Coordinates of square vertices

I have coordinates of square vertices A / -3; 1/and B/1; 4 /. Find coordinates of vertices C and D, C and D. Thanks, Peter.

Correct answer:

x₂ = 2.9696
y₂ = -0.5957
x₃ = -1.0304
y₃ = -3.5957
x₄ = -0.9696
y₄ = 8.5957
x₅ = -4.9696
y₅ = 5.5957

x_{0} = - 3 y_{0} = 1 x_{1} = 1 y_{1} = 4 a = (x_{0} - x_{1})^{2} + (y_{0} - y_{1})^{2} = ((- 3) - 1)^{2} + (1 - 4)^{2} = 5 tan α = \frac{y _{0} - y _{1}}{x _{0} - y _{1}} = \frac{1 - 4}{( - 3 ) - 4} = \frac{3}{7} ≐ 0.4286 α = arctan (\frac{y _{0} - y _{1}}{x _{0} - y _{1}}) = arctan (\frac{1 - 4}{( - 3 ) - 4}) ≐ 0.4049 rad d_{x} = a \cdot cos α = 5 \cdot cos 0.4049 ≐ 4.5957 d_{y} = a \cdot sin α = 5 \cdot sin 0.4049 ≐ 1.9696 x_{2} = x_{1} + d_{y} = 1 + 1.9696 = 2.9696

y_{2} = y_{1} - d_{x} = 4 - 4.5957 = - 0.5957

x_{3} = x_{0} + d_{y} = (- 3) + 1.9696 = - 1.0304

y_{3} = y_{0} - d_{x} = 1 - 4.5957 = - 3.5957

x_{4} = x_{1} - d_{y} = 1 - 1.9696 = - 0.9696

y_{4} = y_{1} + d_{x} = 4 + 4.5957 = 8.5957

x_{5} = x_{0} - d_{y} = (- 3) - 1.9696 = - 4.9696

y_{5} = y_{0} + d_{x} = 1 + 4.5957 ≐ 5.5957 a_{2} = (x_{0} - x_{3})^{2} + (y_{0} - y_{3})^{2} = ((- 3) - (- 1.0304))^{2} + (1 - (- 3.5957))^{2} = 5 a_{3} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = (1 - 2.9696)^{2} + (4 - (- 0.5957))^{2} = 5 a_{4} = (x_{2} - x_{3})^{2} + (y_{3} - y_{2})^{2} = (2.9696 - (- 1.0304))^{2} + ((- 3.5957) - (- 0.5957))^{2} = 5 b_{2} = (x_{0} - x_{5})^{2} + (y_{0} - y_{5})^{2} = ((- 3) - (- 4.9696))^{2} + (1 - 5.5957)^{2} = 5 b_{3} = (x_{1} - x_{4})^{2} + (y_{1} - y_{4})^{2} = (1 - (- 0.9696))^{2} + (4 - 8.5957)^{2} = 5 b_{4} = (x_{4} - x_{5})^{2} + (y_{4} - y_{5})^{2} = ((- 0.9696) - (- 4.9696))^{2} + (8.5957 - 5.5957)^{2} = 5

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