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Question
Three vertices of parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2)
1) the coordinate of the fourth vertex D
2) length of diagonal BD
3) equation of the side AD of the parallelogram ABCD
Solution
Three vertices of a parallelogram taken in order are A(3, 6), B(5, 10) and C(3, 2)
1) We need to find the coordinates of D.
We know that the diagonals of a parallelogram bisect each other.
Let x, y be the coordinates of D.
∴ Mid-point of diagonal AC = `((3+3)/2, (6+2)/2) = (3,4)`
And midpoint of diagonal BD = `((5+x)/2, (10 + y)/2)`
Thus we have
`(5 + x)/2 = 3` and `(10 + y)/2 = 4`
`=> 5 + x = 6` and `10 + y = 8`
=> x = 1 and y = -2
`:. D = (1,-2)`
2) Lenght of diagonal BD = `sqrt((1 - 5)^2 + (-2-10)^2)`
`= sqrt((-4)^2 + (-12)^2)`
`= sqrt(16 + 144)`
`= sqrt(160)`
`= 4sqrt10`
3) A(3,6) = `(x_1. y_1)` and B(5,10) = `(x_2, y_2)`
Slope of line AB = `m(y_2 - y_1)/(x_2 - x_1) = (10 - 6)/(5-3) = 4/2 = 2`
∴ Equation of line AB is given by
`y - y_1 = m(x - x_1)`
`=> y - 6 = 2(x - 3)`
`=> y - 6 = 2x - 6`
`=> 2x - y = 0`
`=> 2x = y`
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