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.

Result

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

Solution:

s0=3 s1=2  x0=1 y0=3  d0=x0s0=1(3)=4 d1=y0s1=(3)(2)=1  nd n=d n.d=0  n0=d1=(1)=1 n1=d0=4  x1=s0d0=(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=s1d1=(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=s0n0=(3)1=4x_{ 3 } = s_{ 0 }-n_{ 0 } = (-3)-1 = -4
y3=s1n1=(2)4=6y_{ 3 } = s_{ 1 }-n_{ 1 } = (-2)-4 = -6



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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. Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.

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