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.
▶️Answer/Explanation
(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) \)