Question

Which one of the following points on the line 2x - 3y = 5 is equidistant from (1, 2) and (3, 4)?

• A. (7, 3)
• B. (4, 1)
• C. (1, -1)
• D. (-2, -3)

Answer 1

The correct option is B (4, 1)

Let point $P(x_1,y_1)$ be equidistant from point A(1, 2) and B(3, 4).
$\therefore$ PA = PB
$\Rightarrow P{A}^2=P{B}^2$
$\Rightarrow (1-x_1{)}^2+(2-y_1{)}^2=(3-x_1{)}^2+(4-y_1{)}^2\Rightarrow 1+x_1^2-2x_1+4+y_1^2-4y_1=9+x_1^2-6x_1+16+y_1^2-8y_1\Rightarrow x_1+y_1=5$
As $P(x_1,y_1)$ lies on 2x - 3y = 5
$\therefore 2x_1-3y_1=5$
On solving Eqs. (1) and (2), we get
$x_1=4and{y}_1=1$
$\therefore$ Coordinates of P are (4, 1).