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:

$$\begin{gathered} x_0=-3 \\ {y}_0=1 \\ {x}_1=1 \\ {y}_1=4 \\ a=\sqrt{(x_0-x_1{)}^2+(y_0-y_1{)}^2}=\sqrt{((-3)-1{)}^2+(1-4{)}^2}=5 \\ \tan \alpha ={\displaystyle \frac{y_0-y_1}{x_0-y_1}}={\displaystyle \frac{1-4}{(-3)-4}}={\displaystyle \frac{3}{7}}\doteq 0.4286 \\ \alpha =\arctan ({\displaystyle \frac{y_0-y_1}{x_0-y_1}})=\arctan ({\displaystyle \frac{1-4}{(-3)-4}})\doteq 0.4049\text{ rad } \\ {d}_x=a\cdot \cos (\alpha )=5\cdot \cos (0.4049)\doteq 4.5957 \\ {d}_y=a\cdot \sin (\alpha )=5\cdot \sin (0.4049)\doteq 1.9696 \\ {x}_2=x_1+d_y=1+1.9696=2.9696 \end{gathered}$$

$y_2=y_2=y_1-d_x=4-4.5957=-0.5957$

$x_3=x_0+d_y=(-3)+1.9696=-1.0304$

$y_3=y_0-d_x=1-4.5957=-3.5957$

$x_4=x_4=x_1-d_y=1-1.9696=-0.9696$

$y_4=y_4=y_1+d_x=4+4.5957=8.5957$

$x_5=x_0-d_y=(-3)-1.9696=-4.9696$

$$\begin{gathered} y_5=y_0+d_x=1+4.5957\doteq 5.5957 \\ {a}_2=\sqrt{(x_0-x_3{)}^2+(y_0-y_3{)}^2}=\sqrt{((-3)-(-1.0304){)}^2+(1-(-3.5957){)}^2}=5 \\ {a}_3=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{(1-2.9696{)}^2+(4-(-0.5957){)}^2}=5 \\ {a}_4=\sqrt{(x_2-x_3{)}^2+(y_3-y_2{)}^2}=\sqrt{(2.9696-(-1.0304){)}^2+((-3.5957)-(-0.5957){)}^2}=5 \\ {b}_2=\sqrt{(x_0-x_5{)}^2+(y_0-y_5{)}^2}=\sqrt{((-3)-(-4.9696){)}^2+(1-5.5957{)}^2}=5 \\ {b}_3=\sqrt{(x_1-x_4{)}^2+(y_1-y_4{)}^2}=\sqrt{(1-(-0.9696){)}^2+(4-8.5957{)}^2}=5 \\ {b}_4=\sqrt{(x_4-x_5{)}^2+(y_4-y_5{)}^2}=\sqrt{((-0.9696)-(-4.9696){)}^2+(8.5957-5.5957{)}^2}=5 \end{gathered}$$

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