If x−y+5=0 is ⊥r bisector of AB then B is Image of A(1, −2) =(x1, y1)h−11=k+2−1=−2(1+2+5)1+1h−11=k+2−1=−8B(h k)= (−7, 6)If x+2y=0 is ⊥r bisector of AC thenc is Image of A(1, −2)h−11=k+22=−2(1−4)1+4h−11=k+22=65h=65+1⇒ 115,k=125−2, ⇒25∴ C(h k)=(115, 25)slope of BC¯ =25−6115+7=−2846=−1423Mid point of BC =(115−72, 25+62)⇒ (−125, 165)Equation of ⊥r bisector of BC¯ is y−165=2314(x+125)⇒ 5y−16=2314(5x+12)⇒ 70y−224=115x+276⇒ 115x−70y+500=0

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If the equations of the perpendicular bisectors of the sides AB and AC of a $Δ A B C a r e x - y + 5 = 0 a n d x + 2 y = 0$ respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

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$3 x + 3 y + 5 = 0$

$9 x - 23 y + 40 = 0$

$6 x + 15 y = 5$

$23 x - 14 y + 100 = 0$

answer is D.

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Detailed Solution

$\begin{array}{l} I f x - y + 5 = 0 i s ⊥ r b i \sec t o r o f A B t h e n \\ B i s Im a g e o f A (1, - 2) = (x_{1}, y_{1}) \\ \frac{h - 1}{1} = \frac{k + 2}{- 1} = \frac{- 2 (1 + 2 + 5)}{1 + 1} \\ \frac{h - 1}{1} = \frac{k + 2}{- 1} = - 8 \\ B (h k) = (- 7, 6) \\ I f x + 2 y = 0 i s ⊥ r b i \sec t o r o f A C t h e n \\ c i s Im a g e o f A (1, - 2) \\ \frac{h - 1}{1} = \frac{k + 2}{2} = \frac{- 2 (1 - 4)}{1 + 4} \\ \frac{h - 1}{1} = \frac{k + 2}{2} = \frac{6}{5} \end{array}$

$\begin{array}{l} h = \frac{6}{5} + 1 \Rightarrow \frac{11}{5}, k = \frac{12}{5} - 2, \Rightarrow \frac{2}{5} \\ ∴ C (h k) = (\frac{11}{5}, \frac{2}{5}) \\ s l o p e o f \bar{B C} = \frac{\frac{2}{5} - 6}{\frac{11}{5} + 7} = \frac{- 28}{46} = \frac{- 14}{23} \\ M i d p o int o f B C = (\frac{\frac{11}{5} - 7}{2}, \frac{\frac{2}{5} + 6}{2}) \Rightarrow (\frac{- 12}{5}, \frac{16}{5}) \\ E q u a t i o n o f ⊥ r b i \sec t o r o f \bar{B C} i s \\ y - \frac{16}{5} = \frac{23}{14} (x + \frac{12}{5}) \\ \Rightarrow 5 y - 16 = \frac{23}{14} (5 x + 12) \\ \Rightarrow 70 y - 224 = 115 x + 276 \\ \Rightarrow 115 x - 70 y + 500 = 0 \end{array}$