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Answer in Brief

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).

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#### Solution

Let the coordinates of the vertices P and R of ∆PQR be (*a*_{1}_{, }*b*_{1}) and (*a*_{2}_{, }*b*_{2}), respectively.

Suppose X(2, −1) is the midpoint of PQ.

Then,

`(2,-1)=((a_1+3)/2, (b_1+2)/2)`

`=>(a_1+3)/2=2 `

⇒ a_{1}=1 and b_{1}=−4

Therefore, the coordinates of P are (1, −4).

Again, suppose Y(1, 2) is the midpoint of QR.

Now,

`(1,2)=((a_2+3)/2,(b_2+2)/2)`

`=>(a_2+3)/2=1 `

⇒ a_{2}=−1 and b_{2}=2

Therefore, the coordinates of R are (−1, 2).

Thus, the vertices of ∆PQR are P(1,−4), Q(3, 2) and R(−1, 2).

Now,

Area of ∆PQR =`1/2`×[1(2−2)+3(2+4)−1(−4−2)]

=`1/2`(18+6)

=`1/2`(24)

=12

Thus, the area of ∆PQR is 12 square units.

Concept: Area of a Triangle

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