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(A) The given equation is $18x^2 - 9xy + y^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$ and $y = 6x$.The inte

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$(3, 12)$

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(A) The given equation is $18x^2 - 9xy + y^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$ and $y = 6x$.The intersection point of these two lines is the origin $(0, 0)$.The intersection points of these lines with the line $y = c$ are $(\frac{c}{3}, c)$ and $(\frac{c}{6}, c)$.The area of the triangle formed by these three points $(0, 0)$,$(\frac{c}{3}, c)$,and $(\frac{c}{6}, c)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 27$.$\frac{1}{2} |0(c - c) + \frac{c}{3}(c - 0) + \frac{c}{6}(0 - c)| = 27$.$\frac{1}{2} |\frac{c^2}{3} - \frac{c^2}{6}| = 27$ $\Rightarrow \frac{1}{2} |\frac{c^2}{6}| = 27$ $\Rightarrow c^2 = 324$.Taking $c = 18$ (assuming positive area),the vertices are $(0, 0)$,$(6, 18)$,and $(3, 18)$.The centroid is $(\frac{0 + 6 + 3}{3}, \frac{0 + 18 + 18}{3}) = (3, 12)$.

Suppose a triangle of area $27$ sq. units is formed by $18x^2 - 9xy + y^2 = 0$ and $y = c$. Then the centroid of the triangle is

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