2$ is $2 - (- 3) = 2 + 3 = 5$ , not $- 1$ . Write brackets round every negative number when you substitute it.
So for $A (- 3, 4)$ and $B (2, - 8)$ :
$A B = \sqrt{5^{2} + {(- 12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$
A quick sense check: the hypotenuse is the longest side of a right-angled triangle, so the length must be bigger than each difference on its own. $13$ is bigger than $5$ and bigger than $12$ , so it is believable.

Part 3 of 6

Exact answers and rounded answers

A length need not be a whole number. Take $P (1, - 2)$ and $Q (3, 4)$ . The differences are $2$ and $6$ , so $P Q = \sqrt{4 + 36} = \sqrt{40}$ .
On the non-calculator paper (Paper 2), leave the answer exact, as a surd. $\sqrt{40}$ is already a correct exact answer, but if the question asks for the simplest surd form you must simplify it, and it is good practice anyway. Find the largest square number that divides the number under the root: $40 = 4 \times 10$ .
$\sqrt{40} = \sqrt{4} \times \sqrt{10} = 2 \sqrt{10}$
On a calculator paper, give a non-exact answer to 3 significant figures unless the question asks for something else: $\sqrt{40} = 6.324 \dots$ , so $P Q = 6.32$ .
If a length is used again in a later part, carry the exact value or the full calculator value forward, not the rounded $6.32$ . Rounding early can cost the accuracy mark in the later part.
Never take the root of each square separately. $\sqrt{2^{2} + 6^{2}}$ is not $2 + 6 = 8$ . Add the squares first, then take one square root of the total.

Part 4 of 6

The midpoint is an average

The midpoint of a segment is the point exactly halfway between its endpoints. It is a point, so the answer is a pair of coordinates, not a single number.
Halfway along the segment means halfway across and halfway up at the same time. For $A (1, 2)$ and $B (7, 10)$ : halfway between $x = 1$ and $x = 7$ is $4$ , and halfway between $y = 2$ and $y = 10$ is $6$ . The midpoint is $(4, 6)$ .
graph diagram — not drawn yet
Coordinate axes from 0 to 11 on both axes with grid lines; a segment from A(1, 2) to B(7, 10) with its midpoint M(4, 6) marked as a dot and labelled; dashed lines show a step of 3 across and 4 up from A to M, and an identical step of 3 across and 4 up from M to B.
M is halfway across and halfway up at once, so the step from A to M is the same as the step from M to B.
Halfway between two numbers is their mean: add them and divide by $2$ . Do that separately for the $x$ -coordinates and for the $y$ -coordinates:
$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
This formula is not on the List of formulas either. It needs nothing more than: average the $x$ s, average the $y$ s.
Negative coordinates work the same way. For $A (- 3, 4)$ and $B (2, - 8)$ :
$M = (\frac{- 3 + 2}{2}, \frac{4 + (- 8)}{2}) = (- \frac{1}{2}, - 2)$
Notice that the midpoint uses a plus sign and the length uses a minus sign. The trap is mixing them up. Half the difference, $\frac{2 - (- 3)}{2} = 2.5$ , tells you how far across the midpoint is from each end. It does not tell you where the midpoint is.
A fractional coordinate is fine. Write $- \frac{1}{2}$ or $- 0.5$ , but do not mix fractions and decimals inside one answer.
Check that the midpoint lies between the endpoints in both directions. $- \frac{1}{2}$ is between $- 3$ and $2$ , and $- 2$ is between $- 8$ and $4$ .

Part 5 of 6

Working backwards from a midpoint

Sometimes you are given one endpoint and the midpoint, and you have to find the other endpoint.
Think of it as a journey. Going from $A$ to $M$ is one step. Because $M$ is halfway, the same step again from $M$ takes you to $B$ .
Take $A (2, - 1)$ and $M (5, 3)$ . From $A$ to $M$ , $x$ goes up by $3$ and $y$ goes up by $4$ . Do the same again from $M$ : $B = (5 + 3, 3 + 4) = (8, 7)$ .
The same thing in one line: each coordinate of $B$ is twice the midpoint's coordinate minus $A$ 's coordinate.
$B = (2 \times 5 - 2, 2 \times 3 - (- 1)) = (8, 7)$
Always check by finding the midpoint of $A$ and your answer: $(\frac{2 + 8}{2}, \frac{- 1 + 7}{2}) = (5, 3)$ . It matches $M$ , so $B$ is correct.
The trap is averaging $A$ and $M$ . That gives the point a quarter of the way along $A B$ , not the far end.

