Question 6 - A-Level Maths

Edexcel CP2 2019 June — Question 6 9 marks

Exam Board	Edexcel
Module	CP2 (Core Pure 2)
Year	2019
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Complex Numbers Argand & Loci
Type	Geometric relationships on Argand diagram
Difficulty	Standard +0.8 This question requires understanding of rotation in the complex plane (multiplying by e^(i2π/3) or equivalent), converting between forms, and recognizing that the medial triangle has 1/4 the area of the original. While the concepts are standard for Further Maths, the multi-step nature and need to work with exact surds/complex arithmetic elevates it above routine exercises.
Spec	4.02a Complex numbers: real/imaginary parts, modulus, argument 4.02k Argand diagrams: geometric interpretation

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 \(6 + 2 \mathrm { i }\).
1. Find the complex numbers represented by the points \(B\) and \(C\), giving your answers in the form \(x + \mathrm { i } y\), where \(x\) and \(y\) are real and exact.
The points \(D , E\) and \(F\) are the midpoints of the sides of triangle \(A B C\).
Find the exact area of triangle \(D E F\).

Show mark scheme Show mark scheme source

Question 6:

Part (a) - First rotation (120°)

Answer	Marks	Guidance
Working	Mark	Guidance
\(\begin{pmatrix}\cos120 & -\sin120\\ \sin120 & \cos120\end{pmatrix}\begin{pmatrix}6\\2\end{pmatrix}\) or \((6+2i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\) or \(\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right)+i\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)\)	M1	3.1a - Identifies suitable method to rotate by 120° about origin
\(\left(-3-\sqrt{3}\right)\) or \(\left(3\sqrt{3}-1\right)i\)	A1	1.1b - Correct real or imaginary part
\(\left(-3-\sqrt{3}\right)+\left(3\sqrt{3}-1\right)i\)	A1	1.1b - Completely correct complex number

Second rotation (240°)

Answer	Marks	Guidance
Working	Mark	Guidance
\(\begin{pmatrix}\cos240 & -\sin240\\ \sin240 & \cos240\end{pmatrix}\begin{pmatrix}6\\2\end{pmatrix}\) or \((6+2i)\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\) or \(\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right)+i\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\)	M1	3.1a - Identifies suitable method to rotate by 240° about origin
\(\left(-3+\sqrt{3}\right)\) or \(\left(-3\sqrt{3}-1\right)i\)	A1	1.1b - Correct real or imaginary part
\(\left(-3+\sqrt{3}\right)+\left(-3\sqrt{3}-1\right)i\)	A1	1.1b - Completely correct complex number

Part (b) - Area of DEF

Way 1:

Answer	Marks	Guidance
Working	Mark	Guidance
Area \(ABC = 3\times\frac{1}{2}\sqrt{6^2+2^2}\sqrt{6^2+2^2}\sin120°\) or Area \(AOB = \frac{1}{2}\sqrt{6^2+2^2}\sqrt{6^2+2^2}\sin120°\)	M1	2.1 - Correct strategy for area of relevant triangle
Area \(DEF = \frac{1}{4}ABC\) or \(\frac{3}{4}AOB\)	dM1	3.1a - Completes by linking area of \(DEF\) correctly; dependent on first M1
\(=\frac{3}{8}\times40\times\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}\)	A1	1.1b - Correct exact area

Way 2:

Answer	Marks	Guidance
Working	Mark	Guidance
\(OD=\sqrt{\left(\frac{3-\sqrt{3}}{2}\right)^2+\left(\frac{3\sqrt{3}+1}{2}\right)^2}=\sqrt{10}\); Area \(DOF=\frac{1}{2}\sqrt{10}\sqrt{10}\sin120°\)	M1	2.1
Area \(DEF = 3DOF\)	dM1	3.1a
\(=3\times\frac{1}{2}\times\sqrt{10}\sqrt{10}\times\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}\)	A1	1.1b

