Exam

JEE Main

Year

2018

Date

15 April

Shift

Morning Shift

Mode

Online

Subject

Mathematics

Chapter

Straight Lines

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Straight Lines

In a triangle A B C , coordianates of A are (1,2) and the equations...

In a triangle $A B C$, coordianates 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 $\triangle A B C$ (in sq. units) is

Step-by-step solution

Median through $C$ is $x=4$ So the $x$ coordianate of $C$ is 4. let $C \equiv(4, y)$, then the midpoint of $A(1,2)$ and $C(4, y)$ is $D$ which lies on the median through $B$.
\therefore \quad D $\equiv \left(\frac{1+4}{2}, \frac{2+y}{2}\right)Now \frac{1+4+2+y}{2}=5 \Rightarrow y=3$. So, $C$\equiv(4,3)$. The centroid of the triangle is the intersection of the mesians. Here the medians$x=4$and$x+4$and$x+y=5$intersect at$G(4,1)$. The area of triangle$\triangle A B C=3 \times \triangle$A G C$ =3 \times $\frac{1}{2}$[1(1-3)+4(3-2)+4(2-1)]=9.