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Show that the points (0, -1), (-2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area.

Coordinate Geometry

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Answer

Let A(0, -1), B(-2, 3), C(6, 7) and D(8, 3).

Show that the points (0, -1), (-2, 3), (6, 7) and (8, 3), taken in order, are the vertices of a rectangle. Also find its area. Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By distance formula,

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

AB=(−2−0)2+[3−(−1)]2=(−2)2+[3+1]2=4+42=4+16=20.BC=[6−(−2)]2+(7−3)2=[6+2]2+(4)2=82+42=64+16=80.CD=(8−6)2+(3−7)2=22+(−4)2=4+16=20AD=(8−0)2+[3−(−1)]2=82+[3+1]2=64+42=64+16=80.AB = \sqrt{(-2 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{(-2)^2 + [3 + 1]^2}\\[1em] = \sqrt{4 + 4^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20}. \\[1em] BC = \sqrt{[6 - (-2)]^2 + (7 - 3)^2} \\[1em] = \sqrt{[6 + 2]^2 + (4)^2} \\[1em] = \sqrt{8^2 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}. \\[1em] CD = \sqrt{(8 - 6)^2 + (3 - 7)^2} \\[1em] = \sqrt{2^2 + (-4)^2} \\[1em] = \sqrt{4 + 16} \\[1em] = \sqrt{20} \\[1em] AD = \sqrt{(8 - 0)^2 + [3 - (-1)]^2} \\[1em] = \sqrt{8^2 + [3 + 1]^2} \\[1em] = \sqrt{64 + 4^2} \\[1em] = \sqrt{64 + 16} \\[1em] = \sqrt{80}.

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Area of rectangle ABCD = AB × BC

=20×80=25×45=8×5=40 sq. units.= \sqrt{20} \times \sqrt{80} \\[1em] = 2\sqrt{5} \times 4\sqrt{5} \\[1em] = 8 \times 5 \\[1em] = 40 \text{ sq. units}.

Hence, area of rectangle = 40 sq. units.

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