Coordinate Geometry
This domain tests the distance formula, the section and midpoint formulas, and the area of a triangle from coordinates, including collinearity.
Core rules
- Distance between (x₁, y₁) and (x₂, y₂): √[(x₂ − x₁)² + (y₂ − y₁)²].
- Section formula for division in the ratio m₁ : m₂: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
- Midpoint: ((x₁ + x₂)/2, (y₁ + y₂)/2), the 1 : 1 case of the section formula.
- Area of a triangle: ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
- Three points are collinear exactly when the area of the triangle they form is zero.
Worked example
Find the point dividing the join of A(2, 3) and B(6, 7) in the ratio 1 : 3. Using m₁ = 1 and m₂ = 3: x = (1 × 6 + 3 × 2)/(1 + 3) = 12/4 = 3. y = (1 × 7 + 3 × 3)/4 = 16/4 = 4. The point is (3, 4). Check by distance: AP = √[(1)² + (1)²] = √2 and PB = √[(3)² + (3)²] = 3√2, so AP : PB = 1 : 3.
Traps
- Reversing m₁ and m₂ in the section formula.
- Dropping the absolute value in the area formula and reporting a negative area.
- Confusing the area formula with the distance formula.
- Forgetting that the midpoint is the 1 : 1 case.