We will investigate how to determine Hamilton paths and circuits
Hamilton path: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex can be different.
Hamilton circuit: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex must be the same!
As the starting point we can choose any of the points. We will choose point ( F ) and trace a path as follows:
![F\to D\to E\to C\to A\to B\to F](https://tex.z-dn.net/?f=F%5Cto%20D%5Cto%20E%5Cto%20C%5Cto%20A%5Cto%20B%5Cto%20F)
The above path covers all the vertices/points with the starting and ending point/vertex to be ( F ). Such a path is called a Hamilton circuit per definition.
We will choose a different point now. Lets choose ( E ) as our starting point and trace the path as follows:
![E\to D\to F\to B\to A->C](https://tex.z-dn.net/?f=E%5Cto%20D%5Cto%20F%5Cto%20B%5Cto%20A-%3EC)
The above path covers all the vertices/points with the starting and ending point/vertex are different with be ( E ) and ( C ), respectively. Such a path is called a Hamilton path per definition.
One more thing to note is that all Hamilton circuits can be converted into a Hamilton path like follows:
![F\to D\to E\to C\to A\to B](https://tex.z-dn.net/?f=F%5Cto%20D%5Cto%20E%5Cto%20C%5Cto%20A%5Cto%20B)
The above path is a hamilton path that can be formed from the Hamilton circuit example.
But its not necessary for all Hamilton paths to form a Hamilton circuit! Unfortunately, this is not the case in the network given. Every point is in a closed loop i.e there is no loose end/vertex that is not connected by any other vertex.
Using the distance formula:
Distance = √((x2-x1)^2 + (y2-y1)^2)
Distance = √((5-2)^2 + (-8 - -4)^2
Distance = √(3^2 + -4^2)
Distance = √(9 +16)
Distance = √25
Distance = 5
The answer is letter B. The image shows a relection because it is just the same as the original. Reflection is a transformation that gets the 2nd image just the same as the original. The transformation of DEFG is just the same as D'E'F'G'.
280
Answer:
Step-by-step explanation: