Question

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
?
A) A'(0, 0) B'(3, 0) C'(0, 3)
B) The vertices would not change.
C) A'(0, 0) B'(
2
3
, 0) C'(0,
2
3
)
D) A'(
1
3
,
1
3
) B'(
7
3
,
1
3
) C'(
1
3
,
7
3
)

Answer 1

C would be the right answer!

If you have any more questions please don't be hesitant to ask me:))

Answer 2

The correct option is C.

Explanation:

It is given that the vertices of a triangle on the coordinate plane are A(0, 0), B(2, 0) and C(0, 2).

If a figure dilated by a factor of k about the origin, then

`(x,y)\rightarrow (kx,ky)`

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, then

`(x,y)\rightarrow ((1)/(3)x,(1)/(3)y)`

The vertices of image are

`A(0,0)\rightarrow A'(0,0)`

`B(2,0)\rightarrow B'((2)/(3),0)`

`C(0,2)\rightarrow C'(0,(2)/(3))`

The coordinates of triangle A'B'C' are A'(0,0), B'(2/3,0) and C'(0,2/3).

Therefore, the correct option is C.