Question

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

Answer 1

Step $1:$ Simplification of given data
Let the given points be $A(7,6)\text{and}B(3,4)$
Let $C$ be a point on the $x-$axis
Coordinates of $C=(x,0)$
We know that distance between two points $(x_1,y_1)and(x_2,y_2)$ is
$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Distance between $A(7,6)\&C(x,0)$
$AC=\sqrt{(x-7{)}^2+(0-6{)}^2}$
$=\sqrt{(x-7{)}^2+36}$
Distance between $B(3,4)\&C(x,0)$
$BC=\sqrt{(x-3{)}^2+(0-4{)}^2}$
$=\sqrt{(x-3{)}^2+16}$

Step $2:$ Solve for required point $C$
Given that point $C$ is equidistant from the points $A\&B$
Hence, $\text{Distance}AC=\text{Distance}BC$
$\Rightarrow \sqrt{(x-7{)}^2+36}=\sqrt{(x-3{)}^2+16}$
Squaring both sides
$\Rightarrow {\left(\sqrt{(x-7{)}^2+36}\right)}^2={\left(\sqrt{(x-3{)}^2+16}\right)}^2$
$\Rightarrow (x-7{)}^2+36=(x-3{)}^2+16$
$\Rightarrow (x-7{)}^2-(x-3{)}^2=16-36$
$\Rightarrow x^2+49-14x-(x^2+9-6x)=-20$
$\Rightarrow x^2+49-14x-x^2-9+6x=-20$
$\Rightarrow -8x+40=-20$
$\Rightarrow 8x=40+20$
$\Rightarrow x=\frac{60}{8}=\frac{15}{2}$

Therefore, Required point = $C(x,0)=(\frac{15}{2},0)$