X1+X2 Y1+Y2
--------- , ---------
 2 2
Thats the equation. You write that (the equation above) in parentheses.
        
             
        
        
        
Define
![{x} =   \left[\begin{array}{ccc}x_{1}\\x_{2}\end{array}\right]](https://tex.z-dn.net/?f=%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%5C%5Cx_%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20)
Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)
Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)
That is,
![\dot{x} =   \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right] x(0)](https://tex.z-dn.net/?f=%5Cdot%7Bx%7D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20x%280%29)
Note that
![\left[\begin{array}{ccc}0&1\\-1&09\end{array}\right]   \left[\begin{array}{ccc}cos(t)&sin(t)\\-sin(t)&cos(t)\end{array}\right] =  \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%2609%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28t%29%26sin%28t%29%5C%5C-sin%28t%29%26cos%28t%29%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-sin%28t%29%26cos%28t%29%5C%5C-cos%28t%29%26-sin%28t%29%5Cend%7Barray%7D%5Cright%5D%20)
Therefore
![x(t) =   \left[\begin{array}{ccc}0&1\\-1&0\end{array}\right] x(t)](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D%20x%28t%29) 
 
        
        
        
The answer is true. Exterior Angle Theorem: The measurement of an exterior angle of a triangle is equal to the sum of the two remote interior angles. <span />