## Question 10(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{grad } CD = \dfrac{5-3}{3-(-1)} \left[= \dfrac{2}{4}\right]$ o.e. isw | M1 | NB needs to be obtained independently of grad AB |
| $\text{grad } AB = \dfrac{3-(-1)}{6-(-2)}$ or $\dfrac{4}{8}$ isw | M1 | |
| Same gradient so parallel www | A1 | Must be explicit conclusion mentioning 'same gradient' or 'parallel'; if M0, allow B1 for 'parallel lines have same gradient' o.e. |

## Question 10(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $[BC^2 =] 3^2 + 2^2$; $[BC^2 =] 13$ | M1, A1 | Accept $(6-3)^2 + (3-5)^2$ o.e.; or $[BC =] \sqrt{13}$ |
| Showing $AD^2 = 1^2 + 4^2 [= 17] [\neq BC^2]$ isw | A1 | Or $[AD =] \sqrt{17}$; or equivalent marks for finding AD or $AD^2$ first; alt method: showing $AC \neq BD$ — mark equivalently |

## Question 10(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| [BD equation is] $y = 3$ | M1 | e.g. allow for 'at M, $y = 3$' or for $3$ substituted in equation of AC |
| Equation of AC is $y - 5 = \frac{6}{5}(x - 3)$ o.e. $[y = 1.2x + 1.4$ o.e.$]$ | M2 | Or M1 for grad $AC = \frac{6}{5}$ o.e. (accept unsimplified) and M1 for using their grad of AC with coords of $A(-2, -1)$ or $C(3, 5)$ in equation of line or M1 for 'stepping' method to reach M |
| M is $(\frac{4}{3}, 3)$ o.e. isw | A1 | Allow: at M, $x = \frac{16}{12}$ o.e. [eg $= \frac{4}{3}$] isw; A0 for $1.3$ without a fraction answer seen |

## Question 10(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| Midpoint of $BD = (\frac{5}{2}, 3)$ or equivalent simplified form cao | M1 | Or showing $BM \neq MD$ o.e. $[BM = \frac{14}{3},\ MD = \frac{7}{3}]$ |
| Midpoint of $AC = (\frac{1}{2}, 2)$ or equivalent simplified form cao or 'M is $\frac{2}{3}$ of way from A to C' | M1 | Or showing $AM \neq MC$ or $AM^2 \neq MC^2$ |
| Conclusion 'neither diagonal bisects the other' | A1 | In these methods A1 is dependent on coords of M having been obtained in part (iii) or in this part; the coordinates of M need not be correct; it is also dependent on midpoints of both AC and BD attempted, at least one correct; alt method: show that mid point of BD does not lie on AC (M1) and vice-versa (M1), A1 for both and conclusion |