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Show that the points (a, b), (a + 3, b + 4), (a - 1, b + 7) and (a - 4, b + 3) are the vertices of a parallelogram.

Section Formula

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Answer

Let the points be A(a, b), B(a + 3, b + 4), C(a - 1, b + 7) and D(a - 4, b + 3).

We know that,

Distance formula = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

By distance formula,

AB=(a+3−a)2+(b+4−b)2=32+42=9+16=25=5 units.BC=[(a−1)−(a+3)]2+[(b+7)−(b+4)]2=[a−a−1−3]2+[b−b+7−4]2=[−4]2+[3]2=16+9=25=5 units.CD=[(a−4)−(a−1)]2+[(b+3)−(b+7)]2=[a−a−4+1]2+[b−b+3−7]2=[−3]2+[−4]2=9+16=25=5 units.DA=[a−(a−4)]2+[b−(b+3)]2=42+(−3)2=16+9=25=5 units.AB = \sqrt{(a + 3 - a)^2 + (b + 4 - b)^2} \\[1em] = \sqrt{3^2 + 4^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5\text{ units}. \\[1em] BC = \sqrt{[(a - 1) - (a + 3)]^2 + [(b + 7) - (b + 4)]^2} \\[1em] =\sqrt{[a - a - 1 - 3]^2 + [b - b + 7 - 4]^2} \\[1em] = \sqrt{[-4]^2 + [3]^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] CD = \sqrt{[(a - 4) - (a - 1)]^2 + [(b + 3) - (b + 7)]^2} \\[1em] = \sqrt{[a - a - 4 + 1]^2 + [b - b + 3 - 7]^2} \\[1em] = \sqrt{[-3]^2 + [-4]^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}. \\[1em] DA = \sqrt{[a - (a - 4)]^2 + [b - (b + 3)]^2} \\[1em] = \sqrt{4^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.

Since, AB = BC = CD = DA.

Hence, proved that ABCD are the vertices of a parallelogram.

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