question_answer

Find the area of the triangle PQR with \[Q(3,2)\] and the mid-points of the sides through Q being \[(2,-1)\] and \[(1,2)\].

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.