Answer:
Let \[A(1,-4),\,\,B({{x}_{1}},{{y}_{1}})\] and \[C\,({{x}_{2}},{{y}_{2}})\] be the vertices of a triangle ABC and let \[P(2,-1)\] and \[Q(0,-1)\]be the mid-points of AB and AC respectively. \[\because \,\,p\]is the mid-point of AB. \[\therefore \frac{1+{{x}_{1}}}{2}=2,\frac{-4+{{y}_{1}}}{2}=-1\] \[{{x}_{1}}=3,\,\,{{y}_{1}}=2\] So, \[B({{x}_{1}},{{y}_{1}})=B(3,2)\] Similar, Q is the mid-point of AC. \[\therefore \frac{1+{{x}_{2}}}{2}=0,\frac{-4+{{y}_{2}}}{2}=-1\] \[{{x}_{2}}=-1,{{y}_{2}}=2\] So, \[C({{x}_{2}},{{y}_{2}})=C(-1,2)\] Thus, Area of \[\Delta \text{ }ABC\] \[=\frac{1}{2}[1(2-2)+3(2+4)-1(-4-2)]\] \[=\frac{1}{2}\times 24=12\,\] sq. units.
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