# CIE A level -Pure Mathematics 1 : Topic :  1.5 Trigonometry: sketch and use graphs of the sine, cosine and tangent functions : Exam Questions Paper 1

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) The midpoint of AB, which lies on the perpendicular bisector.
The midpoint $$M$$ of a line segment with endpoints $$A(x_1, y_1)$$ and $$B(x_2, y_2)$$ has coordinates $$M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$.
For A(-1, 6) and B(7, 2):
$$M\left(\frac{-1 + 7}{2}, \frac{6 + 2}{2}\right) = M(3, 4)$$
The slope $$m$$ is given by $$m = \frac{y_2 – y_1}{x_2 – x_1}$$.
For A and B:
$$m_{AB} = \frac{2 – 6}{7 – (-1)} = \frac{-4}{8} = -\frac{1}{2}$$
The slope of the perpendicular bisector will be the negative reciprocal of $$m_{AB}$$:
$$m_{perpendicular} = -\frac{1}{m_{AB}} = 2$$
The equation of the perpendicular bisector :
$$y – 4 = 2(x – 3)$$
$$y = 2x – 6 + 4$$
$$y = 2x – 2$$
(ii) The point C lies on the perpendicular bisector, so it satisfies its equation:
$$q = 2p – 2$$ (1)
The distance OC (from the origin to point C)
$$\sqrt{(p – 0)^2 + (q – 0)^2} = 2$$
$$p^2 + q^2 = 4$$ (2)
We can now use equation (1) to substitute q in equation (2):
$$p^2 + (2p – 2)^2 = 4$$
$$p^2 + (4p^2 – 8p + 4) = 4$$
$$5p^2 – 8p = 0$$
$$p(5p – 8) = 0$$
$$p = 0 \] or $5p – 8 = 0$$ $$p = \frac{8}{5}$$ For $$p = 0$$: $$q = 2(0) – 2$$ $$q = -2$$ For $$p = \frac{8}{5}$$: $$q = 2\left(\frac{8}{5}\right) – 2$$ $$q = \frac{16}{5} – \frac{10}{5}$$ $$q = \frac{6}{5}$$ Thus, the possible positions for point C are: $$C_1(0, -2)$ and \[ C_2\left(\frac{8}{5}, \frac{6}{5}\right)$$

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