# Question 6(a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{2}$ | B1 | Accept exact equivalents |

# Question 6(a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0, \quad 3x + 5 = 0 \Rightarrow x = -\frac{5}{3}$ | M1A1 | M1: Sets $y = 0$ and attempts to find $x$. Accept as evidence $3x+5=0 \Rightarrow x=..$ or awrt $-1.7$. A1: $x = -\frac{5}{3}$ or exact equivalent including $1.\dot{6}$ **(i.e. a clear dot over the 6)** |

# Question 6(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $l_2 = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$ | M1 | Uses $m_2 = -\frac{1}{m_1}$ to find the gradient of $l_2$ (may be implied by their line equation). Allow an attempt to find $m_2$ from $m_1 \times m_2 = -1$ |
| Point $B$ has $y$ coordinate of 4 | B1 | This may be embedded within the equation of the line but must be seen in part (b). |
| e.g. $y - \text{'4'} = -\frac{2}{3}(x-1)$ or $\frac{y-\text{'4'}}{x-1} = -\frac{2}{3}$ | M1 | A correct straight line method with a changed gradient and their point $(1, \text{'4'})$. There must have been attempt to find the $y$ coordinate of $B$. If using $y = mx + c$, must reach as far as finding a value for $c$. |
| e.g. $y - 4 = -\frac{2}{3}(x-1)$ or $\frac{y-4}{x-1} = -\frac{2}{3}$ | A1 | A correct un-simplified equation |
| $2x + 3y - 14 = 0$ | A1 | Accept $A(2x+3y-14)=0$ where $A$ is an integer. Terms can be in any order but must have '$= 0$'. |

# Question 6(b) Alt:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient $l_2 = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$ | M1 | Uses $m_2 = -\frac{1}{m_1}$ to find the gradient of $l_2$ as before |
| $\frac{3}{2}x + \frac{5}{2} = -\frac{2}{3}x + c$ | B1 | A correct statement for $l_1 = l_2$ |
| $x = 1 \Rightarrow c = \frac{14}{3}$ | M1 | Substitutes $x = 1$ to find a value for $c$ |
| $y = -\frac{2}{3}x + \frac{14}{3}$ | A1 | Correct equation |
| $2x + 3y - 14 = 0$ | A1 | Accept $A(2x+3y-14)=0$ where $A$ is an integer. |

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# Question 6(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 2x - 14 = 0 \Rightarrow x = 7$ | M1 | Attempts to find $C$ using $y = 0$ in the equation obtained in part (b) |
| $= \frac{1}{2} \times \left(\text{'7'} + \text{'}\frac{5}{3}\text{'}\right) \times \text{'4'}$ | M1 | Attempts area of triangle using $\frac{1}{2} \times AC \times (y \text{ coord of } B)$ |
| $= \frac{52}{3}$ | A1 | Area $= \frac{52}{3}$ or exact equivalent e.g. $17\frac{1}{3}$ or $17.\dot{3}$ **(i.e. a clear dot over the 3)** |

# Question 6(c) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 2x - 14 = 0 \Rightarrow x = 7$ | M1 | Attempts to find $C$ using $y = 0$ in the equation obtained in part (b) |
| $\frac{1}{2}AB \times BC = \frac{1}{2}\sqrt{\frac{208}{9}} \times \sqrt{52}$ | M1 | A complete method for the area including correct attempts at finding $AB$ and $BC$ using their values. |
| $= \frac{52}{3}$ | A1 | Area $= \frac{52}{3}$ or exact equivalent e.g. $17\frac{1}{3}$ or $17.\dot{3}$ **(i.e. a clear dot over the 3)** |

# Question 6(c) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 2x - 14 = 0 \Rightarrow x = 7$ | M1 | Attempts to find $C$ using $y = 0$ in the equation obtained in part (b) |
| $\frac{1}{2}\begin{vmatrix} 1 & 7 & -\frac{5}{3} & 1 \\ 4 & 0 & 0 & 4 \end{vmatrix} = \frac{1}{2}\left\vert -\frac{20}{3} - 28\right\vert$ | M1 | Uses shoelace method. Must see a correct method including $\frac{1}{2}$. |
| $= \frac{52}{3}$ | A1 | Area $= \frac{52}{3}$ or exact equivalent e.g. $17\frac{1}{3}$ or $17.\dot{3}$ **(i.e. a clear dot over the 3)** |

# Question 6(c) Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow 2x - 14 = 0 \Rightarrow x = 7$ | M1 | Attempts to find $C$ using $y = 0$ in the equation obtained in part (b) |
| $\int_{-\frac{5}{3}}^{1}\left(\frac{3x}{2}+\frac{5}{2}\right)dx + \int_{1}^{7}\left(-\frac{2x}{3}+\frac{14}{3}\right)dx = \left[\frac{3x^2}{4}+\frac{5}{2}x\right]_{-\frac{5}{3}}^{1} + \left[-\frac{2x^2}{6}+\frac{14}{3}x\right]_{1}^{7}$ | M1 | A complete method using their values with correct integration on $l_1$ and their $l_2$ |
| $= \frac{52}{3}$ | A1 | Area $= \frac{52}{3}$ or exact equivalent e.g. $17\frac{1}{3}$ or $17.\dot{3}$ **(i.e. a clear dot over the 3)** |

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