The equations of the pairs of opposite sides | vedclass

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(C) The given equations are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$.Solving these,we get $x = 2, 3$ and $y = 1, 5$.The vertices of the parallelogram

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$4x + y = 13$ and $y = 4x - 7$

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(C) The given equations are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$.Solving these,we get $x = 2, 3$ and $y = 1, 5$.The vertices of the parallelogram are $(2, 1), (3, 1), (3, 5),$ and $(2, 5)$.The diagonal $d_1$ connects $(2, 1)$ and $(3, 5)$. Its equation is $y - 1 = \frac{5 - 1}{3 - 2}(x - 2)$ $\Rightarrow y - 1 = 4(x - 2)$ $\Rightarrow y = 4x - 7$.The diagonal $d_2$ connects $(3, 1)$ and $(2, 5)$. Its equation is $y - 1 = \frac{5 - 1}{2 - 3}(x - 3)$ $\Rightarrow y - 1 = -4(x - 3)$ $\Rightarrow 4x + y = 13$.Thus,the equations of the diagonals are $4x + y = 13$ and $y = 4x - 7$.

The equations of the pairs of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations of its diagonals are:

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