Question

The coordinates of the vertices of triangle ABC are A(3,6), B(6, 3), and C(9, 9). If triangle ABC Is dilated by 1/3x and 1/3y to create triangle A'B'C', which set of coordinates represents the vertices of triangle ABC?
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Answer 1

Answer:

$A' = (1,2)$

$B'=(2,1)$

$C' =(3,3)$

Step-by-step explanation:

Given

$A = (3,6)$

$B = (6,3)$

$C = (9,9)$

$Scale = \frac{1}{3}$

Required

Determine the vertices of ABC

If truly the question asks for the vertices of ABC, then the vertices are:

$A = (3,6)$   $B = (6,3)$   $C = (9,9)$

However, I'm quite sure that's not the requirement of the question. So, I'll solve for A'B'C'

To do this, we simply multiply the vertices of ABC by the scale of dilation.

$(A'B'C') = Scale\ Factor * (ABC)$

$A' = \frac{1}{3} * A$

$A' = \frac{1}{3} * (3,6)$

$A' = (\frac{3}{3},\frac{6}{3})$

$A' = (1,2)$

$B' = \frac{1}{3} * B$

$B' = \frac{1}{3} * (6,3)$

$B' = (\frac{6}{3},\frac{3}{3})$

$B'=(2,1)$

$C' = \frac{1}{3} * C$

$C' = \frac{1}{3} * (9,9)$

$C' = (\frac{9}{3},\frac{9}{3})$

$C' =(3,3)$