Question

The point A has coordinates (−1, 6) and the point B has coordinates (7, 2).

(i) Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(ii) A point C on the perpendicular bisector has coordinates (p, q). The distance OC is 2 units, where O is the origin. Write down two equations involving p and q and hence find the coordinates of the possible positions of C.

(i) mid-point=(3,4)

Gradient AB=$$-\frac{1}{2}\rightarrow$$ gradient of perpendicular =2

y-4=2(x-3)

y=2x+2

(ii)q=2p-2   $$p^{2}+q^{2}=4$$

$$p^{2}+(2p-2)^{2}=4\rightarrow 5p^{2}-8p=0$$

OR

$$\frac{1}{4}\left ( q+2 \right )^{2}+q^{2}=4\rightarrow 5q^{2}+4q-12=0$$

(0,-2) and $$\left ( \frac{8}{5},\frac{6}{5} \right )$$