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