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). [CBSE 2015]

Answer 1

Let P(x₁, y₁), Q(3, 2) and R(x₂, y₂) be the vertices of the ∆PQR.

Suppose S(2, −1) and T(1, 2) be the mid-points of sides QR and PQ, respectively.



Using mid-point formula, we have

$$\begin{gathered} \frac{x_1+3}{2}=1,\frac{y_1+2}{2}=2 \\ \Rightarrow x_1+3=2,y_1+2=4 \\ \Rightarrow x_1=-1,y_1=2 \end{gathered}$$

∴ Coordinates of P = (−1, 2)

Also,

$$\begin{gathered} \frac{x_2+3}{2}=2,\frac{y_2+2}{2}=-1 \\ \Rightarrow x_2+3=4,y_2+2=-2 \\ \Rightarrow x_2=1,y_2=-4 \end{gathered}$$

∴ Coordinates of R = (1, −4)

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

$$\begin{gathered} \therefore \mathrm{ar}\left(∆\mathrm{PQR}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|-1\left[2-\left(-4\right)\right]+3\left(-4-2\right)+1\left(2-2\right)\right| \\ =\frac{1}{2}\left|-6-18+0\right| \\ =\frac{1}{2}\left|-24\right| \\ =\frac{1}{2}\times 24 \\ =12\mathrm{square}\mathrm{units} \end{gathered}$$

Hence, the area of the triangle is 12 square units.