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 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=2.97x_{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 } \doteq \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 \ \text{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=2.97
y2=y1−dx=4−4.5957=−0.596y_{2}=y_{1} - dx=4 - 4.5957=-0.596
x3=x0+dy=(−3)+1.9696=−1.03x_{3}=x_{0} + dy=(-3) + 1.9696=-1.03
y3=y0−dx=1−4.5957=−3.596y_{3}=y_{0} - dx=1 - 4.5957=-3.596
x4=x1−dy=1−1.9696=−0.97x_{4}=x_{1} - dy=1 - 1.9696=-0.97
y4=y1+dx=4+4.5957=8.596y_{4}=y_{1} + dx=4 + 4.5957=8.596
x5=x0−dy=(−3)−1.9696=−4.97x_{5}=x_{0} - dy=(-3) - 1.9696=-4.97

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