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Q.
If $(1,2)$ is the centroid of an equilateral triangle and $(-2,-2)$ is one of the vertices, then
Straight Lines
Solution:
$\sin 30^{\circ}=\frac{\text { ID }}{\sqrt{9+16}} \Rightarrow ID$ (inradius) $=\frac{5}{2}$
Also, $ AG =\sqrt{9+16}=5$
$\therefore BD$ (length of altitude) $=5+\frac{5}{2}=\frac{15}{2}$
$\therefore$ Area $(\triangle 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 $B C$ is $(y-4)=\left(\frac{-3}{4}\right)\left(x-\frac{5}{2}\right)$
$\Rightarrow 6 x+8 y=47 $
