The centroid of the triangle formed by the pa | vedclass

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(C) The given pair of lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring this,we get $(6x - 7y)(2x - y) = 0$. Thus,the equations of the two lines are $6x -

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$\left(\frac{8}{3}, \frac{8}{3}\right)$

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(C) The given pair of lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring this,we get $(6x - 7y)(2x - y) = 0$. Thus,the equations of the two lines are $6x - 7y = 0$ $(i)$ and $2x - y = 0$ (ii). The third line is $2x - 3y + 4 = 0$ (iii). Solving $(i)$ and (ii): $6x - 7y = 0$ and $2x - y = 0 \implies x=0, y=0$. Vertex $A = (0, 0)$. Solving $(i)$ and (iii): $6x - 7y = 0$ and $2x - 3y = -4$. Multiplying (iii) by $3$: $6x - 9y = -12$. Subtracting from $(i)$: $2y = 12 \implies y = 6, x = 7$. Vertex $B = (7, 6)$. Solving (ii) and (iii): $2x - y = 0$ and $2x - 3y = -4$. Subtracting: $2y = 4 \implies y = 2, x = 1$. Vertex $C = (1, 2)$. The centroid is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+7+1}{3}, \frac{0+6+2}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

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:

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