Straight Lines: Review & Extension

This lesson revises and extends the key results for straight lines that underpin all of coordinate geometry at A-Level. You must be completely fluent with gradients, equations of lines, midpoints, distances, and the conditions for parallel and perpendicular lines. These results are used constantly — in circle problems, parametric questions, and coordinate proofs.

Gradient of a Line

The gradient (or slope) of the straight line passing through two points ((x_1, y_1)) and ((x_2, y_2)) is

m = (y₂ − y₁) / (x₂ − x₁)

The gradient measures the rate of change of y with respect to x. A positive gradient means the line slopes upward from left to right; a negative gradient means it slopes downward.

Example 1: Find the gradient of the line through A(−3, 5) and B(1, −7).

m = (−7 − 5) / (1 − (−3)) = −12 / 4 = −3

The gradient is −3.

Exam Tip: Always be careful with signs when subtracting negative coordinates. Write out the subtraction in full to avoid errors.

Midpoint of a Line Segment

The midpoint M of the line segment joining ((x_1, y_1)) and ((x_2, y_2)) is

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Example 2: Find the midpoint of the segment from P(2, −4) to Q(8, 6).

M = ((2 + 8)/2, (−4 + 6)/2) = (5, 1)

Distance Between Two Points

The distance between ((x_1, y_1)) and ((x_2, y_2)) is given by

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

This is a direct application of Pythagoras' theorem in the coordinate plane.

Example 3: Find the distance between A(−1, 3) and B(5, −5).

d = √[(5 − (−1))² + (−5 − 3)²]
  = √[6² + (−8)²]
  = √[36 + 64]
  = √100
  = 10

Equations of Straight Lines

The form y = mx + c

This is the gradient-intercept form. Here m is the gradient and c is the y-intercept.

Example 4: A line has gradient 2 and passes through (0, −3). Its equation is y = 2x − 3.

The form y − y₁ = m(x − x₁)

This is useful when you know the gradient m and a point ((x_1, y_1)) on the line.

Example 5: Find the equation of the line with gradient −4 through (3, 7).

y − 7 = −4(x − 3)
y − 7 = −4x + 12
y = −4x + 19

The form ax + by + c = 0

This is the general form. At A-Level, you are often asked to give your answer in this form with integer coefficients.

Example 6: Write y = ³⁄₄x − 2 in the form ax + by + c = 0.

y = (3/4)x − 2
4y = 3x − 8
3x − 4y − 8 = 0

Exam Tip: When asked for the equation "in the form ax + by + c = 0", always clear fractions so that a, b, and c are integers.

Parallel and Perpendicular Lines

Parallel Lines

Two lines are parallel if and only if they have the same gradient.

If line (l_1) has gradient (m_1) and line (l_2) has gradient (m_2), then

l₁ ∥ l₂  ⟺  m₁ = m₂

Example 7: The line y = 3x − 1 is parallel to the line y = 3x + 5, since both have gradient 3.

Perpendicular Lines

Two lines are perpendicular if and only if the product of their gradients is −1.

l₁ ⊥ l₂  ⟺  m₁ × m₂ = −1

Equivalently, the gradient of a line perpendicular to a line with gradient m is −1/m (the negative reciprocal).

Example 8: A line has gradient 2/5. Find the gradient of a perpendicular line.

m₂ = −1 / (2/5) = −5/2

Example 9: Show that the line through A(1, 4) and B(3, 10) is perpendicular to the line through C(2, 1) and D(8, −1).

Gradient of AB = (10 − 4)/(3 − 1) = 6/2 = 3
Gradient of CD = (−1 − 1)/(8 − 2) = −2/6 = −1/3

Product = 3 × (−1/3) = −1  ✓

The product is −1, so the lines are perpendicular.

Finding the Equation of a Perpendicular Bisector

The perpendicular bisector of a line segment AB is the line that passes through the midpoint of AB and is perpendicular to AB.

Method:

Find the midpoint of AB.
Find the gradient of AB.
Find the negative reciprocal to get the gradient of the perpendicular bisector.
Use y − y₁ = m(x − x₁) with the midpoint.

Example 10: Find the equation of the perpendicular bisector of the segment from A(2, 1) to B(6, 5).

Midpoint = ((2 + 6)/2, (1 + 5)/2) = (4, 3)

Gradient of AB = (5 − 1)/(6 − 2) = 4/4 = 1

Perpendicular gradient = −1

Equation: y − 3 = −1(x − 4)
          y = −x + 7

Intersection of Two Lines

To find where two lines intersect, solve their equations simultaneously.

Example 11: Find the intersection of y = 2x + 1 and 3x + y = 16.

Substitute y = 2x + 1 into 3x + y = 16:

3x + (2x + 1) = 16
5x + 1 = 16
5x = 15
x = 3

y = 2(3) + 1 = 7

The lines meet at (3, 7).

Collinearity

Three points are collinear (lie on the same straight line) if the gradient between any two pairs of points is the same.

Example 12: Show that A(1, 3), B(3, 7), and C(6, 13) are collinear.

Gradient of AB = (7 − 3)/(3 − 1) = 4/2 = 2
Gradient of AC = (13 − 3)/(6 − 1) = 10/5 = 2

Since gradient AB = gradient AC, the points are collinear.

Worked Examination-Style Problem

Problem: The line (l_1) passes through A(−2, 5) and B(4, 2). The line (l_2) is perpendicular to (l_1) and passes through B.

(a) Find the equation of (l_1) in the form ax + by + c = 0.

(b) Find the equation of (l_2).

(d) Find the area of triangle ABC.

Solution:

(a)

Gradient of l₁ = (2 − 5)/(4 − (−2)) = −3/6 = −1/2

y − 2 = −1/2 (x − 4)
2(y − 2) = −(x − 4)
2y − 4 = −x + 4
x + 2y − 8 = 0

(b)

Gradient of l₂ = −1/(−1/2) = 2

y − 2 = 2(x − 4)
y = 2x − 6

0 = 2x − 6
x = 3

C = (3, 0).

(d)

AB = √[(4−(−2))² + (2−5)²] = √[36 + 9] = √45 = 3√5

The perpendicular distance from l₁ to C is the distance from C(3, 0) to line x + 2y − 8 = 0:

d = |3 + 0 − 8| / √(1² + 2²) = |−5| / √5 = 5/√5 = √5

Area = ½ × base × height = ½ × 3√5 × √5 = ½ × 15 = 7.5

The area of triangle ABC is 7.5 square units.

Summary

Gradient: m = (y₂ − y₁)/(x₂ − x₁).
Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2).
Distance: √[(x₂ − x₁)² + (y₂ − y₁)²].
Line equations: y = mx + c, y − y₁ = m(x − x₁), or ax + by + c = 0.
Parallel: equal gradients. Perpendicular: product of gradients = −1.
Perpendicular bisector: passes through midpoint, gradient is the negative reciprocal.
Collinearity: equal gradients between pairs of points.

Exam Tip: Always show your gradient calculations explicitly. When asked for a line equation in a specific form, make sure you rearrange fully. For perpendicular bisectors, state both the midpoint and the perpendicular gradient clearly before writing the equation — this makes your working easy for the examiner to follow and earns method marks even if you make an arithmetic slip.