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प्रश्न
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.
उत्तर
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 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 - 10 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
`("y" - 10)/("x" - 1) = (10 - 4)/(1 - 6) = 6/-5`
∴ 5y + 50 = 6x − 6
∴ 6x + 5y − 56 = 0
and equation of the diagonal BD is
`("y" - 4)/("x" - 1) = (4 - 10)/(1 - 6) = (-6)/-5 = 6/5`
∴ 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
∴ `36"x"^2 - 30"xy" + 84"x" + 30"xy" - 25"y"^2 + 70"y" - 336"x" + 280"y" - 784 = 0`
∴ `36"x"^2 - 25"y"^2 - 252"x" + 350"y" - 784 = 0`
[Note: Answer in the textbook is incorrect]
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