Mathematics 9709 · Pure Mathematics 1

Parallel and perpendicular lines

The gradient is how steep the line is: $m=\dfrac{y_2-y_1}{x_2-x_1}$ , that is rise over run.
This chapter has two rules. The first is easy. The second is the perpendicular rule. It is used in almost every tangent, normal and perpendicular-bisector question.

Parallel lines point the same way, so they have the same gradient: $m_1=m_2$ .

Perpendicular lines (lines that meet at a right angle) have gradients that multiply to $-1$ , so $m_2=-\dfrac{1}{m_1}$ . You turn the fraction upside down and change the sign (the negative reciprocal). Perpendicular to $3$ is $-\tfrac13$ . Perpendicular to $-\tfrac25$ is $\tfrac52$ .

The two rules

$m_1\times m_2=-1$

Why it works: a gradient of $m$ means “go right 1, then up $m$ .”
Turn that step a quarter-turn. Now it says “go up 1, then left $m$ .” That is a step of $-\tfrac1m$ , the negative reciprocal.

Same gradient means parallel. Turn the gradient upside down and change its sign for perpendicular.

The perpendicular bisector (drawn above) goes through the midpoint of $AB$ and crosses it at a right angle.
Its gradient is the gradient of $AB$ turned upside down with the sign changed (the negative reciprocal).

Worked examples

Worked example

$A(k-5,\ -15)$ , $B(10,\ k)$ , $C(6,\ -k)$ are collinear. Find $k$

Cambridge Pure Mathematics 1, Worked example 3.4. “Collinear” means all on one line, so it is just a gradient match.

Collinear means the gradient of $AB$ equals the gradient of $BC$ : $\dfrac{k-(-15)}{10-(k-5)}=\dfrac{-k-k}{6-10}$ .
Tidy each side, then cross-multiply: $\dfrac{k+15}{15-k}=\dfrac{k}{2}\;\Rightarrow\;2(k+15)=k(15-k)$ .
Expand and collect to a quadratic: $k^2-13k+30=0\;\Rightarrow\;(k-3)(k-10)=0$ .

Answer

k=3\quad\text{or}\quad k=10

Worked example

The midpoint of $P(-4,\ 5)$ and $Q(6,\ 1)$ is $M$ . $R$ is $(-3,\ -7)$ . Show $RM \perp PQ$

Cambridge Pure Mathematics 1, Exercise 3B Q2. To prove a right angle, show that the two gradients multiply to $-1$ .

Midpoint first: $M=\left(\tfrac{-4+6}{2},\tfrac{5+1}{2}\right)=(1,3)$ .
Two gradients: $m_{PQ}=\dfrac{1-5}{6-(-4)}=-\dfrac{2}{5},\quad m_{RM}=\dfrac{3-(-7)}{1-(-3)}=\dfrac{5}{2}$ .
Multiply — the test for a right angle: $-\tfrac25\times\tfrac52=-1$ .

Answer

m_{PQ}\times m_{RM}=-1\ \Rightarrow\ RM\perp PQ

When two gradients give a quadratic in k

The angle ABC = 90° pattern

A harder version gives a point with an unknown $k$ in it. It then says “angle $ABC=90^\circ$ .” You use the perpendicular rule: $m_{AB}\times m_{BC}=-1$ . Because $k$ is in both gradients, multiplying them out gives a quadratic in $k$ . So expect two values. Each one is a position of $B$ that makes a right angle.

For example, with $A(11,3)$ , $B(2k,k)$ , $C(-1,-11)$ , the rule gives $5k^2-12k-44=0$ , so $k=4.4$ or $k=-2$ . (Cambridge P1 WE 3.5)

Where the marks go

Common mistake

The perpendicular gradient is the negative reciprocal. Perpendicular to gradient

3

-\tfrac13

. It is not

\tfrac13

(you forgot the sign). It is not

-3

(you forgot to turn it upside down).

Common mistake

Keep the coordinates in the same order on top and bottom:

\dfrac{y_2-y_1}{x_2-x_1}

, not

\dfrac{y_2-y_1}{x_1-x_2}

. Mark schemes accept a mixed-up order only if it cancels out. Do not count on that. (9709/12 Nov 2022 Q1)

Now you try

The line $x+2y=10$ is a tangent to a circle at the point $A(4,3)$ . Find the equation of the normal to the circle at $A$ . (9709/12 Nov 2024 Q8(b)(i))

Your turn— tap to reveal the worked answer (9709/12 Nov 2024 Q8)

The normal is the radius, and it is perpendicular to the tangent. The tangent $x+2y=10$ is $y=5-\tfrac12 x$ , with gradient $-\tfrac12$ .

