Answer:
Let \[P({{x}_{1}},{{y}_{1}}),Q(3,2)\] an \[R({{x}_{2}},{{y}_{2}})\] be the vertices of a triangle PQR and let \[A(2,-1)\] and \[B(1,2)\] be the mid-points of PQ and QR respectively. \[\because \] A is the mid-point of PQ \[\therefore \frac{3+{{x}_{1}}}{2}=2,\frac{2+{{y}_{1}}}{2}=-1\] \[\Rightarrow {{x}_{1}}=1,{{y}_{1}}=-4\] So, \[P(1,-4)\] \[\because \] B is the mid-point of QR \[\therefore \frac{3+{{x}_{2}}}{2}=1,\frac{2+{{y}_{2}}}{2}=2\] \[\Rightarrow {{x}_{2}}=-1,{{y}_{2}}=2\] So, \[R(-1,2)\] Thus, Area of \[\Delta \,PQR=\frac{1}{2}|[1(2-2)-1(2+4)+3(-4-2)]|\] \[=\frac{1}{2}|[1(0)-1(6)+3(-6)]|\] \[=\frac{1}{2}|[-6-18]|\] \[=\frac{24}{2}=12\] sq. units.
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