Straight lines · Coordinate geometry & circles

This lesson builds on coordinate geometry. For points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

DistanceAB = √((x₂ − x₁)² + (y₂ − y₁)²)

MidpointM = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Gradientm = (y₂ − y₁) ÷ (x₂ − x₁)

The line through $(x_1, y_1)$ with gradient $m$ is $y - y_1 = m(x - x_1)$ . Three points are collinear (on one line) when the gradient between each pair is the same.

More: distance, gradient and equations

NECO 2023 · Paper 1 · Q26Find the distance between the points $(2, 3)$ and $(-1, 7)$ .
NECO 2023 · Paper 1 · Q27Find the equation of the straight line which passes through $(2, 2)$ and $(3, 6)$ .
WAEC 2016 · Paper 2 · Q2Given that $(5, 2)$ , $(-4, k)$ and $(2, 1)$ lie on a straight line, find the value of $k$ .
WAEC 2022 · Paper 1 · Q40The length of the line joining points $(x, 4)$ and $(-x, 3)$ is 7 units. Find the value of $x$ .
WAEC 2023 · Paper 1 · Q12Given that $M$ is the midpoint of $T(2, 4)$ and $Q(-8, 6)$ , find the length $MQ$ .
WAEC 2020 · Paper 1 · Q25Calculate, correct to two decimal places, the area enclosed by the line $3x - 5y + 4 = 0$ and the axes.

Parallel and perpendicular lines

Parallel lines have equal gradients. Perpendicular lines have gradients that multiply to $-1$ :

Perpendicular linesm₁ × m₂ = −1, so m₂ = −1/m₁

To use the gradient of a line like $3x + 2y - 4 = 0$ , rearrange it as $y = mx + c$ first.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q2

Find the equation of the line which passes through the point $(3, -2)$ and is perpendicular to the line $3x + 2y - 4 = 0$ .

The given gradient
- ${2y = -3x + 4}$ , so ${y = -\frac32x + 2}$ : the gradient is ${-\frac32}$ .
Think first. Make y the subject.
The perpendicular gradient
- ${m = -1 \div \left(-\frac32\right) = \frac23}$ .
The line
- ${y + 2 = \frac23(x - 3)}$ .
- Multiply by 3: ${3y + 6 = 2x - 6}$ , so ${2x - 3y - 12 = 0}$ .

The perpendicular bisector of $AB$ passes through the midpoint of $AB$ , at right angles to it. Every point on it is the same distance from $A$ and $B$ .

The perpendicular bisectorThrough M, with gradient −1 ÷ (gradient of AB)

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q9

The curve $y = 7 - \dfrac6x$ and the line $y + 2x - 3 = 0$ intersect at two points. Find the:

coordinates of the two points;

equation of the perpendicular bisector of the line joining the two points.

Where they meet
- ${7 - \frac6x = 3 - 2x}$ . Multiply by $x$ : ${7x - 6 = 3x - 2x^2}$ .
- So ${2x^2 + 4x - 6 = 0}$ , that is ${x^2 + 2x - 3 = 0}$ .
- ${(x + 3)(x - 1) = 0}$ : the points are ${(1, 1)}$ and ${(-3, 9)}$ .
Think first. From the line, y = 3 − 2x. Put that into the curve.
Midpoint and gradient
- Midpoint: ${\left(\frac{1 - 3}{2}, \frac{1 + 9}{2}\right) = (-1, 5)}$ .
- Gradient of the chord: ${\frac{9 - 1}{-3 - 1} = -2}$ , so the bisector has gradient ${\frac12}$ .
The bisector
- ${y - 5 = \frac12(x + 1)}$ , so ${2y - 10 = x + 1}$ .
- That is ${2y - x - 11 = 0}$ .

More: perpendicular lines

The angle between two lines

If two lines have gradients $m_1$ and $m_2$ , the acute angle $\theta$ between them satisfies

The angle between two linestan θ = |(m₁ − m₂) ÷ (1 + m₁m₂)|

\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|

Gradients and the angleSet the two gradients

77.47°acute angle θ−1/2m₁ × m₂

m₁ = 2m₂ = −1/4

tan θ = |(2 − (−1/4)) ÷ (1 + 2 × (−1/4))| = |9/4 ÷ 1/2| = 4.5, so θ = 77.47°.

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q9

Two lines $L_1$ and $L_2$ pass through the point of intersection of $2y = x - 13$ and $3y + x + 12 = 0$ . $L_1$ passes through $P(-4, -7)$ and $L_2$ is perpendicular to $2x - 5y = 4$ . Find the acute angle between $L_1$ and $L_2$ (2 d.p.).

The common point
- ${3y + (2y + 13) + 12 = 0}$ , so ${y = -5}$ and ${x = 3}$ : the point ${(3, -5)}$ .
Think first. From the first line, x = 2y + 13.
The two gradients
- $L_1$ through $(3, -5)$ and $(-4, -7)$ : ${m_1 = \frac{-7 + 5}{-4 - 3} = \frac27}$ .
- ${2x - 5y = 4}$ has gradient ${\frac25}$ , so $L_2$ has ${m_2 = -\frac52}$ .
The angle
- ${\tan\theta = \left|\frac{\frac27 + \frac52}{1 - \frac57}\right| = \frac{39/14}{2/7} = \frac{39}{4}}$ .
- ${\theta = \tan^{-1} 9.75 \approx 84.14^\circ}$ .

More: the angle between two lines

WAEC 2022 · Paper 1 · Q29Find, correct to the nearest degree, the acute angle formed by the lines $y = 2x + 5$ and $2y = x - 6$ .

Dividing a line in a ratio

The point dividing $AB$ internally in the ratio $m : n$ is $\left(\dfrac{nx_1 + mx_2}{m + n}, \dfrac{ny_1 + my_2}{m + n}\right)$ , as with position vectors (see the ratio figure in vectors). Externally, the point lies beyond $B$ (or beyond $A$ ), and the signs change:

External division((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n))

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q2 (a)

Find the coordinates of the point which divides the line joining $(7, -5)$ and $(-2, 7)$ externally in the ratio $3 : 2$ .

The formula
- ${x = \frac{3(-2) - 2(7)}{3 - 2} = -6 - 14 = -20}$ .
- ${y = \frac{3(7) - 2(-5)}{3 - 2} = 21 + 10 = 31}$ .
Think first. m : n = 3 : 2, with (x₁, y₁) = (7, −5) and (x₂, y₂) = (−2, 7).
The point
- ${(-20, 31)}$ : beyond $B$ , since the ratio is more than 1.

More: dividing lines and parallelograms

Your turn

WAEC 2017 · Paper 2 · Q3

Coordinate geometry & circles

(a)
A line is parallel to $2x + 3y = 5$ . If the line is the perpendicular bisector of the line joining the points $(3, y)$ and $(5, 2)$ , find its equation.
Show the answer
$6y + 4x - 19 = 0$

Worked solution (try it first)

$2x + 3y = 5$ has gradient $-\frac23$ , so the bisector has gradient $-\frac23$ too (parallel).
The segment is perpendicular to it, with gradient $\frac32$ : $\dfrac{2 - y}{5 - 3} = \dfrac32$ , so $y = -1$ .
The midpoint of $(3, -1)$ and $(5, 2)$ is $\left(4, \frac12\right)$ .
$y - \frac12 = -\frac23(x - 4)$ .
Multiply by 6: $6y - 3 = -4x + 16$ .
So $6y + 4x - 19 = 0$ .

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