If we know the three Vertexes of square ... Can we find the fourth Vertex?
what are the equations of theses vertexes?
Sides of the square is not parallel to axes Coordinateness.
1. Let A, B, C denote the known vertices.
2. The point D is the intersection-point of the parallels to $\displaystyle \overline{AB}$ and $\displaystyle \overline{BC}$. This case is drawn in blue.
3. The point D is the image of B by reflection over $\displaystyle \overline{AC}$. This case is drawn in red.
That looks fine to me.
I've found a 3rd way to get the 4th vertex:
Let $\displaystyle A(A_x, A_y),\ B(B_x, B_y),\ C(C_x, C_y)$ denote the known vertices of the square.
The midpoint of the square is M. Then the M has the coordinates:
$\displaystyle M\left(\dfrac{A_x+C_x}2,\ \dfrac{A_y+C_y}2 \right)$
M is simultaneously the midpoint of BD:
$\displaystyle \left(\dfrac{A_x+C_x}2,\ \dfrac{A_y+C_y}2 \right) = \left(\dfrac{B_x+D_x}2,\ \dfrac{B_y+D_y}2 \right)$
Calculate $\displaystyle D_x, D_y$!
I've got: $\displaystyle D\left(A_x+C_x-B_x,\ A_y+C_y-B_y \right)$
In my opinion this way seems to be the most simple one.
Hello, rqeeb!
If you make a sketch, the answer is obvious.
If we know the three vertexes of square, can we find the fourth vertex?
Suppose the three given vertices are: .$\displaystyle A(1,5),\;B(2,2),\;C(5,3)$
Plot points $\displaystyle A,B,C.$
Code:| D | ♠ | (1,5) : 1 | A ♥ - - - - - . | 3 | | | | C | ♥ (5,3) | :1 | B ♥ - - - - - . | (2,2) 3 | - - + - - - - - - - - - - - - |
Going from $\displaystyle B$ to $\displaystyle C$, we move right 3 and up 1.
Since $\displaystyle AD \parallel BC$ and $\displaystyle AD = BC,$
. . going from $\displaystyle A$ to $\displaystyle D$, we again move right 3 and up 1.
Therefore, $\displaystyle D$ is located at $\displaystyle (4,6).$