Question

branliest to correct answer

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

The correct answer is Choice C.

By multiplying the points, we get:

A(0,0) -> A'(1/3*0,1/3*0) = A'(0,0)
B(2,0) -> B'(1/3*2,1/3*0) = B'(2/3,0)
C(0,3) -> C'(1/3*0,1/3*2) = B'(0,2/3)

By close inspection, we see that Choice C matches all these points correctly.

:)