Question

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

• A. $(7,-4)$
• B. $(-10,15)$
• C. $(\frac{1}{7},\frac{8}{7})$
• D. $(0,\frac{5}{4})$

Answer 1

The correct option is B $(-10,15)$

Let the point $(x_1,y_1)$ be on the line $4x+3y=5$.
$\therefore 4x_1+3y_1=5$ ....(i)
Also, we have
${(x_1-1)}^2+{(y_1-2)}^2={(x_1-3)}^3+{(y_1-4)}^2$
$\Rightarrow x_1^2+1-2x_1+y_1^2+4-4y_1=x_1^2+9-6x_1+y_1^2+16-8y_1$
$\Rightarrow 4x_1+4y_1=20$
$\Rightarrow x_1+y_1=5$ ....(ii)
From equations (i) and (ii), we get
$y_1=15$ and $x_1=-10$