The equations to a pair of opposite sides of | 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 $y = 1, 5$.These represent the lines $x=2, x

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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 $y = 1, 5$.These represent the lines $x=2, x=3, y=1, y=5$,which form a rectangle with vertices $(2,1), (3,1), (3,5),$ and $(2,5)$.The diagonal $d_1$ passes through $(2,1)$ 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 $d_2$ passes through $(3,1)$ and $(2,5)$. Its equation is $y - 1 = \frac{5 - 1}{2 - 3}(x - 3)$ $\Rightarrow y - 1 = -4(x - 3)$ $\Rightarrow y - 1 = -4x + 12$ $\Rightarrow 4x + y = 13$.Thus,the equations of the diagonals are $4x + y = 13$ and $y = 4x - 7$.

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

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