# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $PA$ is $-\frac{1}{2}$ | M1 | Uses perpendicular gradients; condone $-\frac{1}{2}x$ if followed by correct work |
| $y - 5 = -\frac{1}{2}(x-7)$ using gradient $-\frac{1}{2}$ and point $(7,5)$ | M1 | Award for method of finding equation of line with changed gradient and point $(7,5)$; if $y=mx+c$ used, must proceed to $c=\ldots$ |
| $2y + x = 17$ (completes proof) | A1* | Completes proof with no errors or omissions |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $2y+x=17$ and $y=2x+1$ simultaneously | M1 | Must be valid attempt to find both coordinates; do not allow starting from $17-x=2x+1$ |
| $P=(3,7)$ | A1 | |
| $PA = \sqrt{(3-7)^2+(7-5)^2} = \sqrt{20}$ | M1 | Uses Pythagoras with their $P=(3,7)$ and $(7,5)$; must attempt difference of coordinates |
| Equation of $C$ is $(x-7)^2+(y-5)^2=20$ | A1 | Do not accept $(x-7)^2+(y-5)^2=(\sqrt{20})^2$ |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find where $y=2x+k$ meets $C$ using $\overrightarrow{OA}+\overrightarrow{PA}$ | M1 | $(11,3)$ or one correct coordinate of $(11,3)$ is evidence of this award |
| Substitutes $(11,3)$ into $y=2x+k$ to find $k$ | M1 | Alternative: solving $(4k-34)^2 - 4\times5\times(k^2-10k+54)=0 \Rightarrow k=\ldots$ |
| $k=-19$ | A1 | $k=-19$ only |

### Alternative I (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $y=2x+k$ into $(x-7)^2+(y-5)^2=20$, forming quadratic $5x^2+(4k-34)x+k^2-10k+54=0$ where both $b$ and $c$ depend on $k$ | M1 | Terms in $x^2$ and $x$ must be collected |
| Uses $b^2-4ac=0$: $(4k-34)^2-4\times5\times(k^2-10k+54)=0$ | M1 | Independent of previous M; may be awarded from one term in $k$ |
| $k=-19$ | A1 | |

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