108k views
5 votes
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
)

User CKT
by
8.2k points

2 Answers

2 votes
C would be the right answer!

If you have any more questions please don't be hesitant to ask me:))
User Cannin
by
8.2k points
3 votes

Answer:

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.

User Mariajose
by
7.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories