Question

Find the value of $x$ for which the distance between $p(2,-3)$ and $q(x,5)$ is $10$ units.

Answer 1

The distance between $p(2,-3)$ and $q(x,5)$ is $\sqrt{(x-2{)}^2+(5+3{)}^2}$ units.

Given,

$\sqrt{(x-2{)}^2+(5+3{)}^2}=10$

Squaring we have,

$(x-2{)}^2+64=100$

or, $(x-2{)}^2=36$

or, $x-2=\pm 6$

or, $x=8,-4$.