question_answer

Find the area of the triangle ABC with \[A\,(1,-4)\] and mid-points of sides through A being \[(2,-1),\] and \[(0,-1)\].

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.