Question

Find a point on the x-axis which is equidistant from the points $(7,6)$ and $(-3,4)$.

Answer 1

Given points $A(7,6)$ and $B(-3,4)$

Let the point be $P(x,0)$ [we know that point on x-axis is $(x,0)$]

So, $AP=PB$ and

$A{P}^2=P{B}^2$

$\Rightarrow (x-7{)}^2+(0-6{)}^2=(x+3{)}^2+(0-4{)}^2$

[Distance between two points is given by distance formula, $d=\sqrt{({x_2-x_1)}^2+(y_2-y_1{)}^2}$]

$\Rightarrow x^2+49-14x+36=x^2+9+6x+16$

$\Rightarrow 49-9+36-16=6x+14x$

$\Rightarrow 40+20=20x$

$\Rightarrow x=\frac{60}{20}=3$

Hence, point on x-axis which is equidistant from the points $A(7,6)$ and $B(-3,4)$ is $(3,0)$