Question

Find a point on the x-axis which is equidistant from $(2,-5)$ and $(-2,9)$.

Answer 1

Find the coordinate of point:

Let a point $P(x,0)$ on x-axis be equidistant from the points $A(2,-5)$ and $B(-2,9)$.

Then, $PA=PB$.

According to the distance formula, the distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\Rightarrow$$\sqrt{{(x_{}-2)}^2+{[0-(-5)]}^2}$ $=\sqrt{{(-2_{}-x)}^2+{(9-0)}^2}$

$\Rightarrow$ ${(x_{}-2)}^2+25$ $={(-2_{}-x)}^2+81$

$\Rightarrow$ ${(x_{}-2)}^2+25$ $={[-(2_{}+x\left)\right]}^2+81$

$\Rightarrow$ ${x_{}}^2-4x+4+25$ $=x^2+4x+4+81$ [ as ${(a-b)}^2=a^2-2ab+b^2$ and ${(a+b)}^2=a^2+2ab+b^2$ ]

$\Rightarrow$ $-4x+25$ $=4x+81$

$\Rightarrow$ $-4x-4x$ $=81-25$

$\Rightarrow$ $-8x$ $=56$

$\Rightarrow$ $x$$=-7$

Hence, the point $P(-7,0)$ is equidistant from the points $A(2,-5)$ and $B(-2,9)$.