## Question 7:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Single straight line with negative gradient through both axes | M1 | Must pass through both axes; ignore line joining $(3,0)$ and $(0,-6)$ |
| Line has negative $y$-intercept, passing through both axes with negative gradient | A1 | Ignore any intercept marked; ignore line joining $(3,0)$ and $(0,-6)$ |
| Shading above the line | B1 | Must shade above line, not a bounded triangular region |

### Part (ii) — Main Method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $m = \tan\!\left(\frac{\pi}{3}\right) = \sqrt{3}$ and $y - 0 = m(x-2)$, giving $y = \sqrt{3}x - 2\sqrt{3}$ | M1 | Find Cartesian equations for both loci using gradient $= \tan(\text{argument})$; must attempt both equations but one correct scores M1 |
| One equation correct | A1 | Need not be simplified |
| $m = \tan\!\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$ and $y - 0 = m(x-(-1))$, giving $y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}$ | A1 | Both equations correct, need not be simplified |
| $\sqrt{3}x - 2\sqrt{3} = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3} \Rightarrow x = \ldots$ | M1 | Solve simultaneously to find real or imaginary component |
| $y = \sqrt{3}\!\left(\frac{7}{2}\right) - 2\sqrt{3} = \ldots$ | M1 | Find the other component |
| $w = \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\,\mathrm{i}$ | A1 | Correct exact answer; if left as coordinate scores A0; if defines $w=a+b\mathrm{i}$ and states $a=\frac{7}{2}$, $b=\frac{3\sqrt{3}}{2}$ scores A1 |

### Part (ii) — Alternative Method (Cartesian with triangles):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\!\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3} = \frac{y}{x_{-1}}$ and $\tan\!\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{y}{x_2}$, giving $x_{-1} = \sqrt{3}\,y$, $x_2 = \frac{\sqrt{3}}{3}\,y$ | M1, A1, A1 | Use both arguments to form equations in $x$ and $y$; one then two correct triangle expressions |
| $y\sqrt{3} = y\dfrac{\sqrt{3}}{3} + 3 \Rightarrow y = \ldots$ | M1 | Must come from $x_2 = x_{-1} + 3$ |
| Uses $x = y\sqrt{3} - 1$ or $x = \frac{\sqrt{3}}{3}y + 2$ with their $y$ | M1 | Find $x$ value |
| $w = \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\,\mathrm{i}$ | A1 | Correct exact answer |

### Part (ii) — Alternative 3 (Modulus-argument geometry):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = 3\sin\!\left(\frac{\pi}{3}\right)$ and $c = 3\cos\!\left(\frac{\pi}{3}\right)$ | M1, A1, A1 | Use correct geometry to form equations in $a$ and $c$; one then two correct equations |
| $b = 3\sin\!\left(\frac{\pi}{3}\right) = \ldots$ | M1 | Finds imaginary component |
| $a = 2 + 3\cos\!\left(\frac{\pi}{3}\right) = \ldots$ | M1 | Uses $2 +$ their $c$ to find real component |
| $w = \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\,\mathrm{i}$ | A1 | Correct exact answer |

### Part (ii) — Alternative 4 (Sine rule):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{3}{\sin 30} = \dfrac{AB}{\sin 120}$, giving $AB = 3\sqrt{3}$ | M1, A1 | Use sine rule to find $AB$; correct length |
| $\sin 30 = \dfrac{BC}{3\sqrt{3}} \Rightarrow BC = \dfrac{3}{2}\sqrt{3}$ and $\sin 60 = \dfrac{AC}{3\sqrt{3}} \Rightarrow AC = \dfrac{7}{2}$ | M1, A1 | Use trigonometry to find real or imaginary component; correct value |
| Uses trigonometry to find the other component | M1 | |
| $w = \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\,\mathrm{i}$ | A1 | Correct exact answer |