## Question 6:

### Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(AB=)\sqrt{(-3-5)^2+(1-0)^2}$ or $(BC=)\sqrt{(5-9)^2+(0-7)^2}$ | M1 | Correct formula for distance between two points for either $AB$ or $BC$. 3 out of 4 values correct for either distance |
| $AB = BC = \sqrt{65}$ | A1 | Correctly showing $AB = BC$, exact values needed |

### Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(AC=)\sqrt{(-3-9)^2+(1-7)^2}$ | M1 | Attempt to find $AC$ (or its square) – 3 out of 4 values correct. Or find gradients of both line segments $m_{AB}=\frac{1-0}{-3-5}$ and $m_{BC}=\frac{7-0}{9-5}$ |
| $\left(\sqrt{65}\right)^2 + \left(\sqrt{65}\right)^2 (=130) \neq 180\left(=AC^2\right)$ so angle $ABC$ is not a right angle. Or $\cos ABC = -\frac{5}{13}$, which is not $=0$ therefore angle $ABC$ is not a right angle. Or Angle $ABC = 112.62...°$ which is not a right angle | A1 | Show correctly that Pythagoras does not hold in triangle $ABC$. Using cosine rule. Or $-\frac{1}{8}\times\frac{7}{4}=\left(-\frac{7}{32}\right)\neq -1$, so angle $ABC$ is not a right angle o.e. |

### Part (c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3, 4)$ | B1 | |

### Part (d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{y-0}{4-0} = \frac{x-5}{3-5}$ | M1 | Correct formula for equation of line between $B$ and their midpoint of $AC$ from (c). Or using $y-y_1 = m_{BM}(x-x_1)$, or using $y = m_{BM}x + c$ |
| $2x + y = 10$ | A1 | o.e. required form |

### Part (e):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x+3)^2 + (y-1)^2 = 65$ | B1, B1FT | B1 for correct LHS. B1FT for their $AB^2$ on RHS. Must be an equation to gain marks |

### Part (f):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1, 8)$ | B1 | |

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