Coordinates of square vertices

The ABCD square has the center S [−3, −2] and the vertex A [1, −3]. Find the coordinates of the other vertices of the square.

Correct result:

x1 =  -7
y1 =  -1
x2 =  -2
y2 =  2
x3 =  -4
y3 =  -6

Solution:

s0=−3 s1=−2  x0=1 y0=−3  d0=x0−s0=1−(−3)=4 d1=y0−s1=(−3)−(−2)=−1  n⊥d ∣n∣=∣d∣ n.d=0  n0=−d1=−(−1)=1 n1=d0=4  x1=s0−d0=(−3)−4=−7s_{0}=-3 \ \\ s_{1}=-2 \ \\ \ \\ x_{0}=1 \ \\ y_{0}=-3 \ \\ \ \\ d_{0}=x_{0}-s_{0}=1-(-3)=4 \ \\ d_{1}=y_{0}-s_{1}=(-3)-(-2)=-1 \ \\ \ \\ n \perp d \ \\ |n|=|d| \ \\ n . d=0 \ \\ \ \\ n_{0}=-d_{1}=-(-1)=1 \ \\ n_{1}=d_{0}=4 \ \\ \ \\ x_{1}=s_{0}-d_{0}=(-3)-4=-7
y1=s1−d1=(−2)−(−1)=−1y_{1}=s_{1}-d_{1}=(-2)-(-1)=-1
x2=s0+n0=(−3)+1=−2x_{2}=s_{0}+n_{0}=(-3)+1=-2
y2=s1+n1=(−2)+4=2y_{2}=s_{1}+n_{1}=(-2)+4=2
x3=s0−n0=(−3)−1=−4x_{3}=s_{0}-n_{0}=(-3)-1=-4
y3=s1−n1=(−2)−4=−6y_{3}=s_{1}-n_{1}=(-2)-4=-6

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