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(D) The given equation of two sides is $x^2 - 7xy + 6y^2 = 0$. Factoring this,we get $(x - 6y)(x - y) = 0$. So,the two sides are $x - 6y = 0$ and $x -

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$2x - 7y - 3 = 0$

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(D) The given equation of two sides is $x^2 - 7xy + 6y^2 = 0$. Factoring this,we get $(x - 6y)(x - y) = 0$. So,the two sides are $x - 6y = 0$ and $x - y = 0$,which intersect at the origin $A(0, 0)$. Let the vertices be $A(0, 0)$,$B(x_1, y_1)$,and $C(x_2, y_2)$. The centroid $G$ is given by $\left(\frac{0 + x_1 + x_2}{3}, \frac{0 + y_1 + y_2}{3}\right) = (1, 0)$. This implies $x_1 + x_2 = 3$ and $y_1 + y_2 = 0$. Since $B(x_1, y_1)$ lies on $x - 6y = 0$,we have $x_1 = 6y_1$. Since $C(x_2, y_2)$ lies on $x - y = 0$,we have $x_2 = y_2$. Substituting these into the centroid equations: $6y_1 + y_2 = 3$ and $y_1 + y_2 = 0 \Rightarrow y_2 = -y_1$. Substituting $y_2 = -y_1$ into $6y_1 + y_2 = 3$,we get $5y_1 = 3$,so $y_1 = \frac{3}{5}$ and $x_1 = \frac{18}{5}$. Then $y_2 = -\frac{3}{5}$ and $x_2 = -\frac{3}{5}$. The third side $BC$ passes through $B\left(\frac{18}{5}, \frac{3}{5}\right)$ and $C\left(-\frac{3}{5}, -\frac{3}{5}\right)$. The slope $m = \frac{-\frac{3}{5} - \frac{3}{5}}{-\frac{3}{5} - \frac{18}{5}} = \frac{-\frac{6}{5}}{-\frac{21}{5}} = \frac{6}{21} = \frac{2}{7}$. The equation is $y - \left(-\frac{3}{5}\right) = \frac{2}{7}(x - (-\frac{3}{5}))$,which simplifies to $y + \frac{3}{5} = \frac{2}{7}(x + \frac{3}{5})$. $7(5y + 3) = 10(x + \frac{3}{5})$ $\Rightarrow 35y + 21 = 10x + 6$ $\Rightarrow 10x - 35y - 15 = 0$. Dividing by $5$,we get $2x - 7y - 3 = 0$.

If two sides of a triangle are represented by $x^2 - 7xy + 6y^2 = 0$ and the centroid is $(1, 0)$,then the equation of the third side is

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