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Show that the following points are vertices of a square:(i) (5, 6), (1,5), (2, 1) and (6,2)(3, 2), (0, 5), (-3, 2) and (0, -1). |
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Answer» AB=BC=CD=AD=AC=BD= ii)Given:A(3, 2) = (x₁, y₁)B(0,5) = (x₂, y₂)C(-3, 2) = (x₃, y₃)D(0, -1) = (x₄, y₄)To Prove:Given POINTS (ABCD) are the vertices of SQUARE.Solution:By applying the distance formula,Distance of AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} (x 2 −x 1 ) 2 +(y 2 −y 1 ) 2 On substituting the values, we GET = \sqrt{(0-3)^2+(5-2)^2} (0−3) 2 +(5−2) 2 = \sqrt{9+9} 9+9 = \sqrt{18} 18 Distance of BC = \sqrt{(x_3-x_2)^2+(y_3-y_2)^2} (x 3 −x 2 ) 2 +(y 3 −y 2 ) 2 On substituting the value, we get = \sqrt{(-3-0)^2+(2-5)^2} (−3−0) 2 +(2−5) 2 = \sqrt{9+9} 9+9 = \sqrt{18} 18 Distance of CD = \sqrt{(x_4-x_3)^2+(y_4-y_3)^2} (x 4 −x 3 ) 2 +(y 4 −y 3 ) 2 On substituting the values, we get = \begin{gathered}\sqrt{(0-(-3))^2\\+(-1-2)^2}\end{gathered} = \sqrt{9+9} 9+9 = \sqrt{18} 18 Distance of DA = \sqrt{(x_4-x_1)^2+(y_4-y_1)^2} (x 4 −x 1 ) 2 +(y 4 −y 1 ) 2 On substituting the values, we get = \sqrt{(0-3)^2+(-1-2)^2} (0−3) 2 +(−1−2) 2 = \sqrt{9+9} 9+9 = \sqrt{18} 18 All the four SIDES are equal. Thus, the given points are the vertices of a square.Step-by-step explanation:Since AB=BC=CD=AD and diagonals AC= BDTherefore, ABCD is a square. |
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