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


x0=3 y0=1  x1=1 y1=4  a=(x0x1)2+(y0y1)2=((3)1)2+(14)2=5  tanα=y0y1x0y1=14(3)4370.4286  α=arctan(y0y1x0y1)=arctan(14(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=y1dx=44.5957=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=y0dx=14.5957=3.596y_{3}=y_{0} - dx=1 - 4.5957=-3.596
x4=x1dy=11.9696=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=x0dy=(3)1.9696=4.97x_{5}=x_{0} - dy=(-3) - 1.9696=-4.97

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For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
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See also our trigonometric triangle calculator.

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