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

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(7, -4)

(15, -10)

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

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

answer is B.

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Detailed Solution

PA = PB

${(x - 1)}^{2} + {(y - 2)}^{2} = {(x - 3)}^{2} + {(y - 4)}^{2}$

$x + y - 5 = 0 \to (1)$

Given, $3 x + 4 y - 5 = 0 \to (2)$

Solve (1) and (2), we get

$x = 15, y = - 10$

Required point== $(15, - 10)$

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