Question

Find sum of values of $a$ for which the distance between the points $P(11,-2)$ and $Q(a,1)$ is $5$ units.

Answer 1

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 on both sides we get
${(a-11)}^2=25-9=16\Rightarrow a-11$=$\pm \sqrt{16}=a-11=\pm 4$
$\Rightarrow a=15$ or 7