Question

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

• A. $5$
• B. $9$
• C. $12$
• D. $4$

Answer 1

Answer: (B)

$9$

Median through $C$ is $x=4$

So clearly the $x$ coordinate 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, $D=(\frac{1+4}{2},\frac{2+y}{2})$.

Now, we have, $\frac{3+4+y}{2}=5\Rightarrow y=3$. So, $C=(4,3)$.

The centroid of the triangle is the intersection of the medians. It is easy to see that the medians $x=4$ and $x+y=5$ intersect at $G=(4,1)$.

The area of triangle $\Delta ABC=3\times \Delta AGC=3\times \frac{1}{2}\times 3\times 2=9$.

(In this case, it is easy as the points $G$ and $C$ lie on the same vertical line $x=4$. So the base $GC=2$ and the altitude from $A$ is $3$ units.)

So the answer is option B.