Coordinate Geometry: Distance, Section Formula and Area of a Triangle
Coordinate Geometry tests the distance formula, section formula, midpoint and the area of a triangle - usually as 1-mark and 3-mark questions.
Formulas
- Distance: √[(x₂ − x₁)² + (y₂ − y₁)²]; distance from the origin is √(x² + y²).
- Midpoint of AB: ((x₁ + x₂)/2, (y₁ + y₂)/2).
- Section formula, P dividing AB in m₁ : m₂: ((m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)).
- Centroid: ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3).
- Area of a triangle: ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Three points are collinear exactly when this area is 0.
Worked example
Find the point dividing the join of A(2, 3) and B(6, 7) in the ratio 1 : 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 = √(1² + 1²) = √2 and PB = √(3² + 3²) = 3√2, a ratio of 1 : 3 ✓.
Useful techniques
To find an unknown ratio, take it as k : 1 and apply the section formula. To identify a quadrilateral, compute all four sides and both diagonals with the distance formula; equal sides with equal diagonals indicate a square, equal diagonals only a rectangle.
Traps that cost marks
- In the section formula m₁ multiplies the coordinates of the second point - mixing the order is the standard error.
- Do not drop the absolute value or the factor ½ in the area formula.
- Subtracting a negative coordinate needs brackets written out, or signs go wrong.
- "Find k so that the points are collinear" means set the area equal to zero.