Answer:
The correct option is;
Jason's statement is correct. RST is the same orientation, shape, and size as ABC
Step-by-step explanation:
Here we have
ABC = (2, 1), (3, 3), (4, 1)
RST = (-4, -2), (-3, 0), (-2, -2)
Therefore the length of the sides are as follows
AB = ![\sqrt{(2-3)^2+(1-3)^2} = \sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B%282-3%29%5E2%2B%281-3%29%5E2%7D%20%3D%20%5Csqrt%7B5%7D)
AC = ![\sqrt{(2-4)^2+(1-1)^2} =2](https://tex.z-dn.net/?f=%5Csqrt%7B%282-4%29%5E2%2B%281-1%29%5E2%7D%20%3D2)
BC = ![\sqrt{(3-4)^2+(3-1)^2} = \sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B%283-4%29%5E2%2B%283-1%29%5E2%7D%20%3D%20%5Csqrt%7B5%7D)
For triangle SRT we have
RS = ![\sqrt{(-4-(-3))^2+(-2-0)^2} = \sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-4-%28-3%29%29%5E2%2B%28-2-0%29%5E2%7D%20%3D%20%5Csqrt%7B5%7D)
RT = ![\sqrt{(-4-(-2))^2+(-2-(-2))^2} = 2](https://tex.z-dn.net/?f=%5Csqrt%7B%28-4-%28-2%29%29%5E2%2B%28-2-%28-2%29%29%5E2%7D%20%3D%202)
ST = ![\sqrt{(-3-(-2))^2+(0-(-2))^2} = \sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-3-%28-2%29%29%5E2%2B%280-%28-2%29%29%5E2%7D%20%3D%20%5Csqrt%7B5%7D)
Therefore their dimensions are equal
However the side with length 2 occurs between (2, 1) and (4, 1) in triangle ABC and between (-4, -2) and (-2, -2) in triangle RST
That is Jason's statement is correct. RST is the same orientation, shape, and size as ABC.