Question

The centroid of the triangle formed by the pair of straight lines $12x^2-20xy+7y^2=0$ and the line $2x-3y+4=0$ is $(\alpha ,\beta )$. Then, $\alpha +2\beta =$

• A. $-\frac{4}{3}$
• B. 2
• C. 8
• D. $-\frac{8}{3}$

Answer 1

Answer: (C)

Centroid at $12x^2-20xy+7y^2=0$ (pair at straight lines) and $2x-3y+4=0$ is $(\alpha ,\beta )$.
So, $12x^2-14xy-6xy+7y^2=0$
$2x(6x-7y)-y(6x-7y)=0$
$2x-y=0,6x-7y=0$
and $2x-3y+4=0$

So, centroid is $\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$
$\Rightarrow \frac{0+1+7}{3},\frac{0+2+6}{3}$
$\Rightarrow \left(\frac{8}{3,}\frac{8}{3}\right)$
$\therefore \alpha =\beta =\frac{8}{3}$
Now, $\alpha +2\beta =\frac{8}{3}+\frac{16}{3}=\frac{24}{3}=8$