Mathematics 9709 · Pure Mathematics 1

Length of a line segment and midpoint

This whole topic is really just Pythagoras.
Take two points and join them with a line. From that one line you can find how steep it is, where its middle is, and how long it is. They all come from one right-angled triangle.

Distance is the long side of the triangle (the hypotenuse): $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Every “find the length” or “find the distance” part uses this one square root.

Midpoint is just the average of the two points: $M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$ . There is no square root. For each coordinate, add the two values and halve.

Where the two formulas come from

Take two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ . The gap across is $x_2-x_1$ . The gap up is $y_2-y_1$ .
Those two gaps are the short sides of a right-angled triangle. $PQ$ is the long side. So the length comes from Pythagoras:

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The midpoint sits halfway along that long side, so each coordinate is the average of the two ends, with no square root.

$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

PQ is the long side of a right-angled triangle. M is the point exactly halfway along it.

Move $A$ and $B$ below and watch the gradient, midpoint and length change together: the three numbers every question here asks for.

A xA yB xB y

gradient m = 0.7·midpoint (0, 1)·length AB = 7.21

Worked examples

Worked example

$M(\tfrac{3}{2},\ -11)$ is the midpoint of $P(-7,\ 4)$ and $Q(a,\ b)$ . Find $a$ and $b$

Cambridge Pure Mathematics 1, Worked example 3.1. Read “midpoint = average” as two small equations, one for each coordinate.

Set the $x$ -average equal to $\tfrac32$ : $\dfrac{-7+a}{2}=\dfrac{3}{2}\;\Rightarrow\; -7+a=3$ .
Set the $y$ -average equal to $-11$ : $\dfrac{4+b}{2}=-11\;\Rightarrow\; 4+b=-22$ .
Solve each: $a=10$ , $b=-26$ .

Answer

a=10,\quad b=-26

Worked example

The distance between $P(-2,\ a)$ and $Q(a-2,\ -7)$ is $17$ . Find $a$

Cambridge Pure Mathematics 1, Worked example 3.3. A length with an unknown in it almost always gives a quadratic, and a quadratic usually has two answers.

Put both points into the distance formula and square both sides at once: $(a-2+2)^2+(-7-a)^2=17^2$ .
Expand and collect — most of the marks are in this step: $a^2+(49+14a+a^2)=289\;\Rightarrow\;2a^2+14a-240=0$ .
Divide by 2, then factorise: $a^2+7a-120=0\;\Rightarrow\;(a-8)(a+15)=0$ .

Answer

a=8\quad\text{or}\quad a=-15

Worked example

Real question: $A(-1,\ -2)$ , $B(7,\ 4)$ are ends of a diameter. Find the circle

9709/11 May/June 2020 Q10(a). This uses both formulas at once, which is a common way the diameter idea is tested.

Centre = midpoint of $AB$ : $C=\left(\tfrac{-1+7}{2},\tfrac{-2+4}{2}\right)=(3,1)$ .
Radius = distance from centre to an end; keep it as $r^2$ : $r^2=(7-3)^2+(4-1)^2=16+9=25$ .
Write it straight in centre–radius form (next topic): $(x-3)^2+(y-1)^2=25$ .

Answer

(x-3)^2+(y-1)^2=25

Notice we never took a square root. The circle equation needs $r^2$ , so stopping at 25 is faster and safer.

Equidistant, isosceles, right-angled: same method

How the distance formula answers shape questions

Many marks come from questions that never say the word “distance.” They hide it inside another word:

Equidistant from two points means two distances are equal. Set $PA^2=PB^2$ and solve. You can square both, so you need no roots.
Isosceles means two sides have the same length. Work out all three sides and compare them.
Right-angled means the three squared lengths fit Pythagoras, $a^2+b^2=c^2$ . A faster way: two gradients multiply to $-1$ (next topic).

In every case, do the same thing. Work with the squared distance for as long as you can. Only take the square root at the end, and only if the question asks for a length.

Where the marks go

Common mistake

The distance formula squares the gaps and adds them under one root. It is not

(x_2-x_1)+(y_2-y_1)

. The square root covers the whole sum, not each part on its own.

Common mistake

When a length gives a quadratic, you usually get two answers. Give both, unless the question or a diagram rules one out. (Cambridge P1 WE 3.3)

Common mistake

Cambridge will not accept an answer measured off a drawing. You must show the calculation. The words “by calculation” in the question are a clear warning. (9709/12 Mar 2022 Q8)

Now you try

The circle $(x-2)^2+y^2=8$ has a chord $AB$ . It meets the positive $y$ -axis at $A$ and is parallel to the $x$ -axis. Find the coordinates of $A$ and $B$ . (9709/12 Mar 2022 Q8(a))

Your turn— tap to reveal the worked answer (9709/12 Mar 2022 Q8)

$A$ is on the $y$ -axis, so put $x=0$ : $4+y^2=8\Rightarrow y=\pm2$ . Take the positive one: $A=(0,2)$ .

$AB$ is horizontal, so $B$ also has $y=2$ : $(x-2)^2+4=8\Rightarrow (x-2)^2=4$ , so $x=0$ or $4$ .

