Coordinate Geometry
The exam tests the distance formula, the section and midpoint formulas, area of a triangle from coordinates, and collinearity, all on the Cartesian plane.
Core rules
- Distance between (x₁, y₁) and (x₂, y₂): √[(x₂ − x₁)² + (y₂ − y₁)²].
- Distance from the origin: √(x² + y²).
- Section formula for dividing AB in ratio m₁ : m₂: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
- Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2), the case m₁ = m₂ = 1.
- Area of a triangle: ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
- Points are collinear when this area is zero.
Worked example
Find the distance between A(3, −2) and B(−1, 4). Subtract coordinates: (−1 − 3) = −4 and (4 − (−2)) = 6. Then AB = √[(−4)² + 6²] = √(16 + 36) = √52 = 2√13 units.
Traps
Reversing the order in the section formula, which swaps the ratio; dropping the absolute value in the area formula and reporting a negative area; confusing distance from the x-axis (|y|) with distance from the y-axis (|x|); forgetting to square both differences under the root; not simplifying surds.