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.

Result

x2 =  2.97
y2 =  -0.596
x3 =  -1.03
y3 =  -3.596
x4 =  -0.97
y4 =  8.596
x5 =  -4.97
y5 =  5.596

Solution:

x0=−3 y0=1  x1=1 y1=4  a=(x0−x1)2+(y0−y1)2=((−3)−1)2+(1−4)2=5  tan⁡α=y0−y1x0−y1=1−4(−3)−4=37≐0.4286  α=arctan⁡(y0−y1x0−y1)=arctan⁡(1−4(−3)−4)≐0.4049 rad  dx=a⋅ cos⁡(α)=5⋅ cos⁡(0.4049)≐4.5957 dy=a⋅ sin⁡(α)=5⋅ sin⁡(0.4049)≐1.9696  x2=x1+dy=1+1.9696≐2.9696=2.97x_{ 0 } = -3 \ \\ y_{ 0 } = 1 \ \\ \ \\ x_{ 1 } = 1 \ \\ y_{ 1 } = 4 \ \\ \ \\ a = \sqrt{ (x_{ 0 }-x_{ 1 })^2+(y_{ 0 }-y_{ 1 })^2 } = \sqrt{ ((-3)-1)^2+(1-4)^2 } = 5 \ \\ \ \\ \tan α = \dfrac{ y_{ 0 }-y_{ 1 } }{ x_{ 0 }-y_{ 1 } } = \dfrac{ 1-4 }{ (-3)-4 } = \dfrac{ 3 }{ 7 } \doteq 0.4286 \ \\ \ \\ α = \arctan (\dfrac{ y_{ 0 }-y_{ 1 } }{ x_{ 0 }-y_{ 1 } } ) = \arctan (\dfrac{ 1-4 }{ (-3)-4 } ) \doteq 0.4049 \ rad \ \\ \ \\ dx = a \cdot \ \cos(α) = 5 \cdot \ \cos(0.4049) \doteq 4.5957 \ \\ dy = a \cdot \ \sin(α) = 5 \cdot \ \sin(0.4049) \doteq 1.9696 \ \\ \ \\ x_{ 2 } = x_{ 1 } + dy = 1 + 1.9696 \doteq 2.9696 = 2.97
y2=y2=y1−dx=4−4.5957≐−0.5957=−0.596y_{2} =y_{ 2 } = y_{ 1 } - dx = 4 - 4.5957 \doteq -0.5957 = -0.596
x3=x0+dy=(−3)+1.9696≐−1.0304=−1.03x_{ 3 } = x_{ 0 } + dy = (-3) + 1.9696 \doteq -1.0304 = -1.03
y3=y0−dx=1−4.5957≐−3.5957=−3.596y_{ 3 } = y_{ 0 } - dx = 1 - 4.5957 \doteq -3.5957 = -3.596
x4=x4=x1−dy=1−1.9696≐−0.9696=−0.97x_{4} =x_{ 4 } = x_{ 1 } - dy = 1 - 1.9696 \doteq -0.9696 = -0.97
y4=y4=y1+dx=4+4.5957≐8.5957=8.596y_{4} =y_{ 4 } = y_{ 1 } + dx = 4 + 4.5957 \doteq 8.5957 = 8.596
x5=x0−dy=(−3)−1.9696≐−4.9696=−4.97x_{ 5 } = x_{ 0 } - dy = (-3) - 1.9696 \doteq -4.9696 = -4.97
y5=y0+dx=1+4.5957≐5.5957=5.596  a2=(x0−x3)2+(y0−y3)2=((−3)−(−1.0304))2+(1−(−3.5957))2≐5.0004 a3=(x1−x2)2+(y1−y2)2=(1−2.9696)2+(4−(−0.5957))2≐5.0004 a4=(x2−x3)2+(y3−y2)2=(2.9696−(−1.0304))2+((−3.5957)−(−0.5957))2=5  b2=(x0−x5)2+(y0−y5)2=((−3)−(−4.9696))2+(1−5.5957)2≐5.0002 b3=(x1−x4)2+(y1−y4)2=(1−(−0.9696))2+(4−8.5957)2≐5.0004 b4=(x4−x5)2+(y4−y5)2=((−0.9696)−(−4.9696))2+(8.5957−5.5957)2≐5.0002y_{ 5 } = y_{ 0 } + dx = 1 + 4.5957 \doteq 5.5957= 5.596 \ \\ \ \\ a_{ 2 } = \sqrt{ (x_{ 0 }-x_{ 3 })^2+(y_{ 0 }-y_{ 3 })^2 } = \sqrt{ ((-3)-(-1.0304))^2+(1-(-3.5957))^2 } \doteq 5.0004 \ \\ a_{ 3 } = \sqrt{ (x_{ 1 }-x_{ 2 })^2+(y_{ 1 }-y_{ 2 })^2 } = \sqrt{ (1-2.9696)^2+(4-(-0.5957))^2 } \doteq 5.0004 \ \\ a_{ 4 } = \sqrt{ (x_{ 2 }-x_{ 3 })^2+(y_{ 3 }-y_{ 2 })^2 } = \sqrt{ (2.9696-(-1.0304))^2+((-3.5957)-(-0.5957))^2 } = 5 \ \\ \ \\ b_{ 2 } = \sqrt{ (x_{ 0 }-x_{ 5 })^2+(y_{ 0 }-y_{ 5 })^2 } = \sqrt{ ((-3)-(-4.9696))^2+(1-5.5957)^2 } \doteq 5.0002 \ \\ b_{ 3 } = \sqrt{ (x_{ 1 }-x_{ 4 })^2+(y_{ 1 }-y_{ 4 })^2 } = \sqrt{ (1-(-0.9696))^2+(4-8.5957)^2 } \doteq 5.0004 \ \\ b_{ 4 } = \sqrt{ (x_{ 4 }-x_{ 5 })^2+(y_{ 4 }-y_{ 5 })^2 } = \sqrt{ ((-0.9696)-(-4.9696))^2+(8.5957-5.5957)^2 } \doteq 5.0002

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