Part 6 of 6

When a coordinate is unknown

A question can give you the midpoint or the length and leave one coordinate as a letter. Put the letter into the method exactly as if it were a number, and you get an equation to solve.
Midpoint first, because it gives a simple linear equation. The midpoint of $(a, 3)$ and $(6, b)$ is $(2, - 1)$ . Each coordinate gives its own equation:
$\frac{a + 6}{2} = 2 \Rightarrow a = - 2 \frac{3 + b}{2} = - 1 \Rightarrow b = - 5$
A length gives a squared term, so expect up to two answers. Suppose the distance from $(2, k)$ to $(5, 1)$ is $5$ . Square the length so that there is no square root to deal with:
${(5 - 2)}^{2} + {(1 - k)}^{2} = 5^{2}$
So $9 + {(1 - k)}^{2} = 25$ , which gives ${(1 - k)}^{2} = 16$ . Both $4$ and $- 4$ square to $16$ , so $1 - k = 4$ or $1 - k = - 4$ .
That gives $k = - 3$ or $k = 5$ . Both are correct: $(2, - 3)$ is below the line $y = 1$ and $(2, 5)$ is above it, and each is $5$ units from $(5, 1)$ . Give both values unless the question tells you which one it wants.

Worked examples

Example 1. Calculate the length of the line segment joining $C (- 5, 1)$ and $D (3, - 14)$ . Do not use a calculator.

Show the working

Find the change in $x$ , with brackets round the negative coordinate: $3 - (- 5) = 3 + 5 = 8$ .
Find the change in $y$ : $- 14 - 1 = - 15$ . The minus sign only says the segment slopes downwards. It disappears when you square.
Square both differences, add them, then take one square root of the total.
$C D = \sqrt{8^{2} + {(- 15)}^{2}} = \sqrt{64 + 225} = \sqrt{289}$
$17 \times 17 = 289$ , so $C D = 17$ . Sense check: $17$ is longer than both $8$ and $15$ , as the hypotenuse must be.

Answer: $C D = 17$ units.

Example 2. $P$ is the point $(- 1, 3)$ and $Q$ is the point $(5, - 9)$ . (a) Find the exact length of $P Q$ , in its simplest surd form. (b) Give the length of $P Q$ correct to 3 significant figures.

Show the working

The change in $x$ is $5 - (- 1) = 6$ . The change in $y$ is $- 9 - 3 = - 12$ .
Square, add, take one root.
$P Q = \sqrt{6^{2} + {(- 12)}^{2}} = \sqrt{36 + 144} = \sqrt{180}$
Simplify. The largest square number that divides $180$ is $36$ , because $180 = 36 \times 5$ .
$\sqrt{180} = \sqrt{36} \times \sqrt{5} = 6 \sqrt{5}$
If you had spotted only $180 = 9 \times 20$ , you would get $3 \sqrt{20}$ . That is not the simplest form, because $20$ still contains the square number $4$ . Keep going until no square factor is left: $3 \sqrt{20} = 3 \times 2 \sqrt{5} = 6 \sqrt{5}$ .
For (b), use the calculator on the exact value: $6 \sqrt{5} = 13.416 \dots$ , which is $13.4$ to 3 significant figures.

Answer: (a) $P Q = 6 \sqrt{5}$ (b) $P Q = 13.4$ (3 s.f.)

Example 3. Find the coordinates of the midpoint of the line segment joining $R (- 7, 2)$ and $S (3, - 5)$ .

Show the working

Average the $x$ -coordinates. Add, then halve:
$\frac{- 7 + 3}{2} = \frac{- 4}{2} = - 2$
Average the $y$ -coordinates in the same way:
$\frac{2 + (- 5)}{2} = \frac{- 3}{2} = - 1.5$
Check that each coordinate lies between the endpoints: $- 2$ is between $- 7$ and $3$ , and $- 1.5$ is between $- 5$ and $2$ .

Answer: $(- 2, - 1.5)$ , or $(- 2, - \frac{3}{2})$ .

Example 4. $A (- 1, 2)$ and $B (7, 8)$ are the ends of a diameter of a circle. (a) Find the coordinates of the centre of the circle. (b) Find the radius of the circle. (c) Determine whether the point $C (6, 1)$ lies on the circle.

