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

Step-by-step explanation:

x0=3 y0=1  x1=1 y1=4  a=(x0x1)2+(y0y1)2=((3)1)2+(14)2=5  tanα=y0y1x0y1=14(3)4=370.4286  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=2.9696
y2=y1dx=44.5957=0.5957
x3=x0+dy=(3)+1.9696=1.0304
y3=y0dx=14.5957=3.5957
x4=x1dy=11.9696=0.9696
y4=y1+dx=4+4.5957=8.5957
x5=x0dy=(3)1.9696=4.9696
y5=y0+dx=1+4.59575.5957=5.5957  a2=(x0x3)2+(y0y3)2=((3)(1.0304))2+(1(3.5957))2=5 a3=(x1x2)2+(y1y2)2=(12.9696)2+(4(0.5957))2=5 a4=(x2x3)2+(y3y2)2=(2.9696(1.0304))2+((3.5957)(0.5957))2=5  b2=(x0x5)2+(y0y5)2=((3)(4.9696))2+(15.5957)2=5 b3=(x1x4)2+(y1y4)2=(1(0.9696))2+(48.5957)2=5 b4=(x4x5)2+(y4y5)2=((0.9696)(4.9696))2+(8.59575.5957)2=5



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