Question

If the points $(2a,\,a),\,(a,\,2a)$ and $(a,\,\,a)$ form a triangle of area $32\text{ }sq$ units, then the centroid of the triangle is

• A. $(32,\,\,32)$
• B. $(-32,\,\,-32)$
• C. $(3,\,3)$
• D. $\left( \frac{32}{3},\,\frac{32}{3} \right)$

Answer 1

Answer: (D)

$\left( \frac{32}{3},\,\frac{32}{3} \right)$

Answer 2

$\because$ Area of triangle,
$=\frac{1}{2}\left\| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right\|$
$\therefore$ $32=\frac{1}{2}\left\| \begin{matrix} 2a & a & 1 \\\ a & 2a & 1 \\\ a & a & 1 \\\ \end{matrix} \right\|$
$\Rightarrow$ $64={{a}^{2}}\left\| \begin{matrix} 2 & 1 & 1 \\\ 1 & 2 & 1 \\\ 1 & 1 & 1 \\\ \end{matrix} \right\|$
$\Rightarrow$ $64={{a}^{2}}|2(2-1)-1(1-1)+1(1-2)|$
$\Rightarrow$ $64={{a}^{2}}|2-0-1|={{a}^{2}}$
$\Rightarrow$ $a=\pm 8$
$\therefore$ The given points becomes.
$(\pm \,\,16,\,\pm 8),\,(\pm \,\,8,\,\,\pm 16)$
and $(\pm \,\,8,\,\,\pm 8)$ .
$\therefore$ Centroid $=\left[ \pm \frac{(16+8+8)}{3},\pm \frac{(8+16+8)}{3} \right]$
$=\left( \pm \frac{32}{3},\pm \frac{32}{3} \right)$