Points A and B have coordinates (h, h) and (4h + 6, 5h) respectively. The equation of the perpendicular bisector of AB is 3x + 2y = k. Find the values of the constants h and k.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1] Gradient of the perpendicular bisector is given by 3x + 2y = k → 2y = -3x + k → y = -3/2 x + k/2. So, the gradient m_perp = -3/2.
2. [*M1] EITHER: The gradient of AB, m_AB = (y₂ - y₁) / (x₂ - x₁) = (5h - h) / ((4h + 6) - h) = 4h / (3h + 6).
3. [*M1] EITHER: For perpendicular lines, m_AB × m_perp = -1. So (4h / (3h + 6)) × (-3/2) = -1. This leads to -12h / (6h + 12) = -1, or 12h = 6h + 12. Arrive at an equation in h. (This mark is implied by a correct 'h' value or a correct equation that leads to it, e.g., 4h / (3h+6) = -2/3).
4. [A1] EITHER: 6h = 12, so h = 2.
5. [*M1] OR alternative start: Obtain equation of AB using gradient from m₁m₂ = − 1 and a point. The gradient of AB = -1 / (-3/2) = 2/3. Equation of AB: y - h = (2/3)(x - h).
6. [*M1] OR alternative start: Arrive at an equation in h. Substitute co-ordinates of one of the other points. Substitute (4h + 6, 5h) into the equation of AB: 5h - h = (2/3)(4h + 6 - h) → 4h = (2/3)(3h + 6).
7. [A1] OR alternative start: 12h = 6h + 12, so h = 2.
8. [B1FT] The midpoint M of AB has coordinates ((x₁ + x₂)/2, (y₁ + y₂)/2). M = ((h + 4h + 6)/2, (h + 5h)/2) = ((5h + 6)/2, 3h). Substitute h = 2: M = ((5(2) + 6)/2, 3(2)) = (16/2, 6) = (8, 6). Algebraic expression or FT for numerical answer from 'their h'.
9. [DM1] The midpoint M(8, 6) lies on the perpendicular bisector 3x + 2y = k. Substitute these coordinates: 3(8) + 2(6) = k. Substitutes ‘their midpoint' into 3x + 2y = k. If y = -3/2 x + c is used (expect c = 18) the method mark should be withheld until they ×2.
10. [A1] 24 + 12 = k, so k = 36. k = 36 soi.