Way 3:

Answer	Marks	Guidance
Working	Mark	Guidance
\(AB=\sqrt{(9+\sqrt{3})^2+(3-3\sqrt{3})^2}=\sqrt{120}\); Area \(ABC=\frac{1}{2}\sqrt{120}\sqrt{120}\sin60°\left(=30\sqrt{3}\right)\)	M1	2.1
Area \(DEF=\frac{1}{4}ABC\)	dM1	3.1a
\(=\frac{1}{4}\times30\sqrt{3}=\frac{15\sqrt{3}}{2}\)	A1	1.1b

Way 4:

Answer	Marks	Guidance
Working	Mark	Guidance
\(D\left(\frac{3-\sqrt{3}}{2},\frac{3\sqrt{3}+1}{2}\right)\), \(E(-3,-1)\), \(F\left(\frac{3+\sqrt{3}}{2},\frac{-3\sqrt{3}+1}{2}\right)\); \(DE=\sqrt{\left(\frac{3-\sqrt{3}}{2}+3\right)^2+\left(\frac{3\sqrt{3}+1}{2}+1\right)^2}\left(=\sqrt{30}\right)\)	M1	2.1
Area \(DEF=\frac{1}{2}\sqrt{30}\sqrt{30}\sin60°\)	dM1	3.1a
\(=\frac{15\sqrt{3}}{2}\)	A1	1.1b

Way 5:

Answer	Marks	Guidance
Working	Mark	Guidance
Area \(ABC=\frac{1}{2}\begin{vmatrix}6 & -3-\sqrt{3} & \sqrt{3}-3 & 6\\ 2 & 3\sqrt{3}-1 & -3\sqrt{3}-1 & 2\end{vmatrix}=30\sqrt{3}\)	M1	2.1 - Shoelace method
Area \(DEF=\frac{1}{4}ABC\)	dM1	3.1a
\(=\frac{1}{4}\times30\sqrt{3}=\frac{15\sqrt{3}}{2}\)	A1	1.1b

> Note: Marks in (b) can be scored using inexact answers from (a) and A1 scored if exact area obtained.

# Question 6:

## Part (a) - First rotation (120°)

| Working | Mark | Guidance |
|---------|------|----------|
| $\begin{pmatrix}\cos120 & -\sin120\\ \sin120 & \cos120\end{pmatrix}\begin{pmatrix}6\\2\end{pmatrix}$ or $(6+2i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$ or $\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right)+i\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)$ | M1 | 3.1a - Identifies suitable method to rotate by 120° about origin |
| $\left(-3-\sqrt{3}\right)$ or $\left(3\sqrt{3}-1\right)i$ | A1 | 1.1b - Correct real or imaginary part |
| $\left(-3-\sqrt{3}\right)+\left(3\sqrt{3}-1\right)i$ | A1 | 1.1b - Completely correct complex number |

**Second rotation (240°)**

| Working | Mark | Guidance |
|---------|------|----------|
| $\begin{pmatrix}\cos240 & -\sin240\\ \sin240 & \cos240\end{pmatrix}\begin{pmatrix}6\\2\end{pmatrix}$ or $(6+2i)\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$ or $\sqrt{40}\left(\cos\arctan\left(\frac{2}{6}\right)+i\sin\arctan\left(\frac{2}{6}\right)\right)\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)$ | M1 | 3.1a - Identifies suitable method to rotate by 240° about origin |
| $\left(-3+\sqrt{3}\right)$ or $\left(-3\sqrt{3}-1\right)i$ | A1 | 1.1b - Correct real or imaginary part |
| $\left(-3+\sqrt{3}\right)+\left(-3\sqrt{3}-1\right)i$ | A1 | 1.1b - Completely correct complex number |

