Question

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

• A. $12$
• B. $4$
• C. $5$
• D. $9$

Answer 1

The correct option is D $9$

Median through $C$ is $x=4$

So, clearly the $x$ coordinates of $C$ is $4$.

So let $C=(4,y)$ then

The midpoint of $A(1,2)$ and $C(2,y)$

which is $D$ lies on the median through $B$ by definition clearly.

$D=\left(\frac{1+4}{2},\frac{2+y}{2}\right)$

Now we have,

$\begin{array}{c}\frac{3+4+y}{2}=5\\ \Rightarrow y=3\\ So,C=\left(4,3\right)\end{array}$

The centro id of the triangle is the intersection of the medians.

The medians $x=4$ and $x+y=5$

Intersect at $G=(4,1)$

The area of triangle $\Delta ABC=3\times \Delta AGC$

$\begin{array}{c}=3\times \frac{1}{2}\times 3\times 2\\ =9\end{array}$

Hence, option $D$ is correct answer.