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Q.9. Find the equation to the straight line which bisects, and is perpendicular to the straight line joining the points A (a, b) and B (a', b').

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Solution

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$The slope of the line joining the points A (a, b) and B (a', b') = \frac{b' - b}{a' - a} Hence slope of the line ⊥ to it will be = - \frac{a' - a}{b' - b} The mid point of the line A (a, b) and B (a', b') = (\frac{a + a'}{2}, \frac{b + b'}{2}) Hence required equation is given by : (y - \frac{b + b'}{2}) = - \frac{a' - a}{b' - b} (x - \frac{a + a'}{2}) \Rightarrow (\frac{2 y - b - b'}{2}) = \frac{a - a'}{b' - b} (\frac{2 x - a - a'}{2}) \Rightarrow 2 y - b - b' = \frac{a - a'}{b' - b} (2 x - a - a') \Rightarrow (b' - b) (2 y - b - b') = (a - a') (2 x - a - a') \Rightarrow 2 yb' - bb' - b'^{2} - 2 yb + b^{2} + bb' = 2 ax - a^{2} - aa' - 2 xa' + aa' + a'^{2} \Rightarrow 2 (a - a') x + 2 (b - b') y = a^{2} + b^{2} - a'^{2} - b'^{2}$
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