Question

Find the value of x such that the distance between the points $(2,5)$ and $(x,-7)$ is $13$.

Answer 1

$d=13,(x_1,y_1)=(2,5)$ and $(x_2,y_2)=(x,7)$
$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$13=\sqrt{(x-2{)}^2+(-7-5^2{)}^2}=\sqrt{(x-2{)}^2+144}$
Squaring both sides, ${13}^2=x^2-4x+148$
$169=x^2-4x+148$
$x^2-4x+148-169=0$
$x^2-4x-21=0$
$(x-7)(x+3)=0$
$\therefore x=7$ or $x=-3$.