# Question 1:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z-3-4i| = |z+1+i| \Rightarrow |x+iy-3-4i| = |x+iy+1+i|$ $(x-3)^2+(y-4)^2=(x+1)^2+(y+1)^2$ | M1 | Introduces $z=x+iy$, attempts correct modulus method. Allow sign slips in $i$ coordinates. Must be $+$ between brackets. |
| $x^2-6x+9+y^2-8y+16=x^2+2x+1+y^2+2y+1 \Rightarrow ax+by+c=0$ | dM1 | Expands and attempts required form |
| $8x+10y-23=0$ | A1 | This equation or any integer multiple |

## Part (a) ALT (perpendicular bisector):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Mid-point $\left(\frac{3-1}{2}, \frac{4-1}{2}\right)$ and $m=\frac{4+1}{3+1}$ | M1 | Uses points $(3,\pm4)$ and $(-1,\pm1)$ to find midpoint and gradient |
| $y-\frac{3}{2}=-\frac{4}{5}(x-1) \Rightarrow ax+by+c=0$ | dM1 | Correct attempt at perpendicular bisector equation using negative reciprocal gradient |
| $8x+10y-23=0$ | A1 | This equation or any integer multiple |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| [Diagram showing shaded region above line with negative gradient and positive $y$-intercept, in quadrants 1, 2 and 4] | B1 | Shades area above appropriate line, must be in quadrants 1, 2 and 4. Accept solid, dashed or dotted line. Allow any line with negative gradient and positive $y$ intercept. Part (a) need not be correct. |

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