The centroid of the triangle formed by the pa | vedclass

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(C) The pair of straight lines is given by $12x^2 - 20xy + 7y^2 = 0$. Factorizing this,we get: $12x^2 - 14xy - 6xy + 7y^2 = 0$ $2x(6x - 7y) - y(6x - 7

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$8$

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(C) The pair of straight lines is given by $12x^2 - 20xy + 7y^2 = 0$. Factorizing this,we get: $12x^2 - 14xy - 6xy + 7y^2 = 0$ $2x(6x - 7y) - y(6x - 7y) = 0$ $(2x - y)(6x - 7y) = 0$ So,the two lines are $2x - y = 0$ and $6x - 7y = 0$. The third line is $2x - 3y + 4 = 0$. To find the vertices of the triangle,we solve the lines in pairs: $1$. Intersection of $2x - y = 0$ and $6x - 7y = 0$: Solving these,we get $x = 0, y = 0$. So,vertex $A = (0, 0)$. $2$. Intersection of $2x - y = 0$ and $2x - 3y + 4 = 0$: Subtracting the equations: $(2x - y) - (2x - 3y + 4) = 0$ $\Rightarrow 2y - 4 = 0$ $\Rightarrow y = 2$. Substituting $y = 2$ in $2x - y = 0$,we get $x = 1$. So,vertex $B = (1, 2)$. $3$. Intersection of $6x - 7y = 0$ and $2x - 3y + 4 = 0$: From $6x - 7y = 0$,$x = \frac{7y}{6}$. Substituting in $2x - 3y + 4 = 0$: $2(\frac{7y}{6}) - 3y + 4 = 0$ $\Rightarrow \frac{7y}{3} - 3y + 4 = 0$ $\Rightarrow -\frac{2y}{3} = -4$ $\Rightarrow y = 6$. Then $x = \frac{7(6)}{6} = 7$. So,vertex $C = (7, 6)$. The centroid $(\alpha, \beta)$ is given by $(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$: $\alpha = \frac{0 + 1 + 7}{3} = \frac{8}{3}$ $\beta = \frac{0 + 2 + 6}{3} = \frac{8}{3}$ Thus,$\alpha + 2\beta = \frac{8}{3} + 2(\frac{8}{3}) = \frac{8 + 16}{3} = \frac{24}{3} = 8$.

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,$\alpha + 2\beta =$

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