If P, Q, R, S are represented by the complex | vedclass

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(B) Let the complex numbers be $z_P = 4 + i$,$z_Q = 1 + 6i$,$z_R = -4 + 3i$,and $z_S = -1 - 2i$. Calculating the side lengths: $|PQ| = |(1 + 6i) - (4

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Square

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(B) Let the complex numbers be $z_P = 4 + i$,$z_Q = 1 + 6i$,$z_R = -4 + 3i$,and $z_S = -1 - 2i$. Calculating the side lengths: $|PQ| = |(1 + 6i) - (4 + i)| = |-3 + 5i| = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$. $|QR| = |(-4 + 3i) - (1 + 6i)| = |-5 - 3i| = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$. $|RS| = |(-1 - 2i) - (-4 + 3i)| = |3 - 5i| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$. $|SP| = |(4 + i) - (-1 - 2i)| = |5 + 3i| = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$. Since all sides are equal,it is a rhombus. Now check the diagonals: $|PR| = |(-4 + 3i) - (4 + i)| = |-8 + 2i| = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$. $|QS| = |(-1 - 2i) - (1 + 6i)| = |-2 - 8i| = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$. Since the diagonals are equal and all sides are equal,$PQRS$ is a square.

If $P, Q, R, S$ are represented by the complex numbers $4 + i, 1 + 6i, -4 + 3i, -1 - 2i$ respectively,then $PQRS$ is a

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