Answer:
The distance CC' is \(\sqrt5units\)
Step-by-step explanation:
Given the transformation T: (x, y) (x + 2, y + 1)
we have to find the distance CC'
Let coordinate of C are (a,b).
Now, by using transformation T the coordinates of C' are (a+2,a+1)
By using distance formula,
\(CC'=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\= \sqrt{(a+2-a)^2+(b+1-b)^2}\\\\=\sqrt{4+1}=\sqrt5 units\)
Hence, the distance CC' is \(\sqrt5units\)
