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

Coordinate Geometry

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Answer

Let A(2, 1), B(0, 3), C(-2, 1) and D(0, -1) be the four points.

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

By distance formula,

d=(x2−x1)2+(y2−y1)2∴AB=(0−2)2+(3−1)2=(−2)2+(2)2=4+4=8 units.∴BC=(−2−0)2+(1−3)2=(−2)2+(−2)2=4+4=8 units.∴CD=[0−(−2)]2+[−1−1]2=[0+2]2+[−2]2=4+4=8 units.∴AD=(0−2)2+(−1−1)2=(−2)2+(−2)2=4+4=8 units.d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(0 - 2)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore BC = \sqrt{(-2 - 0)^2 + (1 - 3)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.\\[1em] \therefore CD = \sqrt{[0 - (-2)]^2 + [-1 - 1]^2} \\[1em] = \sqrt{[0 + 2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore AD = \sqrt{(0 - 2)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] =\sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.

Since, AB = BC = CD = AD i.e. all sides are equal so, ABCD can be a rhombus or a square.

Calculating diagonals,

AC=(−2−2)2+(1−1)2=(−4)2+02=16=4 units.BD=(0−0)2+(−1−3)2=02+(−4)2=16=4 units.AC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} \\[1em] =\sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] BD = \sqrt{(0 - 0)^2 + (-1 - 3)^2} \\[1em] = \sqrt{0^2 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = \sqrt{4} \text{ units}.

Since, diagonals are equal.

∴ ABCD is a square.

Area of square = (side)2

= (AB)2

= (8)2(\sqrt{8})^2

= 8 sq. units.

Hence, proved that (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square and area = 8 sq. units.

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