If a square ABCD,where A(0,0), B(2,0), C(2,2) | vedclass

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(D) Given,square $ABCD$ with vertices $A(0,0), B(2,0), C(2,2), D(0,2)$.Applying $f_1(x, y) \longrightarrow (y, x)$:$A(0,0)$ $\longrightarrow A'(0,0),

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parallelogram

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(D) Given,square $ABCD$ with vertices $A(0,0), B(2,0), C(2,2), D(0,2)$.Applying $f_1(x, y) \longrightarrow (y, x)$:$A(0,0)$ $\longrightarrow A'(0,0), B(2,0)$ $\longrightarrow B'(0,2), C(2,2)$ $\longrightarrow C'(2,2), D(0,2)$ $\longrightarrow D'(2,0)$.Applying $f_2(x, y) \longrightarrow (x+3y, y)$:$A'(0,0)$ $\longrightarrow A''(0,0), B'(0,2)$ $\longrightarrow B''(6,2), C'(2,2)$ $\longrightarrow C''(8,2), D'(2,0)$ $\longrightarrow D''(2,0)$.Applying $f_3(x, y) \longrightarrow \left(\frac{x-y}{2}, \frac{x+y}{2}\right)$:$A''(0,0)$ $\longrightarrow A'''(0,0), B''(6,2)$ $\longrightarrow B'''(2,4), C''(8,2)$ $\longrightarrow C'''(3,5), D''(2,0)$ $\longrightarrow D'''(1,1)$.Final vertices: $A(0,0), B(2,4), C(3,5), D(1,1)$.Calculating side lengths:$AB = \sqrt{(2-0)^2 + (4-0)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.$BC = \sqrt{(3-2)^2 + (5-4)^2} = \sqrt{1+1} = \sqrt{2}$.$CD = \sqrt{(1-3)^2 + (1-5)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$.$DA = \sqrt{(0-1)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$.Since opposite sides are equal ($AB=CD$ and $BC=DA$) and adjacent sides are not equal,the figure is a parallelogram.

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