Turn it upside down and change the sign: the normal has gradient $2$ . Through $A(4,3)$ : $y-3=2(x-4)$ .

Answer

y=2x-5

PreviousLength of a line segment and midpoint NextEquations of straight lines

Parallel and perpendicular lines

The gradient is how steep the line is: $m=\dfrac{y_2-y_1}{x_2-x_1}$ , that is rise over run.
This chapter has two rules. The first is easy. The second is the perpendicular rule. It is used in almost every tangent, normal and perpendicular-bisector question.

Parallel lines point the same way, so they have the same gradient: $m_1=m_2$ .

The two rules

$m_1\times m_2=-1$

Why it works: a gradient of $m$ means “go right 1, then up $m$ .”
Turn that step a quarter-turn. Now it says “go up 1, then left $m$ .” That is a step of $-\tfrac1m$ , the negative reciprocal.

Same gradient means parallel. Turn the gradient upside down and change its sign for perpendicular.

The perpendicular bisector (drawn above) goes through the midpoint of $AB$ and crosses it at a right angle.
Its gradient is the gradient of $AB$ turned upside down with the sign changed (the negative reciprocal).

Worked examples

Worked example

$A(k-5,\ -15)$ , $B(10,\ k)$ , $C(6,\ -k)$ are collinear. Find $k$

Cambridge Pure Mathematics 1, Worked example 3.4. “Collinear” means all on one line, so it is just a gradient match.

Collinear means the gradient of $AB$ equals the gradient of $BC$ : $\dfrac{k-(-15)}{10-(k-5)}=\dfrac{-k-k}{6-10}$ .
Tidy each side, then cross-multiply: $\dfrac{k+15}{15-k}=\dfrac{k}{2}\;\Rightarrow\;2(k+15)=k(15-k)$ .
Expand and collect to a quadratic: $k^2-13k+30=0\;\Rightarrow\;(k-3)(k-10)=0$ .

Answer

k=3\quad\text{or}\quad k=10

Worked example

The midpoint of $P(-4,\ 5)$ and $Q(6,\ 1)$ is $M$ . $R$ is $(-3,\ -7)$ . Show $RM \perp PQ$

Cambridge Pure Mathematics 1, Exercise 3B Q2. To prove a right angle, show that the two gradients multiply to $-1$ .

Midpoint first: $M=\left(\tfrac{-4+6}{2},\tfrac{5+1}{2}\right)=(1,3)$ .
Two gradients: $m_{PQ}=\dfrac{1-5}{6-(-4)}=-\dfrac{2}{5},\quad m_{RM}=\dfrac{3-(-7)}{1-(-3)}=\dfrac{5}{2}$ .
Multiply — the test for a right angle: $-\tfrac25\times\tfrac52=-1$ .

Answer

m_{PQ}\times m_{RM}=-1\ \Rightarrow\ RM\perp PQ

When two gradients give a quadratic in k

The angle ABC = 90° pattern

For example, with $A(11,3)$ , $B(2k,k)$ , $C(-1,-11)$ , the rule gives $5k^2-12k-44=0$ , so $k=4.4$ or $k=-2$ . (Cambridge P1 WE 3.5)

Where the marks go

Common mistake

The perpendicular gradient is the negative reciprocal. Perpendicular to gradient

3

-\tfrac13

. It is not

\tfrac13

(you forgot the sign). It is not

-3

(you forgot to turn it upside down).

Common mistake

Keep the coordinates in the same order on top and bottom:

\dfrac{y_2-y_1}{x_2-x_1}

, not

\dfrac{y_2-y_1}{x_1-x_2}

. Mark schemes accept a mixed-up order only if it cancels out. Do not count on that. (9709/12 Nov 2022 Q1)

Now you try

The line $x+2y=10$ is a tangent to a circle at the point $A(4,3)$ . Find the equation of the normal to the circle at $A$ . (9709/12 Nov 2024 Q8(b)(i))

Your turn— tap to reveal the worked answer (9709/12 Nov 2024 Q8)

The normal is the radius, and it is perpendicular to the tangent. The tangent $x+2y=10$ is $y=5-\tfrac12 x$ , with gradient $-\tfrac12$ .

Turn it upside down and change the sign: the normal has gradient $2$ . Through $A(4,3)$ : $y-3=2(x-4)$ .

Answer

y=2x-5