## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{Grad AB}=]\frac{7-3}{2-0}$ or $\frac{4}{2}$ | M1 | Allow just a simplified version of $2$ or $-\frac{1}{2}$ for one method mark, but for both to be gained, there must be evidence that the gradients have been obtained independently |
| $[\text{Grad BC}=]\frac{-1-3}{8-0}$ or $\frac{-4}{8}$ | M1 | |
| Product of gradients $=-1$ [when lines are at right angles] | A1 | Or 'negative reciprocal [so perpendicular]' oe; may be implied by correct calculation; correct working must support the statement |
| **or** $AB^2=2^2+4^2\ [=20]$, $BC^2=8^2+4^2\ [=80]$ and $AC^2=6^2+8^2\ [=100]$ | M2 | Or equiv for AB etc; allow at unsimplified stage; or M1 for just one correct expression for one of the sides; allow just a simplified version e.g. $AB^2=20$ for one method mark, but for both to be gained, there must be evidence that the lengths or their squares have been obtained independently |
| $AB^2+BC^2=AC^2$ [so by Pythagoras, angle $ABC=90°$] oe | A1 | May be implied by correct calculation; correct working must support the statement; another possible method: M1 for finding midpt of AC as $(5,3)$, M1 for showing dist from midpt to A, B and C is 5 and M1 for using angle in a semicircle to show that $ABC=90°$ |

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## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre $D=\left(\frac{2+8}{2},\frac{7+-1}{2}\right)$ or $(5,3)$ soi | B1 | May be implied by circle eqn; if already found in (i), must be used in (ii) to get the mark here |
| Radius $=5$ or $r^2=25$ or for finding dist between A, B or C and their centre D oe | B1 | May be implied by circle eqn; if already found in (i), must be used in (ii) to get the mark here |
| $(x-a)^2+(y-b)^2=r^2$ soi | M1 | General formula may be quoted or implied by eqn using their values, but it must be clear that they are using their $r^2$ rather than their $r$ or their $d$ or $d^2$; for this method mark, allow use of their values, even if obtained from AB or BC as diameter |
| $(x-5)^2+(y-3)^2=25$ or $5^2$ isw | A1 | Alternative method: allow B4 for $(y-7)(y+1)+(x-2)(x-8)=0$ |

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## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{grad AD}=]\frac{7-3}{2-5}$ isw or $-\frac{4}{3}$ oe | B1 | Or may use $CD\left(\frac{-1-3}{8-5}\right)$, $AC\left(\frac{7-(-1)}{2-8}\right)$; or ft their D from (ii); or B1 for correct differentiation: $2x+2y\frac{dy}{dx}-10-6\frac{dy}{dx}=0$ oe |
| Grad tgt $=\frac{3}{4}$ oe www or $-1/$ their grad AD oe | M1 | M0 if grad AD used; M0 for a spurious gradient used; perp gradient to AB or BC used: may earn 2nd M1 only |
| $y-7=$ their $\frac{3}{4}(x-2)$ or $7=$ their $\frac{3}{4}\times2+c$ | M1 | |
| $4y=3x+22$ oe where $a,b,c$ are integers, isw | A1 | Allow correct answer to imply 3rd M1, provided first two Ms have been earned |

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