Question

In a triangle ABC, coordinates of A are $(1,2)$ and the equations of the median through B and C are respectively, $x+y=5$ and $x=4$. Then area of $\Delta$ABC (in sq. units) is?

Answer 1

$A(1,2)$

Parametric equation of line

through $B\Rightarrow x=4$

$y=t$

Parametric equation of line

through $C\Rightarrow x=s$

$y=5-s$

In line segment $AB,C^{\prime }$ is the midpoint.

Let $C^{\prime }=(s_1,5-s_1)$

$B=(t,t_1)$

$\Rightarrow \left(\frac{1+4}{2}\right)\left(\frac{2+t_1}{2}\right)=(s_1,5-s_1)$

$\Rightarrow S_1=\frac{5}{2}$ and $\frac{t_1+2}{2}=5-s_1$

$\Rightarrow t_1=3$

$\Rightarrow =(4,3)$

In line segment $AC,B^{\prime }$ vis the midpoint

Let $B^{\prime }=(4,t_2)$

$C=(S_2,5-s_2)$

$\Rightarrow (4,t_2)=(\frac{1+s_2}{2},\frac{2+5-s_2}{2})$

$\Rightarrow S_2=7$

$\Rightarrow C=(7,-2)$

Area of triangle $ABC=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 4& 3& 1\\ 7& -2& 1\end{array}\right|$

$=|\frac{5+6-(8+21)}{2}|=|\frac{11-21-8}{2}|=|-9|=9sq.units.$