A2 June 2019 Paper 2 Q6
6. In an Argand diagram, the points \(A\), \(B\) and \(C\) are the vertices of an equilateral triangle with its centre at the origin. The point \(A\) represents the complex number \(6 + 2\mathrm{i}\).
The points \(D\), \(E\) and \(F\) are the midpoints of the sides of triangle \(ABC\).
| Scheme | Marks | AO |
|---|---|---|
| Examples: \(\begin{pmatrix}\cos 120 & -\sin 120\\ \sin 120 & \cos 120\end{pmatrix}\begin{pmatrix}6\\ 2\end{pmatrix} = \ldots\) or \((6 + 2\mathrm{i})\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\right)\) or \(\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right) + \mathrm{i}\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\left(\dfrac{2\pi}{3}\right) + \mathrm{i}\sin\left(\dfrac{2\pi}{3}\right)\right)\) or \(\sqrt{40}\left(\cos\left(\arctan\left(\frac{2}{6}\right) + \frac{2\pi}{3}\right) + \mathrm{i}\sin\left(\arctan\left(\frac{2}{6}\right) + \frac{2\pi}{3}\right)\right)\) or \(\sqrt{40}\mathrm{e}^{\mathrm{i}\arctan\left(\frac{2}{6}\right)}\mathrm{e}^{\mathrm{i}\left(\frac{2\pi}{3}\right)}\) | M1 | 3.1a |
| \(\left(-3 - \sqrt{3}\right)\) or \(\left(3\sqrt{3} - 1\right)\mathrm{i}\) | A1 | 1.1b |
| \(\left(-3 - \sqrt{3}\right) + \left(3\sqrt{3} - 1\right)\mathrm{i}\) | A1 | 1.1b |
| Examples: \(\begin{pmatrix}\cos 240 & -\sin 240\\ \sin 240 & \cos 240\end{pmatrix}\begin{pmatrix}6\\ 2\end{pmatrix} = \ldots\) or \((6 + 2\mathrm{i})\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\right)\) or \(\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right) + \mathrm{i}\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\left(\dfrac{4\pi}{3}\right) + \mathrm{i}\sin\left(\dfrac{4\pi}{3}\right)\right)\) or \(\sqrt{40}\left(\cos\left(\arctan\left(\frac{2}{6}\right) + \frac{4\pi}{3}\right) + \mathrm{i}\sin\left(\arctan\left(\frac{2}{6}\right) + \frac{4\pi}{3}\right)\right)\) or \(\sqrt{40}\mathrm{e}^{\mathrm{i}\arctan\left(\frac{2}{6}\right)}\mathrm{e}^{\mathrm{i}\left(\frac{4\pi}{3}\right)}\) | M1 | 3.1a |
| \(\left(-3 + \sqrt{3}\right)\) or \(\left(-3\sqrt{3} - 1\right)\mathrm{i}\) | A1 | 1.1b |
| \(\left(-3 + \sqrt{3}\right) + \left(-3\sqrt{3} - 1\right)\mathrm{i}\) | A1 | 1.1b |
| (6) |
Notes
M1: Identifies a suitable method to rotate the given point by 120° (or equivalent) about the origin. May see equivalent work with modulus/argument or exponential form e.g. an attempt to multiply by \(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\) or \(\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\)
A1: Correct real part or correct imaginary part
A1: Completely correct complex number
M1: Identifies a suitable method to rotate the given point by 240° (or equivalent e.g. rotate their \(B\) by 120°) about the origin
May see equivalent work with modulus/argument or exponential form e.g. an attempt to multiply \(6 + 2\mathrm{i}\) by \(\cos\dfrac{4\pi}{3} + \mathrm{i}\sin\dfrac{4\pi}{3}\) or \(\mathrm{e}^{\frac{4\pi}{3}\mathrm{i}}\) or their \(B\) by \(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\) or \(\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\)
A1: Correct real part or correct imaginary part
A1: Completely correct complex number
Way 1
| Scheme | Marks | AO |
|---|---|---|
| Area \(ABC = 3 \times \dfrac{1}{2}\sqrt{6^2 + 2^2}\sqrt{6^2 + 2^2}\sin 120^\circ\) or Area \(AOB = \dfrac{1}{2}\sqrt{6^2 + 2^2}\sqrt{6^2 + 2^2}\sin 120^\circ\) | M1 | 2.1 |
