Question

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

Answer 1

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

Given : $A(5,4)$ and $B(2,3)$ are equidistant from P , So

$AX=BX$, Then

${\left(AX\right)}^2={\left(BX\right)}^2$ ---- ( A )

we can use distance formula ,

That is Distance $\left(d\right)=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

For AX, $x_1=5,x_2=x,y_1=4,y_2=0$

Substitute all values in distance formula we get

$AX=\sqrt{{(x-5)}^2+{(0-4)}^2}$

$=\sqrt{{\left(x\right)}^2-10x+41}$

$\Rightarrow {\left(AX\right)}^2=x^2-10x+41$

For BX, $x_1=2,x_2=x,y_1=3,y_2=0$

Substitute all values in distance formula we get

$BX=\sqrt{{(x-2)}^2+{(0-3)}^2}$

$=\sqrt{{\left(x\right)}^2-2x+13}$

$\Rightarrow {\left(BX\right)}^2=x^2-2x+13$

From equation A , (1) and (2) we get

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

$-8x=-28$

$x=3.5$

Hence the point on x - axis is $(3.5,0)$