Question

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

• A. (7, - 4)
• B. (15, -10)
• C. (1/7, 8/7)
• D. (0, 5/4)

Answer 1

Answer: (B)

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