Higher November 2021 Paper 2 Q19
19 The straight line \(\mathbf{L}\) has equation \(x - y = 3\)
The curve \(\mathbf{C}\) has equation \(3x^2 - y^2 + xy = 9\)
\(\mathbf{L}\) and \(\mathbf{C}\) intersect at the points \(P\) and \(Q\).
Find the coordinates of the midpoint of \(PQ\).
Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(y = x - 3\) or \(x = y + 3\) | B1 |
| eg \(3x^2 - (x - 3)^2 + x(x - 3) = 9\) or eg \(3(3 + y)^2 - y^2 + y(3 + y) = 9\) | M1 |
| eg \(3x^2 + 3x - 18\ (= 0)\) or \(x^2 + x - 6\ (= 0)\) or eg \(3y^2 + 21y + 18\ (= 0)\) or \(y^2 + 7y + 6\ (= 0)\) | M1ft |
eg \((x - 2)(x + 3)\ (= 0)\) \(x = \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -6}}{2 \times 1}\) eg \(\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2 = 6\) or eg \((y + 1)(y + 6)\ (= 0)\) \(y = \dfrac{-7 \pm \sqrt{7^2 - 4 \times 1 \times 6}}{2 \times 1}\) eg \(\left(y + \dfrac{7}{2}\right)^2 - \left(\dfrac{7}{2}\right)^2 = -6\) | M1 |
\(x = -3\), \(x = 2\) and \(y = -1\), \(y = -6\) or one correct midpoint coordinate ie \(x = -\dfrac{1}{2}\) or \(y = -\dfrac{7}{2}\) | A1 |
Working required Answer: \(\left(-\dfrac{1}{2}, -\dfrac{7}{2}\right)\) | A1 |
| (6) | |
| (6 marks) |
Notes
B1: for correct rearrangement of linear equation
M1: substitution of their linear equation into quadratic in \(x\) or \(y\) alone (even if B0 scored)
M1ft: from their substitution (dep on previous M1) for a complete correct method to get a 3-term or 2-term quadratic expression in the form \(ax^2 + bx\ (+ c)\ (= 0)\) [allow \(ax^2 + bx = c\)]
A1: (dep on M2) for \(x = 2\), \(x = -3\) and \(y = -1\), \(y = -6\) or one correct midpoint ie \(x = -\dfrac{1}{2}\) or \(y = -\dfrac{7}{2}\)
A1: (dep on M2) oe