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, $\alpha+2 \beta=$

• A

  $-\frac{4}{3}$
• B

  2
• Correct answer
  C

  8
• D

  $-\frac{8}{3}$

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, $$ \begin{aligned} 12 x^2-14 x y-6 x y+7 y^2 & =0 \\ 2 x(6 x-7 y)-y(6 x-7 y) & =0 \\ 2 x-y=0,6 x-7 y & =0 \end{aligned} $$ and
So, centroid is $\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}$ $$ \begin{array}{ll} \Rightarrow & \frac{0+1+7}{3}, \frac{0+2+6}{3} \\ \Rightarrow & \left(\frac{8}{3}, \frac{8}{3}\right)^3 \end{array} $$ $\therefore \quad \alpha=\beta=\frac{8}{3}$ Now, $\alpha+2 \beta=\frac{8}{3}+\frac{16}{3}=\frac{24}{3}=8$

Hence, the correct answer is option: 'C'.