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Q.
The centroid of the triangle formed by the pair of straight lines $12 x^{2}-20 x y+7 y^{2}=0$ and the line $2 x-3 y+4=0$ is $(\alpha, \beta)$. Then, $\alpha+2 \beta=$
TS EAMCET 2018
Solution:
Centroid at $12 x^{2}-20 x y+7 y^{2}=0$ (pair at straight lines) and $2 x-3 y+4=0$ is $(\alpha, \beta)$.
So, $12 x^{2}-14 x y-6 x y+7 y^{2}=0$
$2 x(6 x-7 y)-y(6 x-7 y)=0$
$2 x-y=0,6 x-7 y=0$
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, $\alpha+2 \beta=\frac{8}{3}+\frac{16}{3}=\frac{24}{3}=8$
