The point on the line 3x + 4y = 5 which is equidistance from (1, 2) and (3, 4) is

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(7, -4)

$(\frac{1}{7}, \frac{8}{7})$

$(0, \frac{5}{4})$

(15, -10)

answer is D.

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Detailed Solution

: Let A (1, 2) and B (3, 4) and be the point on 3x + 4y - 5 = 0

$\begin{array}{l} \Rightarrow 3 α + 4 β - 5 = 0 - - (1) \\ G i v e n P A = P B \\ S . B . S \\ \Rightarrow P A^{2} = P B^{2} \end{array}$

$\begin{array}{l} ∴ {(α - 1)}^{2} + {(β - 2)}^{2} = {(α - 3)}^{2} + {(β - 4)}^{2} \\ \Rightarrow α^{2} + 1 - 2 α + β^{2} + 4 - 4 β = α^{2} + 9 - 6 α + β^{2} + 16 - 8 β \\ \Rightarrow - 2 α - 4 β + 5 + 6 α + 8 β - 25 = 0 \\ \Rightarrow 4 α + 4 β - 20 \Rightarrow α + β - 5 = 0 - - (2) \\ S o l v i n g (1) & (2) \\ α β 1 \\ \begin{matrix} 4 & - 5 & 3 & 4 \\ 1 & - 5 & 1 & 1 \end{matrix} \end{array}$

$\begin{array}{l} \Rightarrow \frac{α}{- 20 + 5} = \frac{β}{- 5 + 15} = \frac{1}{3 - 4} \\ \Rightarrow \frac{α}{- 15} = \frac{β}{10} = \frac{1}{- 1} \\ ∴ α = \frac{- 15}{- 1}, β = \frac{10}{- 1} \\ = 15, = - 10 \\ ∴ P (α, β) = (15, - 10) \end{array}$