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प्रश्न
ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find
1) Coordinates of A
2) An equation of diagonal BD
उत्तर
1) We know that the diagonals of a parallelogram bisect each other
So, coordinates of mid-point of `BD = ((5+2)/2 , (8 -4)/2) = (7/2, 2) ` .....(1)
Now the mid-point of diagonal AC = `((x + 4)/2 , (y + 7)/2)` .....(2)
From (1) and (2), we get`((x + 4)/2 , (y + 7)/2) = (7/2, 2)`
Comparing, we get x + 4 = 7 and y + 7 = 4
Thus x = 3 and y = -3
So, the coordinates of point A are (3, -3)
2) Equation of a line is given by `y - y_1 = (y_2 - y_1)/(x_2 - x_1) (x - x_1)`
Coordinates of point B and D are (5, 8) and (2, -4) respectively.
Equation of a diagonal BD, y - 8 = `(-4-8)/(2-5) (x - 5)`
`=> y - 8 = 4(x - 5)`
or 4x - y = 12
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Find coordinates of the midpoint of a segment joining point A(–1, 1) and point B(5, –7)
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = –7
Using midpoint formula,
∴ Coordinates of midpoint of segment AB
= `((x_1 + x_2)/2, (y_1+ y_2)/2)`
= `(square/2, square/2)`
∴ Coordinates of the midpoint = `(4/2, square/2)`
∴ Coordinates of the midpoint = `(2, square)`