Show the working

The centre of a circle is the midpoint of any diameter, so (a) is a midpoint question:
$(\frac{- 1 + 7}{2}, \frac{2 + 8}{2}) = (3, 5)$
For (b), the radius is half the diameter. Find the length $A B$ first. The changes are $7 - (- 1) = 8$ and $8 - 2 = 6$ .
$A B = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$
So the radius is $10 \div 2 = 5$ .
For (c), every point on a circle is exactly one radius from the centre. So find the distance from the centre $(3, 5)$ to $C (6, 1)$ :
$\sqrt{{(6 - 3)}^{2} + {(1 - 5)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
The distance equals the radius, so $C$ lies on the circle. A distance less than $5$ would put $C$ inside the circle; a distance more than $5$ would put it outside.

Answer: (a) $(3, 5)$ (b) $5$ (c) Yes: $C$ is $5$ units from the centre, which equals the radius.

Where people go wrong

Mistake: Working out $2 - (- 3)$ as $- 1$ when finding the change in a coordinate.

Instead: Put brackets round a negative coordinate when you substitute it: $2 - (- 3) = 2 + 3 = 5$ .

Subtracting a negative number is the same as adding its size. A wrong difference gives a wrong side of the triangle, and every step after it is then wrong too.

Mistake: Taking the square root of each term separately: $\sqrt{9^{2} + 12^{2}} = 9 + 12 = 21$ .

Instead: Add the squares first, then take one square root: $\sqrt{81 + 144} = \sqrt{225} = 15$ .

The square root of a sum is not the sum of the square roots. A straight line is the shortest route between two points, so the length must be less than $9 + 12$ .

Mistake: Typing $- 12^{2}$ into a calculator and getting $- 144$ .

Instead: Type ${(- 12)}^{2}$ with brackets, or simply square $12$ , since the sign does not affect the square.

The calculator follows the order of operations: it squares $12$ first and then applies the minus sign. You then add a negative number under the root and get an error or a wrong length.

Mistake: Finding the midpoint with a minus sign: $(\frac{x_{2} - x_{1}}{2}, \frac{y_{2} - y_{1}}{2})$ .

Instead: Add the coordinates, then halve. The midpoint is the mean of the two endpoints.

Half the difference tells you how far the midpoint is from each end, not where it is. For $(1, 2)$ and $(7, 10)$ it gives $(3, 4)$ , which is not even on the segment. The midpoint is $(4, 6)$ .

Mistake: Given an endpoint $A$ and the midpoint $M$ , finding the other end by averaging $A$ and $M$ .

Instead: Double the midpoint's coordinates and subtract $A$ 's coordinates. Then check that the midpoint of $A$ and your answer is $M$ .

Averaging $A$ and $M$ gives the point a quarter of the way along $A B$ , not the far end.

Check yourself

Calculate the length of the line segment from $(0, 0)$ to $(5, 12)$ .
Answer
$\sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
Calculate the length of the line segment joining $(- 4, - 1)$ and $(- 1, 3)$ .
Answer
The changes are $- 1 - (- 4) = 3$ and $3 - (- 1) = 4$ , so the length is $\sqrt{9 + 16} = \sqrt{25} = 5$ .
Find the exact length of the line segment joining $(1, 1)$ and $(4, 7)$ . Give it in its simplest surd form.
Answer
$\sqrt{3^{2} + 6^{2}} = \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$ .
Find the coordinates of the midpoint of the line segment joining $(4, - 6)$ and $(- 2, 10)$ .
Answer
$(\frac{4 + (- 2)}{2}, \frac{- 6 + 10}{2}) = (1, 2)$ .
$M (1, - 2)$ is the midpoint of $A B$ , and $A$ is the point $(- 3, 5)$ . Find the coordinates of $B$ .
Answer
$B = (2 \times 1 - (- 3), 2 \times (- 2) - 5) = (5, - 9)$ . Check: $(\frac{- 3 + 5}{2}, \frac{5 + (- 9)}{2}) = (1, - 2)$ .
Why does it not matter which endpoint you call $(x_{1}, y_{1})$ when you calculate a length?
Answer
Swapping the endpoints only changes the sign of each difference, and squaring removes the sign. So the sum under the root, and the length, stay the same.
The distance from $(1, p)$ to $(9, 2)$ is $10$ . Find the two possible values of $p$ .
Answer
$8^{2} + {(2 - p)}^{2} = 10^{2}$ , so ${(2 - p)}^{2} = 36$ and $2 - p = 6$ or $2 - p = - 6$ . So $p = - 4$ or $p = 8$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Do not use a calculator in this question. $A$ is the point $(- 4, 1)$ and $B$ is the point $(5, 13)$ .

