Question

P(3,-3) Q(8,7) R(5,-2)

Given S(11,1) is a point which lies on the extended line PR where QS is a perpendicular line to PRS. Find the area of triangle PQR.​

Answer 1

Answer: the area of triangle PQR is 7.5 square units

Explanation:

`\displaystyle\\A_(PQR)=(PR*QS)/(2) \\\\1.\ (3,-3)\ \ \ \ \ (5,-2)\\\\PR=√((5-3)^2+(-2-(-3))^2)\\\\PR=√(2^2+(-2+3)^2) \\\\PR=√(4+1^2) \\\\PR=√(4+1)\\\\ PR=√(5) \ units`

`\diplaystyle\\2.\ (8,7)\ \ \ \ \ (11,1)\\\\QS=√((11-8)^2+(1-7)^2) \\\\QS=√(3^2+(-6)^2) \\\\QS=√(9+36) \\\\QS=√(45)\\\\QS=√(9*5) \\\\QS=√(3^2*5) \\\\QS=3√(5)`

`\displaystyle\\Hence,\\A_(PQR)=(√(5)*3√(5) )/(2) \\\\A_(PQR)=((√(5)*√(5)) *3 )/(2) \\\\A_(PQR)= (5*3)/(2)\\\\A_(PQR)=(15)/(2)\\\\A_(PQR)=7.5 \ units^2`