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

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

BITSATBITSAT 2009

Solution:

Let point $(x_1, y_1)$ be on the line $3x + 4y = 5$.
$3x_1 + 4y_1 = 5$ ....(i)
Also, $(x_1- 1)^2 + (y_1 - 2)^2 = (x_1 - 3)^2 + (y_1 - 4)^2$
$\Rightarrow \, \, \, x_1 + {y^2}_1 - 2x_1 - 4y_1+ 5$
$ = {x^2}_1 + {y^2}_1 - 6x_1 - 8 \, y_1 + 25$ .....(ii)
$\Rightarrow \, \, \, 4x_1 + 4y_1 = 20 $ .....(iii)
On solving Eqs. (i) and (ii), we get,
$x_1 = 15, y_1 = -10$