Answer

A=(0,2),\quad B=(4,2)

PreviousTransformations of functions NextParallel and perpendicular lines

Length of a line segment and midpoint

This whole topic is really just Pythagoras.
Take two points and join them with a line. From that one line you can find how steep it is, where its middle is, and how long it is. They all come from one right-angled triangle.

Distance is the long side of the triangle (the hypotenuse): $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Every “find the length” or “find the distance” part uses this one square root.

Midpoint is just the average of the two points: $M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$ . There is no square root. For each coordinate, add the two values and halve.

Where the two formulas come from

Take two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ . The gap across is $x_2-x_1$ . The gap up is $y_2-y_1$ .
Those two gaps are the short sides of a right-angled triangle. $PQ$ is the long side. So the length comes from Pythagoras:

$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The midpoint sits halfway along that long side, so each coordinate is the average of the two ends, with no square root.

$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

PQ is the long side of a right-angled triangle. M is the point exactly halfway along it.

Move $A$ and $B$ below and watch the gradient, midpoint and length change together: the three numbers every question here asks for.

A xA yB xB y

gradient m = 0.7·midpoint (0, 1)·length AB = 7.21

Worked examples

Worked example

$M(\tfrac{3}{2},\ -11)$ is the midpoint of $P(-7,\ 4)$ and $Q(a,\ b)$ . Find $a$ and $b$

Cambridge Pure Mathematics 1, Worked example 3.1. Read “midpoint = average” as two small equations, one for each coordinate.

Set the $x$ -average equal to $\tfrac32$ : $\dfrac{-7+a}{2}=\dfrac{3}{2}\;\Rightarrow\; -7+a=3$ .
Set the $y$ -average equal to $-11$ : $\dfrac{4+b}{2}=-11\;\Rightarrow\; 4+b=-22$ .
Solve each: $a=10$ , $b=-26$ .

Answer

a=10,\quad b=-26

Worked example

The distance between $P(-2,\ a)$ and $Q(a-2,\ -7)$ is $17$ . Find $a$

Cambridge Pure Mathematics 1, Worked example 3.3. A length with an unknown in it almost always gives a quadratic, and a quadratic usually has two answers.

Put both points into the distance formula and square both sides at once: $(a-2+2)^2+(-7-a)^2=17^2$ .
Expand and collect — most of the marks are in this step: $a^2+(49+14a+a^2)=289\;\Rightarrow\;2a^2+14a-240=0$ .
Divide by 2, then factorise: $a^2+7a-120=0\;\Rightarrow\;(a-8)(a+15)=0$ .

Answer

a=8\quad\text{or}\quad a=-15

Worked example

Real question: $A(-1,\ -2)$ , $B(7,\ 4)$ are ends of a diameter. Find the circle

9709/11 May/June 2020 Q10(a). This uses both formulas at once, which is a common way the diameter idea is tested.

Centre = midpoint of $AB$ : $C=\left(\tfrac{-1+7}{2},\tfrac{-2+4}{2}\right)=(3,1)$ .
Radius = distance from centre to an end; keep it as $r^2$ : $r^2=(7-3)^2+(4-1)^2=16+9=25$ .
Write it straight in centre–radius form (next topic): $(x-3)^2+(y-1)^2=25$ .

Answer

(x-3)^2+(y-1)^2=25

Notice we never took a square root. The circle equation needs $r^2$ , so stopping at 25 is faster and safer.

Equidistant, isosceles, right-angled: same method

How the distance formula answers shape questions

Many marks come from questions that never say the word “distance.” They hide it inside another word:

Equidistant from two points means two distances are equal. Set $PA^2=PB^2$ and solve. You can square both, so you need no roots.
Isosceles means two sides have the same length. Work out all three sides and compare them.
Right-angled means the three squared lengths fit Pythagoras, $a^2+b^2=c^2$ . A faster way: two gradients multiply to $-1$ (next topic).

In every case, do the same thing. Work with the squared distance for as long as you can. Only take the square root at the end, and only if the question asks for a length.

Where the marks go

Common mistake

The distance formula squares the gaps and adds them under one root. It is not

(x_2-x_1)+(y_2-y_1)

. The square root covers the whole sum, not each part on its own.

Common mistake

When a length gives a quadratic, you usually get two answers. Give both, unless the question or a diagram rules one out. (Cambridge P1 WE 3.3)

Common mistake

Cambridge will not accept an answer measured off a drawing. You must show the calculation. The words “by calculation” in the question are a clear warning. (9709/12 Mar 2022 Q8)

Now you try

The circle $(x-2)^2+y^2=8$ has a chord $AB$ . It meets the positive $y$ -axis at $A$ and is parallel to the $x$ -axis. Find the coordinates of $A$ and $B$ . (9709/12 Mar 2022 Q8(a))

Your turn— tap to reveal the worked answer (9709/12 Mar 2022 Q8)

$A$ is on the $y$ -axis, so put $x=0$ : $4+y^2=8\Rightarrow y=\pm2$ . Take the positive one: $A=(0,2)$ .

$AB$ is horizontal, so $B$ also has $y=2$ : $(x-2)^2+4=8\Rightarrow (x-2)^2=4$ , so $x=0$ or $4$ .

Answer

A=(0,2),\quad B=(4,2)