## Question 6(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| grad $AB = \frac{0-6}{1-(-1)}$ oe $[= -3]$ isw | M1 | for full marks, it should be clear that grads are independently obtained; eg grads of $-3$ and $1/3$ without earlier working earn M1M0 |
| grad $BC = \frac{0-4}{1-13}$ oe $[= 1/3]$ isw | M1 | |
| product of grads $= -1$ [so lines perp] stated or shown numerically | M1 | or 'one grad is neg. reciprocal of other'; or M1 for length of one side (or square of it), M1 for length of other two sides (or their squares) found independently, M1 for showing or stating that Pythag holds [so triangle rt angled]; for M3, must be fully correct, with gradients evaluated at least to $-6/2$ and $-4/-12$ stage; $AB^2 = 6^2+2^2=40,\ BC^2=4^2+12^2=160,\ AC^2=14^2+{}^2=200$ |

## Question 6(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \sqrt{40}$ or $BC = \sqrt{160}$ | M1 | allow M1 for $\sqrt{(1-(-1))^2+(6-0)^2}$ or for $\sqrt{(13-1)^2+(4-0)^2}$ |
| $\frac{1}{2} \times \sqrt{40} \times \sqrt{160}$ oe or ft their $AB$, $BC$ | M1 | or M1 for one of area under $AC\ (=70)$, under $AB\ (=6)$ under $BC\ (=24)$ (accept unsimplified) and M1 for their trap. $-$ two triangles; or for rectangle $-$ 3 triangles method, $[6\times14 - \frac{1}{2}(2)(6) - \frac{1}{2}(4)(12) - \frac{1}{2}(2)(14)] = 84-6-24-14]$; M1 for two of the 4 areas correct and M1 for the subtraction |
| $40$ | A1 | |

## Question 6:

### Part (iii)
| Angle subtended by diameter = $90°$ | B1 | Or angle at centre = twice angle at circumference = $2 \times 90 = 180°$; or showing $BM = AM$ or $CM$, where $M$ is midpoint of $AC$; or showing $BM = \frac{1}{2}AC$ | Condone 'AB and BC are perpendicular' or 'ABC is right angled triangle' provided no spurious extra reasoning |

| Midpoint $M$ of $AC = (6, 5)$ | B2 | Allow if seen in circle equation; M1 for correct working seen for **both** coords | |

| Radius of circle $= \frac{1}{2}\sqrt{14^2 + 2^2} [=] \frac{1}{2}\sqrt{200}$ or equivalent using $r^2$ | M1 | Accept unsimplified; or e.g. $r^2 = 7^2 + 1^2$ or $5^2 + 5^2$; may be implied by correct equation for circle or by correct method for $AM$, $BM$ or $CM$ ft their $M$ | Allow M1 bod intent for $AC = \sqrt{200}$ followed by $r = \sqrt{100}$ |

| $(x-a)^2 + (y-b)^2 = r^2$ seen or $(x - \text{their } 6)^2 + (y - \text{their } 5)^2 = k$ used, with $k > 0$ | M1 | | |

| $(x-6)^2 + (y-5)^2 = 50$ cao | A1 | Or $x^2 + y^2 - 12x - 10y + 11 = 0$ | Must be simplified (no surds) |

### Part (iv)
| $(11, 10)$ | 1 | | |

---