---

## Part (b) - Area of DEF

**Way 1:**

| Working | Mark | Guidance |
|---------|------|----------|
| Area $ABC = 3\times\frac{1}{2}\sqrt{6^2+2^2}\sqrt{6^2+2^2}\sin120°$ or Area $AOB = \frac{1}{2}\sqrt{6^2+2^2}\sqrt{6^2+2^2}\sin120°$ | M1 | 2.1 - Correct strategy for area of relevant triangle |
| Area $DEF = \frac{1}{4}ABC$ or $\frac{3}{4}AOB$ | dM1 | 3.1a - Completes by linking area of $DEF$ correctly; dependent on first M1 |
| $=\frac{3}{8}\times40\times\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}$ | A1 | 1.1b - Correct exact area |

**Way 2:**

| Working | Mark | Guidance |
|---------|------|----------|
| $OD=\sqrt{\left(\frac{3-\sqrt{3}}{2}\right)^2+\left(\frac{3\sqrt{3}+1}{2}\right)^2}=\sqrt{10}$; Area $DOF=\frac{1}{2}\sqrt{10}\sqrt{10}\sin120°$ | M1 | 2.1 |
| Area $DEF = 3DOF$ | dM1 | 3.1a |
| $=3\times\frac{1}{2}\times\sqrt{10}\sqrt{10}\times\frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}$ | A1 | 1.1b |

**Way 3:**

| Working | Mark | Guidance |
|---------|------|----------|
| $AB=\sqrt{(9+\sqrt{3})^2+(3-3\sqrt{3})^2}=\sqrt{120}$; Area $ABC=\frac{1}{2}\sqrt{120}\sqrt{120}\sin60°\left(=30\sqrt{3}\right)$ | M1 | 2.1 |
| Area $DEF=\frac{1}{4}ABC$ | dM1 | 3.1a |
| $=\frac{1}{4}\times30\sqrt{3}=\frac{15\sqrt{3}}{2}$ | A1 | 1.1b |

**Way 4:**

| Working | Mark | Guidance |
|---------|------|----------|
| $D\left(\frac{3-\sqrt{3}}{2},\frac{3\sqrt{3}+1}{2}\right)$, $E(-3,-1)$, $F\left(\frac{3+\sqrt{3}}{2},\frac{-3\sqrt{3}+1}{2}\right)$; $DE=\sqrt{\left(\frac{3-\sqrt{3}}{2}+3\right)^2+\left(\frac{3\sqrt{3}+1}{2}+1\right)^2}\left(=\sqrt{30}\right)$ | M1 | 2.1 |
| Area $DEF=\frac{1}{2}\sqrt{30}\sqrt{30}\sin60°$ | dM1 | 3.1a |
| $=\frac{15\sqrt{3}}{2}$ | A1 | 1.1b |

**Way 5:**

| Working | Mark | Guidance |
|---------|------|----------|
| Area $ABC=\frac{1}{2}\begin{vmatrix}6 & -3-\sqrt{3} & \sqrt{3}-3 & 6\\ 2 & 3\sqrt{3}-1 & -3\sqrt{3}-1 & 2\end{vmatrix}=30\sqrt{3}$ | M1 | 2.1 - Shoelace method |
| Area $DEF=\frac{1}{4}ABC$ | dM1 | 3.1a |
| $=\frac{1}{4}\times30\sqrt{3}=\frac{15\sqrt{3}}{2}$ | A1 | 1.1b |

> **Note:** Marks in (b) can be scored using inexact answers from (a) and A1 scored if exact area obtained.

---

Show LaTeX source

\begin{enumerate}
  \item 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 $6 + 2 \mathrm { i }$.\\
(a) Find the complex numbers represented by the points $B$ and $C$, giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real and exact.
\end{enumerate}

The points $D , E$ and $F$ are the midpoints of the sides of triangle $A B C$.\\
(b) Find the exact area of triangle $D E F$.

\hfill \mbox{\textit{Edexcel CP2 2019 Q6 [9]}}

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