Question

If a point $(p,q)$ lies on the X-axis, is equidistant from the points $(5,4)$ and $(-2,3)$, then find the value of $p+q$.

Answer 1

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

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

Here $(x,0)\equiv (x_1,y_1)$ and $(5,4)\equiv (x_2,y_2)$

Distance between $(x,0)$ and $(5,4)=\sqrt{{(5-x)}^2+{(4-0)}^2}=\sqrt{5^2+x^2-10x+16}=\sqrt{x^2-10x+41}$

Here $(x,0)\equiv (x_1,y_1)$ and $(-2,3)\equiv (x_2,y_2)$

Distance between $(x,0)$ and $(-2,3)=\sqrt{{(-2-x)}^2+{(3-0)}^2}=\sqrt{2^2+x^2+4x+9}=\sqrt{x^2+4x+13}$

As the point $(x,0)$ is equidistant from the two points, both the distances calculated are equal.

$\sqrt{x^2-10x+41}=\sqrt{x^2+4x+13}$

$\Rightarrow x^2-10x+41=x^2+4x+13$

$41-13=10x+4x$

$x=2$

Thus, the point is $(2,0)$ $\Rightarrow p=2$ and $q=0$ $\Rightarrow p+q=2$