Question

Find a point on x-axis which is equidistant from A (2,-5) and B (-2 9)

Answer 1

Let the point of x-axis be $P(x,0)$

Given $A(2,-5)$ and $B(-2,9)$ are equidistant from $P$

That is, $PA=PB$

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

Distance between two points is $\left[\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\right]$ ------distance formula

$\therefore PA=\left[\sqrt{(2-x{)}^2+(-5-0{)}^2}\right]$

$P{A}^2=4-4x+x^2+25=x^2-4x+29$

and $PB=\left[\sqrt{(-2-x{)}^2+(9-y{)}^2}\right]$

$P{B}^2=x^2+4x+85$

Equation (1) becomes

$x^2-4x+29=x^2+4x+85$

$-8x=56$

$x=-7$

Hence the point on x-axis is $(-7,0)$