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Mathematics

The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a

  1. rectangle

  2. square

  3. rhombus

  4. trapezium

Coordinate Geometry

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Answer

By distance formula,

The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a? Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

d=(x2−x1)2+(y2−y1)2∴AB=(9−9)2+(6−0)2=0+62=36=6 units.∴BC=(−9−9)2+(6−6)2=(−18)2+0=324=18 units.∴CD=[−9−(−9)]2+(0−6)2=[−9+9]2+(−6)2=0+36=36=6 units.∴AD=(−9−9)2+(0−0)2=(−18)2+0=324=18 units.d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(9 - 9)^2 + (6 - 0)^2} \\[1em] = \sqrt{0 + 6^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore BC = \sqrt{(-9 - 9)^2 + (6 - 6)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}. \\[1em] \therefore CD = \sqrt{[-9 - (-9)]^2 + (0 - 6)^2} \\[1em] = \sqrt{[-9 + 9]^2 + (-6)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore AD = \sqrt{(-9 - 9)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}.

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Hence, Option 1 is the correct option.

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