Question

Draw triangle PQR with vertices P(1,2),Q(3, 3),R(2,6).find PQ.​

Answer 1

Check the picture below.

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P(\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad Q(\stackrel{x_2}{3}~,~\stackrel{y_2}{3})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ PQ=√((~~3 - 1~~)^2 + (~~3 - 2~~)^2) \implies PQ=√(( 2 )^2 + ( 1 )^2) \\\\\\ \implies PQ=√( 4 + 1 ) \implies PQ=√( 5 )\implies PQ\approx 2.24`