| Area \(DEF = \dfrac{1}{4}ABC\) or \(\dfrac{3}{4}AOB\) | dM1 | 3.1a |
| \(= \dfrac{3}{8} \times 40 \times \dfrac{\sqrt{3}}{2} = \dfrac{15\sqrt{3}}{2}\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
In general, the marks in (b) should be awarded as follows:
M1: Attempts to find the area of a relevant triangle
dM1: completes the problem by multiplying by an appropriate factor to find the area of \(DEF\)
Dependent on the first method mark
A1: Correct exact area
In some cases it may not be possible to distinguish the 2 method marks. In such cases they can be awarded together for a direct method that finds the area of \(DEF\)
Examples:
Way 1
M1: A correct strategy for the area of a relevant triangle such as \(ABC\) or \(AOB\)
dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\) or with \(AOB\)
A1: Correct value
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(D\left(\dfrac{3 - \sqrt{3}}{2}, \dfrac{3\sqrt{3} + 1}{2}\right)\) \(OD = \sqrt{\left(\dfrac{3 - \sqrt{3}}{2}\right)^2 + \left(\dfrac{3\sqrt{3} + 1}{2}\right)^2} = \sqrt{10}\) Area \(DOF = \dfrac{1}{2}\sqrt{10}\sqrt{10}\sin 120^\circ\) | M1 | 2.1 |
| Area \(DEF = 3DOF\) | dM1 | 3.1a |
| \(= 3 \times \dfrac{1}{2} \times \sqrt{10}\sqrt{10} \times \dfrac{\sqrt{3}}{2} = \dfrac{15\sqrt{3}}{2}\) | A1 | 1.1b |
M1: A correct strategy for the area of a relevant triangle such as \(DOF\)
dM1: Completes the problem by linking the area of \(DEF\) correctly with \(DOF\)
A1: Correct value
Alternative: Way 3
| Scheme | Marks | AO |
|---|---|---|
| \(AB = \sqrt{\left(9 + \sqrt{3}\right)^2 + \left(3 - 3\sqrt{3}\right)^2} = \sqrt{120}\) Area \(ABC = \dfrac{1}{2}\sqrt{120}\sqrt{120}\sin 60^\circ\left(= 30\sqrt{3}\right)\) | M1 | 2.1 |
| Area \(DEF = \dfrac{1}{4}ABC\) | dM1 | 3.1a |
| \(= \dfrac{1}{4} \times 30\sqrt{3} = \dfrac{15\sqrt{3}}{2}\) | A1 | 1.1b |
M1: A correct strategy for the area of a relevant triangle such as \(ABC\)
dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\)
A1: Correct value
Alternative: Way 4
| Scheme | Marks | AO |
|---|---|---|
| \(D\left(\dfrac{3 - \sqrt{3}}{2}, \dfrac{3\sqrt{3} + 1}{2}\right),\ E(-3, -1),\ F\left(\dfrac{3 + \sqrt{3}}{2}, \dfrac{-3\sqrt{3} + 1}{2}\right)\) \(DE = \sqrt{\left(\dfrac{3 - \sqrt{3}}{2} + 3\right)^2 + \left(\dfrac{3\sqrt{3} + 1}{2} + 1\right)^2}\ \left(= \sqrt{30}\right)\) Area \(DEF = \dfrac{1}{2}\sqrt{30}\sqrt{30}\sin 60^\circ\) | M1 dM1 | 2.1 3.1a |
| \(= \dfrac{15\sqrt{3}}{2}\) | A1 | 1.1b |
M1dM1: A correct strategy for the area of \(DEF\). Finds 2 midpoints and attempts one side of \(DEF\) and uses a correct triangle area formula. By implication this scores both M marks.
A1: Correct value
Alternative: Way 5
| Scheme | Marks | AO |
|---|---|---|
| Area \(ABC = \dfrac{1}{2}\begin{vmatrix}6 & -3 - \sqrt{3} & \sqrt{3} - 3 & 6\\ 2 & 3\sqrt{3} - 1 & -3\sqrt{3} - 1 & 2\end{vmatrix} = 30\sqrt{3}\) | M1 | 2.1 |
| Area \(DEF = \dfrac{1}{4}ABC\) | dM1 | 3.1a |
| \(= \dfrac{1}{4} \times 30\sqrt{3} = \dfrac{15\sqrt{3}}{2}\) | A1 | 1.1b |
M1: A correct strategy for the area of \(ABC\) using the “shoelace” method.
dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\)
A1: Correct value
Note the marks in (b) can be scored using inexact answers from (a) and the A1 scored if an exact area is obtained.
