Question

The vertices $A$ and $B$ of a triangle $ABC$ are $(2,1)$ and $(3,-2)$ respectively. If the third vertex $C$ lies on the locus $y=x+9$ and the centroid of $△ABC$ lies on $Y$-axis, then find the area.

Answer 1

$A(2,1),B=(3,-2),C=(x,x+9)$

$Centroid=(\frac{2+3+x}{3},\frac{1-2+x+9}{3})$

$\Rightarrow G=(\frac{58+x}{3},\frac{8+x}{3})$

$G$ lies in Y-axis $\Rightarrow \frac{5+x}{3}=0\Rightarrow x=-5$

$\therefore C=(-5,4),G=(0,1)$

$AB=\sqrt{1^2+3^2}=\sqrt{10}$

$BC=\sqrt{8^2+6^2}=10$

$AC=\sqrt{7^2+3^2}=\sqrt{58}$

$\cos \theta =\left|\frac{(\sqrt{10}{)}^2+(10{)}^2-(\sqrt{58}{)}^2}{2\left(10\right(\left(\sqrt{10}\right)}\right|$

$\cos \theta =\frac{13}{5\sqrt{10}}\Rightarrow \sin \theta =\sqrt{1-\frac{169}{250}}$

$\sin \theta =\frac{9}{5\sqrt{10}}$

$\therefore$ Area of $△$

$=\frac{1}{2}\left(\sqrt{10}\right)\left(10\right)\left(\frac{9}{5\sqrt{10}}\right)$

$=9$ units