Analytical Geometry — Grade 11 & 12 Maths (CAPS)

Equation of a line through two points, lines parallel and perpendicular to a given line, and the angle of inclination θ where m = tanθ.

1The Grade 10 toolkit, revised

d = √((x₂ − x₁)² + (y₂ − y₁)²)
m = (y₂ − y₁)/(x₂ − x₁) (x₂ ≠ x₁; a vertical line has no gradient)
M = ( (x₁ + x₂)/2 ; (y₁ + y₂)/2 )

2Equation of a line

y − y₁ = m(x − x₁)

Worked ExampleFind the equation of the line through A(−1 ; 2) and B(3 ; 10)

1
m = (10 − 2)/(3 − (−1)) = 8/4 = 2.
Gradient first, always.
2
y − 2 = 2(x − (−1)).
Substitute either point — A is used here.
3
y − 2 = 2x + 2.
4
y = 2x + 4.
Check with B: 2(3) + 4 = 10 ✓

3Parallel and perpendicular

Parallel: m₁ = m₂
Perpendicular: m₁ × m₂ = −1

💡

Vertical lines are the exception. A vertical line has no gradient, so the product rule cannot be used for it: a line perpendicular to a vertical line is horizontal (gradient 0), and the other way round. For two non-vertical lines, perpendicular means m₁ × m₂ = −1.

Worked ExampleFind the line through (2 ; 5) perpendicular to y = 2x + 4

1
m of the given line = 2.
The coefficient of x.
2
Perpendicular gradient: 2 × m = −1 → m = −1/2.
The negative reciprocal — flip it and change the sign.
3
y − 5 = −1/2(x − 2).
4
y = −1/2x + 6.

4Angle of inclination

m = tanθ where 0° ≤ θ < 180°

θ is the angle the line makes with the positive x-axis, measured anticlockwise.

Worked ExampleDetermine the angle of inclination of y = 2x + 4

1
tanθ = 2.
The gradient is the tangent of the inclination.
2
θ = tan⁻¹(2) = 63,43°.
3
θ ≈ 63,4°.
Positive gradient ⇒ acute angle. Sensible.

Worked ExampleDetermine the angle of inclination of a line with m = −1

1
tanθ = −1 gives ref = 45°.
Take tan⁻¹ of the positive value.
2
A negative gradient means θ is obtuse.
The line slopes down to the right.
3
θ = 180° − 45° = 135°.

⚠️

A calculator gives tan⁻¹(−1) = −45°. That is not an angle of inclination — θ must lie in [0° ; 180°), so add 180° to get 135°.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Determine the gradient of the line through A(−1 ; 2) and B(3 ; 10).
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(10 − 2)/(3 + 1) = 2
2
Determine the equation of the line through A(−1 ; 2) and B(3 ; 10).
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y = 2x + 4
3
Determine the length of AB where A(−1 ; 2) and B(3 ; 10).
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√(4² + 8²) = √80 = 4√5 ≈ 8,94
4
Determine the midpoint of AB where A(−1 ; 2) and B(3 ; 10).
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(1 ; 6)
5
Determine the equation of the line through (2 ; 5) perpendicular to y = 2x + 4.
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m = −1/2 → y = −1/2x + 6
6
Determine the equation of the line through (0 ; 1) parallel to y = 2x + 4.
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y = 2x + 1
7
Determine the angle of inclination of y = 2x + 4 (one decimal).
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tanθ = 2 → θ ≈ 63,4°
8
Determine the angle of inclination of a line with gradient −1.
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ref = 45°, obtuse → θ = 135°