Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Overview
Coordinate geometry is formula-driven — accuracy comes from labelling the points (x₁,y₁), (x₂,y₂) before substituting, and watching signs.
1. Core Formulae
Distance: √[(x₂ − x₁)² + (y₂ − y₁)²]
Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2)
Section (m:n): ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))
Slope: m = (y₂ − y₁)/(x₂ − x₁)
Area of triangle: ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
2. Straight Lines
Slope-intercept: y = mx + c
Point-slope: y − y₁ = m(x − x₁)
Parallel lines: equal slopes | Perpendicular lines: m₁ · m₂ = −1
Worked Examples
Example 1 — Distance
Find the distance between A(2, 3) and B(5, 7).
d = √[(5−2)² + (7−3)²] = √[9 + 16] = √25 = 5 units
Example 2 — Section formula
Find the point dividing (1, 2) and (4, 8) internally in ratio 1:2.
x = (1·4 + 2·1)/(1+2) = 6/3 = 2
y = (1·8 + 2·2)/(1+2) = 12/3 = 4 → (2, 4)
Example 3 — Area of a triangle
Vertices (1, 2), (4, 6), (7, 2).
Area = ½ |1(6−2) + 4(2−2) + 7(2−6)| = ½ |4 + 0 − 28| = ½ × 24 = 12 sq.units
Example 4 — Collinearity
Show (1,1), (2,3), (3,5) are collinear.
Area = ½ |1(3−5) + 2(5−1) + 3(1−3)| = ½ |−2 + 8 − 6| = 0 → collinear
Example 5 — Slope and line
Find the equation of the line through (2, 3) with slope 4.
y − 3 = 4(x − 2) → y = 4x − 5
Common Mistakes
- Swapping x and y in the slope or distance formula.
- Dropping the square root in the distance formula.
- Forgetting the absolute value (modulus) in the area formula.
- Mixing up the section-formula order of m, n.
Exam Tips & Revision Checklist
- Label points before substituting; recheck signs.
- Area = 0 ⇒ collinear is a common 2-mark question.
- Practised: distance, midpoint, section, slope, area, collinearity, line equation.
Frequently Asked Questions
Distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²].