Question

Find the values of $x$ for which the distance between the point $P(2,-3)$ and $Q(x,5)$ is $10.$

Answer 1

It is given that distance between $P(2,-3)$ and $Q(x,5)$ is $10.$
$\Rightarrow PQ=10$
In general, the distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by,
$(AB{)}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2$ ......(i)
Substituting given values, we get
$(10{)}^2=(x-2{)}^2+(5+3{)}^2$
$\Rightarrow 100=(x-2{)}^2+8^2$
$\Rightarrow 100=(x-2{)}^2+64$
$\Rightarrow (x-2{)}^2=100-64$
$\Rightarrow (x-2{)}^2=36$
$\Rightarrow x-2=\pm 36$
$\Rightarrow x-2=6$ and $x-2=-6$
$\Rightarrow x=6+2$ and $x=-6+2$
$\therefore x=8$ and $x=-4$