# 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

#### Solution:

$x_{0}=-3 \ \\ y_{0}=1 \ \\ \ \\ x_{1}=1 \ \\ y_{1}=4 \ \\ \ \\ a=\sqrt{ (x_{0}-x_{1})^2+(y_{0}-y_{1})^2 }=\sqrt{ ((-3)-1)^2+(1-4)^2 }=5 \ \\ \ \\ \tan α=\dfrac{ y_{0}-y_{1} }{ x_{0}-y_{1} }=\dfrac{ 1-4 }{ (-3)-4 } \doteq \dfrac{ 3 }{ 7 } \doteq 0.4286 \ \\ \ \\ α=\arctan (\dfrac{ y_{0}-y_{1} }{ x_{0}-y_{1} } )=\arctan (\dfrac{ 1-4 }{ (-3)-4 } ) \doteq 0.4049 \ \text{rad} \ \\ \ \\ dx=a \cdot \ \cos(α)=5 \cdot \ \cos(0.4049) \doteq 4.5957 \ \\ dy=a \cdot \ \sin(α)=5 \cdot \ \sin(0.4049) \doteq 1.9696 \ \\ \ \\ x_{2}=x_{1} + dy=1 + 1.9696=2.97$
$y_{2}=y_{1} - dx=4 - 4.5957=-0.596$
$x_{3}=x_{0} + dy=(-3) + 1.9696=-1.03$
$y_{3}=y_{0} - dx=1 - 4.5957=-3.596$
$x_{4}=x_{1} - dy=1 - 1.9696=-0.97$
$y_{4}=y_{1} + dx=4 + 4.5957=8.596$
$x_{5}=x_{0} - dy=(-3) - 1.9696=-4.97$

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Tips to related online calculators
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.

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