Answer :

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 (x_{1},y_{1}), (x_{2},y_{2}) and (x_{3},y_{3})

= |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|

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

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