Higher January 2019 Paper 2R Q21
21 The curve with equation \(y = (10x - 3)(x + 1)\) and the line with equation \(y - 6x = 0\) intersect at the points \(A\) and \(B\).
Find the coordinates of the midpoint of \(AB\).
Show your working clearly.
(6)
| Scheme | Marks |
|---|---|
| \((10x - 3)(x + 1) = 6x\) | M1 |
| \(10x^2 + x - 3\) (= 0) | M1 |
\((5x + 3)(2x - 1)\) (= 0) or \(x = \dfrac{-1 \pm \sqrt{1^2 - (4 \times 10 \times -3)}}{2 \times 10}\) or \(10(x + 0.05)^2 - 0.025 - 3 = 0\) | M1 |
| \(x = -0.6\) and \(x = 0.5\) (\(y = -3.6\) and \(y = 3\)) | A1 |
| \(\dfrac{\text{‘}{-0.6}\text{’} + \text{‘}0.5\text{’}}{2}\) or \(\dfrac{\text{‘}{-3.6}\text{’} + \text{‘}3\text{’}}{2}\) oe | M1 |
| (−0.05, −0.3) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for a correct equation to find points \(A\) and \(B\)
M1: for rearranging equation in the form \(ax^2 + bx + c\) (= 0)
M1: dep on M1 for solving the quadratic equation using factorisation or using the formula or by completing the square
A1: Both \(x\) values correct dep on M2
M1: dep on M1