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(B) The given equations are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$. Solving $x^2 - 7x + 6 = 0$,we get $(x - 6)(x - 1) = 0$,so $x = 6$ and $x = 1$

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$6x - 5y + 14 = 0$

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(B) The given equations are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$. Solving $x^2 - 7x + 6 = 0$,we get $(x - 6)(x - 1) = 0$,so $x = 6$ and $x = 1$. Solving $y^2 - 14y + 40 = 0$,we get $(y - 10)(y - 4) = 0$,so $y = 10$ and $y = 4$. The vertices of the parallelogram are the intersection points of these lines: $A(1, 10)$,$B(6, 10)$,$C(6, 4)$,and $D(1, 4)$. The diagonal $BD$ passes through $B(6, 10)$ and $D(1, 4)$. The slope of $BD$ is $m = \frac{4 - 10}{1 - 6} = \frac{-6}{-5} = \frac{6}{5}$. The equation of the line $BD$ is $y - 4 = \frac{6}{5}(x - 1)$,which simplifies to $5(y - 4) = 6(x - 1)$,or $5y - 20 = 6x - 6$,leading to $6x - 5y + 14 = 0$.

If the equations of the opposite sides of a parallelogram are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$,then the equation of one of its diagonals is

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