The equation of perpendicular bisectors of sides AB, BC of $Δ A B C a r e x - y - 5 = 0, x + 2 y = 0$ respectively and A (1, -2), then coordinates of C are

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(0, 0)

(5, 0)

(0, 1)

(1, 0)

answer is C.

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Detailed Solution

$\begin{array}{l} I f x - y - 5 = 0 i s ⊥ r b i \sec t o r s o f A B t h e n B (h k) i s Im a g e o f A (1, - 2) = (x_{1}, y_{1}) \\ \frac{h - 1}{1} = \frac{k + 2}{- 1} = \frac{- 2 (1 + 2 - 5)}{1 + 1} \\ \Rightarrow \frac{h - 1}{1} = \frac{k + 2}{- 1} = 2 \\ h - 1 = 2, k + 2 = - 2 \\ h = 3, k = - 4 \\ ∴ B (3, - 4) \end{array}$

$\begin{array}{l} I f x + 2 y = 0 i s ⊥ r b i \sec t o r o f B C t h e n C (h k) i s Im a g e o f B (3, - 4) = (x_{1} y_{1}) \\ \frac{h - 3}{1} = \frac{k + 4}{2} = \frac{- 2 (3 - 8)}{1 + 4} \\ \Rightarrow \frac{h - 3}{1} = \frac{k + 4}{2} = - 2 \\ \Rightarrow h - 3 = 2, k + 4 = 4 \\ h = 2 + 3, k = 0 \\ ∴ C (5, 0) \end{array}$