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Question
Find the equation of the right bisector of the line segment joining the points (3, 4) and (−1, 2).
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Solution
Let A (3, 4) and B (−1, 2) be the given points.
Let C be the midpoint of AB.
\[\therefore C \equiv \left( \frac{3 - 1}{2}, \frac{4 + 2}{2} \right) \equiv \left( 1, 3 \right)\]
\[\because \text { Slope of AB } = \frac{2 - 4}{- 1 - 3} = \frac{1}{2}\]
\[ \therefore \text { Slope of the perpendicular bisector of AB } = - 2\]
Thus, the equation of the perpendicular bisector of AB is
\[y - 3 = - 2\left( x - 1 \right)\]
\[ \Rightarrow 2x + y - 5 = 0\]
Hence, the required line is \[2x + y - 5 = 0\].
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| Column C1 | Column C2 |
| (a) Through the point (2, 1) is | (i) 2x – y = 4 |
| (b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is |
(ii) x + y – 5 = 0 |
| (c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is |
(iii) x – y –1 = 0 |
| (d) Equally inclined to the axes is | (iv) 3x – 4y – 1 = 0 |