(a) Find the coordinates of the midpoint of $A B$ . 2 marks No calculator

Mark scheme

B1 x-coordinate 0.5 oe
B1 y-coordinate 7

(b) Calculate the length of $A B$ . 3 marks No calculator

Mark scheme

M1 differences 9 and 12 (or −9 and −12) seen, e.g. 5 − (−4) and 13 − 1 oe
M1 √(9² + 12²) oe
A1 15 cao

(c) $B$ is the midpoint of the line segment $A D$ . Find the coordinates of $D$ . 2 marks No calculator

Mark scheme

B1 x-coordinate 14
B1 y-coordinate 25

Question 2. $P$ is the point $(2, - 1)$ , $Q$ is the point $(9, 3)$ and $R$ is the point $(6, 6)$ .

(a) Calculate the length of $P Q$ . Give your answer correct to 3 significant figures. 2 marks

Mark scheme

M1 √(7² + 4²) oe
A1 8.06 awrt

(b) Show that triangle $P Q R$ is isosceles. 3 marks

Mark scheme

M1 correct method for the length of PR or QR, e.g. √(4² + 7²) oe
A1 PR = √65 (or PR² = 65) seen, with PQ = √65 (or PQ² = 65); exact values, since equal 3 s.f. decimals do not prove equality
A1 states PQ = PR, so two sides are equal; dep on M1

(c) Find the coordinates of the midpoint of $Q R$ . 2 marks

Mark scheme

B1 x-coordinate 7.5 oe
B1 y-coordinate 4.5 oe

Question 3. Do not use a calculator in this question. $A$ is the point $(3, - 2)$ and $B$ is the point $(7, k)$ . The length of $A B$ is $5$ units.

(a) Find the two possible values of $k$ . 4 marks No calculator

Mark scheme

M1 4² + (k + 2)² = 5² oe
M1 (k + 2)² = 9 or k² + 4k − 5 = 0 oe
A1 k = 1
A1 k = −5

(b) Given that $k > 0$ , find the coordinates of the midpoint of $A B$ . 2 marks No calculator

Mark scheme

B1 x-coordinate 5
B1 y-coordinate −0.5 oe; FT (−2 + their positive k) ÷ 2

How this is examined

Finding an unknown coordinate when the length is given: A Find part: one endpoint has a letter such as $k$ for a coordinate, the length is stated, and you give the possible values of the letter The length method written out with the letter in it and the length squared, e.g. $4^{2} + {(k + 2)}^{2} = 5^{2}$ , is the first method mark. Reducing it to ${(k + 2)}^{2} = 9$ (or to the expanded quadratic) is the second. Each value of $k$ is a separate accuracy mark, so giving only $k = 1$ loses one.
The length of a segment from two pairs of coordinates: A Calculate part with two given points: an exact answer on the non-calculator paper, or an answer to 3 significant figures on a calculator paper Both differences seen, with any negative coordinate handled correctly ( $5 - (- 4) = 9$ ), then one square root of the sum of their squares, $\sqrt{9^{2} + 12^{2}}$ . Those earn the method marks. The accuracy mark needs the exact value (simplified, if the question asks for simplest surd form), or the value to 3 significant figures.
Using lengths to prove a fact about a shape: A Show (that) part, e.g. show that a triangle is isosceles, or that two points are the same distance from a third point Every length you rely on, worked out in full with its method shown, then a sentence that compares them, e.g. ' $P Q = P R = \sqrt{65}$ , so two sides are equal and the triangle is isosceles'. The result is given in the question, so the marks are for the evidence: a comparison that is not stated is an incomplete answer.
The coordinates of the midpoint of a segment: A short Find part asking for the midpoint of two given points One mark for each coordinate: add the $x$ -coordinates and halve, add the $y$ -coordinates and halve. Write the answer as a coordinate pair in brackets, e.g. $(0.5, 7)$ .
Finding the far endpoint from one endpoint and the midpoint: A Find part that gives the midpoint and one endpoint, and asks for the other endpoint Each coordinate of the far end as twice the midpoint's coordinate minus the known end's, e.g. $2 \times 13 - 1 = 25$ . One mark for each coordinate. Check by averaging your answer with the known end.

Words to know

Line segment: The part of a straight line between two endpoints. It has a definite length, unlike a line, which goes on for ever.
Endpoint: One of the two points at the ends of a line segment.
Length of a line segment: The straight-line distance between its endpoints, $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ . It is never negative.
Midpoint: The point exactly halfway between the endpoints, $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .
Hypotenuse: The longest side of a right-angled triangle, opposite the right angle. A slanted segment on a grid is the hypotenuse of a triangle whose other two sides run along the grid lines.
Surd: A square root that is not a whole number or a fraction, such as $\sqrt{10}$ . Leaving a length as a simplified surd keeps it exact.

All Maths lessons Next: 4.4 Equations of linear graphs