Question

Find the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3).

Answer 1

The given points are A (1, 0) and B (2, 3).
Let M be the midpoint of AB.

\[\therefore \text { Coordinates of } M = \left( \frac{1 + 2}{2}, \frac{0 + 3}{2} \right) = \left( \frac{3}{2}, \frac{3}{2} \right)\]

And, slope of AB = \[\frac{3 - 0}{2 - 1} = 3\]

Let m be the slope of the perpendicular bisector of the line joining the points A (1, 0) and B (2, 3).

\[\therefore m \times \text { Slope of AB } = - 1\]

\[ \Rightarrow m \times 3 = - 1\]

\[ \Rightarrow m = - \frac{1}{3}\]

So, the equation of the line that passes through \[M \left( \frac{3}{2}, \frac{3}{2} \right)\]  and has slope  \[- \frac{1}{3}\] is

\[y - \frac{3}{2} = - \frac{1}{3}\left( x - \frac{3}{2} \right)\]

\[ \Rightarrow x + 3y - 6 = 0\]

Hence, the equation of the right bisector of the line segment joining the points A (1, 0) and B (2, 3) is \[x + 3y - 6 = 0\].