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प्रश्न
(4, 2) and (-1, 5) are the adjacent vertices ofa parallelogram. (-3, 2) are the coordinates of the points of intersection of its diagonals. Find the coordinates of the other two vertices.
उत्तर
Let the coordinates of C and D be (x, y) and (a , b) respectively.
Midpoint of AC is O coordinates of O are ,
O (-3 , 2) = O `((4 + "x")/2 , (2 + "y")/2)`
`-3 = (4 + "x")/2 , 2 = (2 + "y")/2`
- 6 = 4 + x , 4 = 2 + y
x = - 10 , y = 2
C (-10 , 2)
Similarly , coordinates of midpoint of DB , i.e. O are ,
O (-3 , 2) = O `(("a" - 1)/2 , ("b" + 5)/2)`
`-3 = ("a" - 1)/2 , 2 = ("b" + 5)/2`
- 6 = a - 1 , 4 = b + 5
a = -5 , b = -1
D (- 5 , -1)
Thus , the coordinates of each other two vertices are (-10 , 2) and (-5 - 1)
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Solution:
Here A(–1, 1), B(5, – 3), C(3, 5) and suppose D(x, y) are coordinates of point D.
Using midpoint formula,
x = `(5 + 3)/2`
∴ x = `square`
y = `(-3 + 5)/2`
∴ y = `square`
Using distance formula,
∴ AD = `sqrt((4 - square)^2 + (1 - 1)^2`
∴ AD = `sqrt((square)^2 + (0)^2`
∴ AD = `sqrt(square)`
∴ The length of median AD = `square`
