The centroid of the triangle formed by the pa | vedclass

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(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x -

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

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(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x - y) - 7y(2x - y) = 0 \implies (6x - 7y)(2x - y) = 0$. The equations of the sides of the triangle are: $L_1: 6x - 7y = 0$ $L_2: 2x - y = 0$ $L_3: 2x - 3y + 4 = 0$ Solving for the vertices: Intersection of $L_1$ and $L_2$: $(0, 0)$. Intersection of $L_1$ and $L_3$: $6x - 7y = 0 \implies x = \frac{7y}{6}$. Substituting into $L_3$: $2(\frac{7y}{6}) - 3y + 4 = 0 \implies \frac{7y}{3} - 3y = -4 \implies -\frac{2y}{3} = -4 \implies y = 6, x = 7$. Vertex is $(7, 6)$. Intersection of $L_2$ and $L_3$: $2x - y = 0 \implies y = 2x$. Substituting into $L_3$: $2x - 3(2x) + 4 = 0 \implies -4x = -4 \implies x = 1, y = 2$. Vertex is $(1, 2)$. The vertices are $(0, 0), (7, 6), (1, 2)$. The centroid is $\left(\frac{0+7+1}{3}, \frac{0+6+2}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

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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