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Question
A(6, -2), B(3, -2) and C(S, 6) are the three vertices of a parallelogram ABCD. Find the coordinates of the fourth vertex c.
Solution
We know that in a parallelogram, diagonals bisect each other .
∴ midpoint of AC = midpoint of BD
`"O" ((6 + 8)/2 , (-2 + 6)/2) = "O" (("x" + 3)/2 , ("y" - 2)/2)`
`therefore (6 + 8)/2 = ("x" + 3)/2 , (-2 + 6)/2 = ("y" - 2)/2`
14 = x + 3 , 4 = y - 2
x = 11 , y = 6
the coordinates of the fourth vertex Dare ( 11,6)
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Solution :
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∴ According to the midpoint theorem,
x = `(x_1 + x_2)/2 = (square + 6)/2 = square/2 = square`
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∴ Co-ordinates of midpoint P are `square`.
