Question

Centroid of the triangle formed by the lines $(x^2+7xy+2y^2)(y-1)=0$ is

• A. $(-\frac{7}{3},\frac{2}{3})$
• B. $(\frac{7}{3},\frac{2}{3})$
• C. $(\frac{7}{3},-\frac{2}{3})$
• D. $(-\frac{7}{3},-\frac{2}{3})$

Answer 1

The correct option is A $(-\frac{7}{3},\frac{2}{3})$

Lines of triangle are given by
$(x^2+7xy+2y^2)(y-1)=0$
$\Rightarrow x^2+7xy+2y^2=0$ and $y=1$
Putting $y=1$ in $x^2+7xy+2y^2=0$
$x^2+7x+2=0$ if $x_1,x_2$ are its roots then $x_1+x_2=-7$ and lines $x^2+7xy+2y^2=0$ passes through $(0,0)$
$\therefore$ Third vertex of the triangle (say $x_3)$ must be zero
$\therefore$ x-coordinate of centroid be
$\frac{x_1+x_2+x_3}{3}=\frac{-7+0}{3}=\frac{7}{3}$
And $y$ co-ordinate of centroid be $\frac{1+0+1}{3}=\frac{2}{3}$
$\therefore$ Required point of centroid be $(-\frac{7}{3},\frac{2}{3})$
Hence (a) is correct choice.