Question

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 1

Let the coordinates of the vertices P and R of ∆PQR be (a_1, b₁) and (a_2, b₂), 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 `

⇒ a₁ = 1 and b₁ = −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 `

⇒ a₂ = −1 and b₂ = 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.