Q84 / 160 · TS EAMCET 2009 · Mathematics
The point on the line $3 x+4 y=5$ which is equidistant from $(1,2)$ and $(3,4)$ is
- A$(7,-4)$
- Correct answerB$(15,-10)$
- C$(1 / 7,8 / 7)$
- D$(0,5 / 4)$
Solution
Let point $\left(x_1, y_1\right)$ be on the line $3 x+4 y=5$.
Also, $\quad\left(x_1-1\right)^2+\left(y_1-2\right)^2=\left(x_1-3\right)^2$ $\begin{aligned} &+\left(y_1-4\right)^2 \\ \Rightarrow x_1^2+y_1^2-2 x_1-4 y_1+5=x_1^2+ & y_1^2-6 x_1 \\ & -8 y_1+25 \end{aligned}$
On solving Eqs. (i) and (ii), we get $x_1=15, y_1=-10$
Also, $\quad\left(x_1-1\right)^2+\left(y_1-2\right)^2=\left(x_1-3\right)^2$ $\begin{aligned} &+\left(y_1-4\right)^2 \\ \Rightarrow x_1^2+y_1^2-2 x_1-4 y_1+5=x_1^2+ & y_1^2-6 x_1 \\ & -8 y_1+25 \end{aligned}$
On solving Eqs. (i) and (ii), we get $x_1=15, y_1=-10$
Hence, the correct answer is option: 'B'.