Question

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

• A. $12$
• B. $4$
• C. $5$
• D. $9$

Answer 1

The correct option is D $9$


The point of intersection of the lines will be centriod
$x+y=5x=4\therefore x=4,y=1$
Let coordinates of $C$
$=(4,a)$
Let coordinates of $B$
$=(b,5-b)$
So,
$\frac{1+4+b}{3}=4\Rightarrow b=7\frac{2+a+5-b}{3}=1\Rightarrow a=3$
Therefore the coordinates of the triangle will be,
$A=(1,2);B=(7,-2);C=(4,3)$
So the area of the triangle will be,
$=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 2\\ 1& 7& -2\\ 1& 4& 3\end{array}\right|$
Using row operations, $R_1\to R_1-R_3,R_2\to R_2-R_3$
$=\frac{1}{2}\left|\begin{array}{ccc}0& -3& -1\\ 0& 3& -5\\ 1& 4& 3\end{array}\right|=\frac{1}{2}\times 1(15+3)=9$
Hence the area will be $9$ sq. units