Answer:
![A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})](https://tex.z-dn.net/?f=A%280%2C0%29%5Crightarrow%20A%27%280%2C0%29%20%5C%5C%20B%282%2C0%29%5Crightarrow%20B%27%28%5Cfrac%7B2%7D%7B3%7D%2C0%29%20%5C%5C%20C%280%2C2%29%5Crightarrow%20C%27%280%2C%5Cfrac%7B2%7D%7B3%7D%29)
Step-by-step explanation:
The question is as following:
The verticies of a triangle on the coordinate plane are
A(0, 0), B(2, 0) and C(0, 2).
What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?
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Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).
IF the triangle is dilated by a factor of k about the origin, then
(x,y) → (kx , ky)
that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.
If a figure dilated by a factor of 1/3 about the origin
So, ![(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Crightarrow%20%28%5Cfrac%7B1%7D%7B3%7Dx%2C%5Cfrac%7B1%7D%7B3%7Dy%29)
<u>So, The coordinates of the triangle A'B'C' are:</u>
![A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})](https://tex.z-dn.net/?f=A%280%2C0%29%5Crightarrow%20A%27%280%2C0%29%20%5C%5C%20B%282%2C0%29%5Crightarrow%20B%27%28%5Cfrac%7B2%7D%7B3%7D%2C0%29%20%5C%5C%20C%280%2C2%29%5Crightarrow%20C%27%280%2C%5Cfrac%7B2%7D%7B3%7D%29)