Gradient and Line Relationships | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Calculate rise over run

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\text{rise}}{\text{run}}$

A positive gradient rises from left to right. A negative gradient falls from left to right.
A horizontal line has gradient $0$ because its vertical change is zero.
A vertical line has undefined gradient because its horizontal change is zero and division by zero is not defined.

Subtract the coordinates in the same order in the numerator and denominator. Reversing both differences is safe; reversing only one changes the sign.

Parallel, perpendicular and collinear

lines have the same direction. Non-vertical parallel lines have equal gradients; distinct vertical lines are also parallel. lines meet at $90^\circ$ . When both gradients are defined, one is the negative reciprocal of the other.

$m_1=m_2\quad\text{(parallel)}$

$m_1m_2=-1\quad\text{(perpendicular)}$

A means invert the fraction and change its sign. Perpendicular to $\frac{2}{3}$ is $-\frac{3}{2}$ . A horizontal line and a vertical line form the special perpendicular pair; do not use $m_1m_2=-1$ because the vertical gradient is undefined.

Points are when they lie on one straight line. Gradients between successive pairs are then equal, unless the common line is vertical.

Swipe left or right to see the whole diagram →

These parallel lines have the same gradient. A perpendicular bisector crosses AB at its midpoint and at 90 degrees.

A parameter can give more than one line

A parameter, such as $k$ , is a fixed number whose value is not yet known.

Worked example

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

Coursebook Worked example 3.4.

If $k=15$ , $AB$ is vertical but $C$ does not have $x=10$ , so that case is not collinear. Otherwise compare gradients.
$m_{AB}=\dfrac{k+15}{15-k}$ and $m_{BC}=\dfrac{-2k}{-4}=\dfrac{k}{2}$ .
Set them equal and multiply both sides by $2(15-k)$ : $2(k+15)=k(15-k)$ , so $k^2-13k+30=0$ .
The calculator solver gives the convenient roots 3 and 10, so write $(k-3)(k-10)=0$ .

Answer

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

Check the method

Your turnIndependent check

Find the gradient of the line through $A(-2,5)$ and $B(4,2)$ . Then find the gradient of a perpendicular line.

Show worked answer

Answer

$m=\dfrac{2-5}{4-(-2)}=-\dfrac12$

The negative reciprocal is $2$ .

Move the two endpoints below. The displayed gradient, midpoint and length are calculated from the coordinates. The perpendicular control also shows what happens when a line is horizontal or vertical, or the two points coincide (occupy the same position).

A endpointB endpointM midpoint

A x-3A y-1B x3B y3

gradient m = 2/3midpoint (0, 1)length AB ≈ 7.21

Calculate rise over run

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\text{rise}}{\text{run}}$

A positive gradient rises from left to right. A negative gradient falls from left to right.
A horizontal line has gradient $0$ because its vertical change is zero.
A vertical line has undefined gradient because its horizontal change is zero and division by zero is not defined.

Subtract the coordinates in the same order in the numerator and denominator. Reversing both differences is safe; reversing only one changes the sign.

Parallel, perpendicular and collinear

$m_1=m_2\quad\text{(parallel)}$

$m_1m_2=-1\quad\text{(perpendicular)}$

Points are when they lie on one straight line. Gradients between successive pairs are then equal, unless the common line is vertical.

Swipe left or right to see the whole diagram →

These parallel lines have the same gradient. A perpendicular bisector crosses AB at its midpoint and at 90 degrees.

A parameter can give more than one line

A parameter, such as $k$ , is a fixed number whose value is not yet known.

Worked example

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

Coursebook Worked example 3.4.

If $k=15$ , $AB$ is vertical but $C$ does not have $x=10$ , so that case is not collinear. Otherwise compare gradients.
$m_{AB}=\dfrac{k+15}{15-k}$ and $m_{BC}=\dfrac{-2k}{-4}=\dfrac{k}{2}$ .
Set them equal and multiply both sides by $2(15-k)$ : $2(k+15)=k(15-k)$ , so $k^2-13k+30=0$ .
The calculator solver gives the convenient roots 3 and 10, so write $(k-3)(k-10)=0$ .

Answer

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

Check the method

Your turnIndependent check

Find the gradient of the line through $A(-2,5)$ and $B(4,2)$ . Then find the gradient of a perpendicular line.

Show worked answer

Answer

$m=\dfrac{2-5}{4-(-2)}=-\dfrac12$

The negative reciprocal is $2$ .

A endpointB endpointM midpoint

A x-3A y-1B x3B y3

gradient m = 2/3midpoint (0, 1)length AB ≈ 7.21