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.

$\therefore \mathrm{Coordinates}\mathrm{of}M=\left(\frac{1+2}{2},\frac{0+3}{2}\right)=\left(\frac{3}{2},\frac{3}{2}\right)$



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).

$$\begin{gathered} \therefore m\times \mathrm{Slope}\mathrm{of}AB=-1 \\ \Rightarrow m\times 3=-1 \\ \Rightarrow m=-\frac{1}{3} \end{gathered}$$
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

$$\begin{gathered} y-\frac{3}{2}=-\frac{1}{3}\left(x-\frac{3}{2}\right) \\ \Rightarrow x+3y-6=0 \end{gathered}$$

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