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Find the area of the triangle $PQR$ with $Q (3, 2)$ and the mid-points of the sides through $Q$ being $(2, -1)$ and $(1, 2)$.
Given:
The mid-points of the sides through $Q(3,2)$ in a triangle $PQR$ are $(2, -1)$ and $(1, 2)$.
To do:
We have to find the area of $\triangle PQR$.
Solution:
Let $P (x_1, y_1), R (x_3, y_3)$ be the other two vertices of the $\triangle PQR$.
$A(2,-1)$ is the mid-point of $QR$.
This implies,
\( (\frac{3+x_{3}}{2}, \frac{2+y_3}{2})=(2,-1) \)
On comparing, we get,
\( \frac{3+x_3}{2}=2 \) and \( \frac{2+y_3}{2}=-1 \)
\( 3+x_3=2(2) \) and \( 2+y_3=-1(2) \)
\( x_3=4-3=1 \) and \( y_3=-2-2=-4 \)
Similarly,
\( B(1,2) \) is the mid-point of \( PQ \).
\( (\frac{3+x_{1}}{2}, \frac{2+y_1}{2})=(1,2) \)
On comparing, we get,
\( \frac{3+x_1}{2}=1 \) and \( \frac{2+y_1}{2}=2 \)
\( 3+x_1=2(1) \) and \( 2+y_1=2(2) \)
\( x_1=2-3=-1 \) and \( y_1=4-2=2 \)
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( PQR=\frac{1}{2}[3(2+4)+(-1)(-4-2)+1(2-2)] \)
\( =\frac{1}{2}[3(6)+(-1)(-6)+1(0)] \)
\( =\frac{1}{2}(18+6+0) \)
\( =\frac{1}{2}(24) \)
\( =12 \) sq. units
The area of $\triangle PQR$ is $12$ sq. units.
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