The points A and B have coordinates (–2, 15) and (3, 5) respectively. The perpendicular to the line AB at the point A (–2, 15) crosses the y-axis at the point C. Find the area of the triangle ABC.

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

1. [1] B1 for gradient of AB = (5-15)/(3-(-2)) = -10/5 = -2.
2. [1] M1 for using m₁m₂ = -1 to find the gradient of the perpendicular line AC = 1/2.
3. [1] M1 for finding the equation of the line AC: y - 15 = (1/2)(x + 2).
4. [1] A1 for finding coordinates of C: when x=0, y=16. So C(0, 16).
5. [1] M1 for a valid method to find the area of triangle ABC (e.g., using determinant method or base x height / 2). Base AB = sqrt((3-(-2))² + (5-15)²) = sqrt(5² + (-10)²) = sqrt(25 + 100) = sqrt(125). Height AC = sqrt((0-(-2))² + (16-15)²) = sqrt(2² + 1²) = sqrt(4+1) = sqrt(5).
6. [1] A1 for Area = 1/2 * sqrt(125) * sqrt(5) = 1/2 * sqrt(625) = 1/2 * 25 = 12.5.