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.

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=(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+1.96962.9696=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 } = \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 \ 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 \doteq 2.9696 = 2.97
y2=y2=y1dx=44.59570.5957=0.596y_{2} =y_{ 2 } = y_{ 1 } - dx = 4 - 4.5957 \doteq -0.5957 = -0.596
x3=x0+dy=(3)+1.96961.0304=1.03x_{ 3 } = x_{ 0 } + dy = (-3) + 1.9696 \doteq -1.0304 = -1.03
y3=y0dx=14.59573.5957=3.596y_{ 3 } = y_{ 0 } - dx = 1 - 4.5957 \doteq -3.5957 = -3.596
x4=x4=x1dy=11.96960.9696=0.97x_{4} =x_{ 4 } = x_{ 1 } - dy = 1 - 1.9696 \doteq -0.9696 = -0.97
y4=y4=y1+dx=4+4.59578.5957=8.596y_{4} =y_{ 4 } = y_{ 1 } + dx = 4 + 4.5957 \doteq 8.5957 = 8.596
x5=x0dy=(3)1.96964.9696=4.97x_{ 5 } = x_{ 0 } - dy = (-3) - 1.9696 \doteq -4.9696 = -4.97
y5=y0+dx=1+4.59575.5957=5.596  a2=(x0x3)2+(y0y3)2=((3)(1.0304))2+(1(3.5957))25.0004 a3=(x1x2)2+(y1y2)2=(12.9696)2+(4(0.5957))25.0004 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)25.0002 b3=(x1x4)2+(y1y4)2=(1(0.9696))2+(48.5957)25.0004 b4=(x4x5)2+(y4y5)2=((0.9696)(4.9696))2+(8.59575.5957)25.0002y_{ 5 } = y_{ 0 } + dx = 1 + 4.5957 \doteq 5.5957= 5.596 \ \\ \ \\ a_{ 2 } = \sqrt{ (x_{ 0 }-x_{ 3 })^2+(y_{ 0 }-y_{ 3 })^2 } = \sqrt{ ((-3)-(-1.0304))^2+(1-(-3.5957))^2 } \doteq 5.0004 \ \\ a_{ 3 } = \sqrt{ (x_{ 1 }-x_{ 2 })^2+(y_{ 1 }-y_{ 2 })^2 } = \sqrt{ (1-2.9696)^2+(4-(-0.5957))^2 } \doteq 5.0004 \ \\ a_{ 4 } = \sqrt{ (x_{ 2 }-x_{ 3 })^2+(y_{ 3 }-y_{ 2 })^2 } = \sqrt{ (2.9696-(-1.0304))^2+((-3.5957)-(-0.5957))^2 } = 5 \ \\ \ \\ b_{ 2 } = \sqrt{ (x_{ 0 }-x_{ 5 })^2+(y_{ 0 }-y_{ 5 })^2 } = \sqrt{ ((-3)-(-4.9696))^2+(1-5.5957)^2 } \doteq 5.0002 \ \\ b_{ 3 } = \sqrt{ (x_{ 1 }-x_{ 4 })^2+(y_{ 1 }-y_{ 4 })^2 } = \sqrt{ (1-(-0.9696))^2+(4-8.5957)^2 } \doteq 5.0004 \ \\ b_{ 4 } = \sqrt{ (x_{ 4 }-x_{ 5 })^2+(y_{ 4 }-y_{ 5 })^2 } = \sqrt{ ((-0.9696)-(-4.9696))^2+(8.5957-5.5957)^2 } \doteq 5.0002



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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. Pythagorean theorem is the base for the right triangle calculator. Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.

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