Let the co-ordinates of P and R be (a,b) and (c,d) and coordinates of Q are (3, 2)

By midpoint formula.

x = , y =

(2 , - 1) is the mid-point of PQ.

∴ 2 = and -1 =

∴ a = 1 and b = -4

∴ Coordinates of P are (1, -4)

(1 , 2) is the mid-point of QR.

∴ 1 = and 2 =

∴ c = -1 and d = 2

∴ Coordinates of P are (-1, 2)

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆PQR = | 3( − 4 – 2 ) + 2( − 1 – 1) + 1( 2 − 4) |

= | − 18 − 4 − 2 |

= 12 sq. units

Hence the area of ∆PQR is 12 sq. units