Question

The point on the x-axis which is equidistant from the points $(5,4)$ and $(-2,3)$ is

• A. $(-2,0)$
• B. $(2,0)$
• C. $(0,2)$
• D. $(2,2)$

Answer 1

The correct option is B $(2,0)$

Let the point of x-axis be $P(x,0)\equiv (x_1,y_1)$

Given: $A(5,4)\equiv (x_2,y_2)$ and $B(-2,3)$ are equidistant from $P$

That is $PA=PB$

Hence $P{A}^2=P{B}^2$ ................. (1)

Distance between two points is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$PA=\sqrt{(5-x{)}^2+(4-0{)}^2}=\sqrt{x^2+25-10x+16}=\sqrt{x^2-10x+41}$

Therefore, $P{A}^2=x^2-10x+41$

Here $P(x,0)\equiv (x_1,y_1)$ $B(-2,3)\equiv (x_2,y_2)$

Similarly, $PB=\sqrt{(-2-x{)}^2+(3-0{)}^2}=\sqrt{x^2+4+4x+9}=\sqrt{x^2+4x+13}$

Therefore, $P{B}^2=x^2+4x+13$

Equation (1) becomes

$x^2-10x+41=x^2+4x+13$

$-10x-4x=13-41$

$-14x=-28$

$x=2$

Hence the point on x-axis is $(2,0)$.