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Equations of pairs of opposite sides of a parallelogram are x2 – 7x+ 6 = 0 and y2 – 14y + 40 = 0. Find the joint equation of its diagonals.

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Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x2 – 7x + 6 = 0 and the combined equation of sides BC and AD is y2 – 14y + 40 = 0.

The separate equations of the lines 

represented by x2 – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.

Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.

The separate equations of the lines

represented by y2 – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.

Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).

∴ equation of the diagonal AC is

∴ -5y + 50 = 6x – 6

∴ 6x + 5y – 56 = 0

and equation of the diagonal BD is

∴ 5y – 20 = 6x – 6

∴ 6x – 5y + 14 = 0

Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.

∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0

∴ 36x2 – 30xy + 84x + 30xy – 25y2 + 70y – 336x + 280y – 784 = 0

∴ 36x2 – 25y2 – 252x + 350y – 784 = 0.

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