Higher November 2024 Paper 2 Q21
21 The point \(P\) has coordinates \((-4, 5)\)
The point \(Q\) has coordinates \((6, -6)\)
The point \(R\) has coordinates \((k, k + 3)\)
Given that angle \(PRQ\) is a right angle,
find the possible values of \(k\).
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-6\), 3.5 | P1 | for process to find gradient of \(PR\) or \(QR\), eg \(\dfrac{k + 3 - 5}{k - -4}\left(= \dfrac{k - 2}{k + 4}\right)\) or \(\dfrac{k + 3 - -6}{k - 6}\left(= \dfrac{k + 9}{k - 6}\right)\) or \(\dfrac{5 - (k + 3)}{-4 - k}\left(= \dfrac{2 - k}{-4 - k}\right)\) or \(\dfrac{-6 - (k + 3)}{6 - k}\left(= \dfrac{-9 - k}{6 - k}\right)\) OR for start of process to use Pythagoras, eg \(PR^2 = (k - -4)^2 + (k + 3 - 5)^2\) or \(RP^2 = (-4 - k)^2 + (5 - (k + 3))^2\) or \(QR^2 = (k - 6)^2 + (k + 3 - -6)^2\) or \(RQ^2 = (6 - k)^2 + (-6 - (k + 3))^2\) |
| P1 | for forming a correct equation, eg \(\dfrac{k + 3 - 5}{k - -4} \times \dfrac{k + 3 - -6}{k - 6} = -1\) or \(\dfrac{k - 2}{k + 4} \times \dfrac{k + 9}{k - 6} = -1\) or \(\dfrac{5 - (k + 3)}{-4 - k} \times \dfrac{-6 - (k + 3)}{6 - k} = -1\) or \(\dfrac{2 - k}{-4 - k} \times \dfrac{-9 - k}{6 - k} = -1\) oe OR \((k - -4)^2 + (k + 3 - 5)^2 + (k - 6)^2 + (k + 3 - -6)^2 = (-6 - 5)^2 + (6 - -4)^2\) or \((k + 4)^2 + (k - 2)^2 + (k - 6)^2 + (k + 9)^2 = (-11)^2 + 10^2\) | |
| P1 | for writing in the form \(ak^2 + bk + c\ (= 0)\), eg \(2k^2 + 5k - 42\ (= 0)\) or \(4k^2 + 10k - 84\ (= 0)\) | |
| P1 | (dep P3) for factorising, eg \((k + 6)(2k - 7)\ (= 0)\) or \((2k + 12)(2k - 7)\) or \((k + 6)(4k - 14)\) OR use of formula, eg \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 2 \times -42}}{2 \times 2}\) or \(\dfrac{-10 \pm \sqrt{10^2 - 4 \times 4 \times -84}}{2 \times 4}\) | |
| A1 | cao SCB4 for correct answer coming from consistent use of reciprocals of gradients |
Additional guidance
Condone missing bracket for first P1 only
Stating eg Grad \(PR = \dfrac{k - -4}{k + 3 - 5}\)
Or grad \(QR = \dfrac{k - 6}{k + 3 - -6}\) scores P0 but check SC