Question

Pls solve this problem​

Answer 1

That's how you do it

Answer 2

$PQ=5 \\ \\ = Distance\ between\ P\ and\ Q=5 \\ \\ =\sqrt{(Difference\ of\ x\ coordinates)^2+(Difference\ of \ y\ coordinates)^2}=5 \\ \\ = \sqrt{(5-2)^2+(1-(y-1))^2}=5 \\ \\ = \sqrt{3^2+(1-y+1)^2}=5 \\ \\ = \sqrt{9+(2-y)^2}=5$

$=\sqrt{9+4-4y+y^2}=5 \\ \\ =\sqrt{y^2-4y+13}=5 \\ \\ \\ \\ \Rightarrow\ y^2-4y+13=5^2 \\ \\ \Rightarrow\ y^2-4y+13=25 \\ \\ \Rightarrow\ y^2-4y+13-25=0 \\ \\ \Rightarrow\ y^2-4y-12=0$

$\Rightarrow\ y^2+2y-6y-12=0 \\ \\ \Rightarrow\ y(y+2)-6(y+2)=0 \\ \\ \Rightarrow\ (y-6)(y+2)=0 \\ \\ \\ \therefore\ y=\bold{6}\ \ \ ; \ \ \ y=\bold{-2}$

$Hope\ this \ helps.\ Plz\ ask\ me\ if\ you \ have\ any \ doubts. \\ \\ \\ Thank\ you.$