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Question
Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram
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Solution
The vertices A(−2, −1), B(4, 0), C(3, 3) and D(−3, 2)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(0 + 1)/(4 + 2) = 1/6`
Slope of BC = `(3 - 0)/(3 - 4) = 3/(-1)` = −3
Slope of CD = `(2 - 3)/(-3 - 3) = (-1)/(-6) = 1/6`
Slope of AD = `(2 + 1)/(-3 + 2) = 3/(-1)` = −3
Slope of AB = Slope of CD = `1/6`
∴ AB || CD ...(1)
Slope of BC = Slope of AD = −3
∴ BC || AD ...(2)
From (1) and (2) we get ABCD is a parallelogram.
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