Question

If area of the triangle formed by the points $(1,-2),(3,4)$ and $(2,0)$ is half of the area of triangle formed by the points $(3,4),(2,0)$ and $(\alpha ,{\alpha }^2)$, then distance between centroids of respective triangles is

• A. $\sqrt{\frac{35}{3}}$ units
• B. $2$ units
• C. $\sqrt{\frac{37}{3}}$ units
• D. $\sqrt{\frac{39}{3}}$ units

Answer 1

The correct option is C $\sqrt{\frac{37}{3}}$ units

$\frac{1}{2}\left|\begin{array}{cccc}1& 3& 2& 1\\ -2& 4& 0& -2\end{array}\right|=\frac{1}{4}\left|\begin{array}{cccc}3& \alpha & 2& 3\\ 4& {\alpha }^2& 0& 4\end{array}\right|$ $\Rightarrow 4=|{\alpha }^2-4\alpha +8|\Rightarrow {\alpha }^2-4\alpha +8=4(\because {\alpha }^2-4\alpha +8=(\alpha -2{)}^2+4>0)$
$\Rightarrow {\alpha }^2-4\alpha +4=0$
$(\alpha -2{)}^2=0$

$\therefore \alpha =2$

$\therefore$ centroid of triangle with coordinates $(1,-2),(3,4)$ and $(2,0)$ is $G_1=(2,\frac{2}{3})$
Centroid of triangle with coordinates $(3,4),(2,0)$ and $(2,4)$ is $G_2=(\frac{7}{3},\frac{8}{3})$
$\therefore G_1{G}_2=\sqrt{{(2-\frac{7}{3})}^2+{\left(\frac{6}{3}\right)}^2}=\sqrt{\frac{1}{9}+\frac{36}{9}}=\frac{\sqrt{37}}{3}$