Question

In the $△ABC,$ the coordinates of $B$ are $(0,0),AB=2,\angle ABC=\frac{\pi }{3}$ and the middle point of $BC$ has the coordinates $(2,0)$. The centroid of the triangle is

• A. $(\frac{1}{2},\frac{\sqrt{3}}{2})$
• B. $(\frac{5}{3},\frac{1}{\sqrt{3}})$
• C. $(\frac{4+\sqrt{3}}{3},\frac{1}{3})$
• D. $Noneofthese$

Answer 1

Answer: (B)

$(\frac{5}{3},\frac{1}{\sqrt{3}})$

Let coordinates of Point A be $(a,b)$
then by distance of AB , we get $a^2+b^2=4$ .....(1)
And by coordinates of midpoint of $BC$, we get coordinates of point $C$ as $(4,0)$
Now apply cosine rule for angle $B$, we get ${(a-4)}^2+b^2=4^2+2^2-2\times 4\times 2\times \cos \frac{\pi }{3}$ ......(2)
From (1) and (2), we get $a$ and $b$ as $1$ and $\sqrt{3}$ respectively.

Then centroid of given triangle will be $(\frac{1+4+0}{3},\frac{\sqrt{3}+0+0}{3})\Rightarrow (\frac{5}{3},\frac{1}{\sqrt{3}})$