Question

If $(1,2)$ is the centroid of an equilateral triangle and $(-2,-2)$ is one of the vertices, then

• A. inradius of triangle is $\frac{5}{2}$.
• B. area of triangle is $\frac{225}{4}$.
• C. equation of side opposite to vertex $(-2,-2)$ is $6x+8y=47$ .
• D. area of triangle is $\frac{225}{4\sqrt{3}}$.

Answer 1

Answer: (A), (C), (D)

$\sin 30^{\circ }=\frac{\text{ ID }}{\sqrt{9+16}}\Rightarrow ID$ (inradius) $=\frac{5}{2}$
Also, $AG=\sqrt{9+16}=5$
$\therefore BD$ (length of altitude) $=5+\frac{5}{2}=\frac{15}{2}$
$\therefore$ Area $(△ABC)=\frac{h^2}{\sqrt{3}}=\frac{225}{4\sqrt{3}}$ Ans.
Let co-ordinates of $E(x,y)$
So, $1=\frac{2(x)+1(-2)}{3}\Rightarrow x=\frac{5}{2}$
Also, $2=\frac{2(y)+1(-2)}{3}\Rightarrow y=4$

So, $E\left(\frac{5}{2},4\right)\Rightarrow m_{BC}=\frac{-1}{m_{AE}}=\frac{-3}{4}$
$\Rightarrow$ Equation of $BC$ is $(y-4)=\left(\frac{-3}{4}\right)\left(x-\frac{5}{2}\right)$
$\Rightarrow 6x+8y=47$