If (a, b) is the centroid of the triangle for | vedclass

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(C) The equation $4x^2 - 17xy + 4y^2 = 0$ can be factored as $(4x - y)(x - 4y) = 0$. Thus,the lines are $L_1: 4x - y = 0$ and $L_2: x - 4y = 0$. The t

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$\frac{65}{6}$

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(C) The equation $4x^2 - 17xy + 4y^2 = 0$ can be factored as $(4x - y)(x - 4y) = 0$. Thus,the lines are $L_1: 4x - y = 0$ and $L_2: x - 4y = 0$. The third line is $L_3: x + y = 5$. To find the vertices of the triangle,we solve the pairs of equations: $1$) $L_1$ and $L_2$: $4x - y = 0$ and $x - 4y = 0$ gives the vertex $V_1 = (0, 0)$. $2$) $L_1$ and $L_3$: $4x - y = 0$ and $x + y = 5$ gives $5x = 5$,so $x = 1, y = 4$. Thus $V_2 = (1, 4)$. $3$) $L_2$ and $L_3$: $x - 4y = 0$ and $x + y = 5$ gives $5y = 5$,so $y = 1, x = 4$. Thus $V_3 = (4, 1)$. The centroid $(a, b)$ is $(\frac{0+1+4}{3}, \frac{0+4+1}{3}) = (\frac{5}{3}, \frac{5}{3})$. So,$a = \frac{5}{3}$ and $b = \frac{5}{3}$. The area $c$ of the triangle with vertices $(0, 0), (1, 4), (4, 1)$ is given by $\frac{1}{2} |0(4-1) + 1(1-0) + 4(0-4)| = \frac{1}{2} |1 - 16| = \frac{15}{2}$. Thus,$a + b + c = \frac{5}{3} + \frac{5}{3} + \frac{15}{2} = \frac{10}{3} + \frac{15}{2} = \frac{20 + 45}{6} = \frac{65}{6}$.

If $(a, b)$ is the centroid of the triangle formed by the lines $4x^2 - 17xy + 4y^2 = 0$ and $x + y - 5 = 0$,and $c$ is the numerical value of the area of the triangle,then $a + b + c =$

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