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Question
Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.
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Solution
Let the vertices of a quadrilateral be A(−2, –1), B(4, 0), C(3, 3), and D(−3, 2).
Slope of AB = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
= `(0 + 1)/(4 + 2)`
= `1/6`
Slope of DC = `(3 - 2)/(3 + 3)`
= `1/6`
Slope of AB = Slope of DC
That means AB || DC
Slope of BC = `(3 - 0)/(3 - 4)`
= `3/(-1)`
= −3
Slope of AD = `(2 + 1)/(-3 + 2)`
= `3/(-1)`
= −3
∴ Slope of BC = Slope of AD
That means BC || AD
Hence, AB || DC, BC || AD
Hence, ABCD is a parallelogram.
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