Question

Find $a$, if the distance between the points $P(11,-2)$ and $Q(a,1)$ is $5$ units.

Answer 1

Solution: Distance between two points $(x_1,y_1)$,
$(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.
Given, PQ = 5,
$\Rightarrow \sqrt{(a-11{)}^2+(1-(-2){)}^2}=5$
Taking square root on both sides, we get
$(a-11{)}^2=25-9=16\Rightarrow a-11$
$=\pm \sqrt{16}=a-11=\pm 4$
$\Rightarrow a=15or7$