Question

Find the area of the triangle $PQR$ if coordinates of $Q$ are $(3,2)$ and the coordinates of mid-points of the sides through $Q$ are $(2,-1)$ and $(1,2)$

Answer 1

Let the coordinate of P and R are $P(x,y)$ and $R(a,b)$

If $(X,Y)$ is the mid-point of $x_1$ and $y_1$ then using mid-point formula

$P(X,Y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

According to the given condition

$\frac{x+3}{2}=2\Rightarrow x=1$

$\frac{y+2}{2}=-1\Rightarrow y=-4$

Thus co-ordinates of $P$ are $(1,-4)$

Also, for point $R$,

$\frac{3+a}{2}=1\Rightarrow a=-1$

$\frac{b+2}{2}=2\Rightarrow b=2$

$\therefore$ Area of $△PQR$$=\frac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$

Here, $x_1=1,y_1=-4,x_2=3,y_2=2,x_3=-1,y_3=2$

$\therefore$ Area of $△PQR$ $=\frac{1}{2}\mid -4(3+1)+2(-1-1)+2(1-3)\mid$

$=\frac{1}{2}\mid -16-4-4\mid$

$=\frac{1}{2}\times 24$

$=12$ Sq.units