Higher November 2020 Paper 2R Q22
22 The curve with equation \(x^2 - x + y^2 = 10\) and the straight line with equation \(x - y = -4\) intersect at the points \(A\) and \(B\).
Work out the exact length of \(AB\).
Show your working clearly and give your answer in the form \(\dfrac{\sqrt{a}}{2}\) where \(a\) is an integer.
(6)
| Scheme | Marks |
|---|---|
| \((y - 4)^2 - (y - 4) + y^2 = 10\) or \(x^2 - x + (x + 4)^2 = 10\) | M1 |
| \(2y^2 - 9y + 10 = 0\) or \(2x^2 + 7x + 6 = 0\) | A1 |
\((2y - 5)(y - 2) = 0\) or \(\dfrac{--9 \pm \sqrt{(-9)^2 - (4 \times 2 \times 10)}}{2 \times 2}\) or \((2x + 3)(x + 2) = 0\) or \(\dfrac{-7 \pm \sqrt{7^2 - (4 \times 2 \times 6)}}{2 \times 2}\) | M1ft |
| (−1.5, 2.5) and (−2, 2) | A1 |
| \(\sqrt{(\text{“}-1.5\text{”} - \text{“}-2\text{”})^2 + (\text{“}2.5\text{”} - \text{“}2\text{”})^2}\) | M1 |
| \(\dfrac{\sqrt{2}}{2}\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for substituting linear equation into the quadratic equation
A1: for a correct equation in the form \(ax^2 + bx + c = 0\)
or \(ax^2 + bx = -c\)
or equations of the same form but in \(y\)
M1ft: For solving their 3 term quadratic equation using any correct method.
If factorising, allow brackets which expanded give 2 out of 3 terms correct (if using formula or completing the square allow one sign error and some simplification – allow as far as eg
\(\dfrac{-7 \pm \sqrt{49 - 48}}{4}\) or eg \(\left(x + \dfrac{7}{4}\right)^2 - \dfrac{1}{16} = 0\) oe
\(\dfrac{9 \pm \sqrt{81 - 80}}{4}\) or eg \(\left(y - \dfrac{9}{4}\right)^2 - \dfrac{1}{16} = 0\) oe
A1: for both pairs of coordinates
oe eg \(\left(\dfrac{-3}{2}, \dfrac{5}{2}\right)\)
accept coordinates listed as pairs, ie \(x_1\), \(y_1\), \(x_2\), \(y_2\)
M1: dep on M1 for finding length of \(AB\)
A1: dep M3