Question

Consider ΔABC
, with vertices A(0,3)
, B(0,1)
, and C(−4,2)
, and ΔDEF
, with vertices D(0,6)
, E(0,2)
, and F(8,4)
. Which transformations could be used to show that the triangles are similar?(1 point)
Responses

ΔABC
is dilated by a scale factor of −2
and is reflected across the y-axis.
triangle upper A upper B upper C is dilated by a scale factor of negative 2 and is reflected across the y -axis.

ΔABC
is dilated by a scale factor of 12
and is reflected across the x-axis.
triangle upper A upper B upper C is dilated by a scale factor of Start Fraction 1 over 2 End Fraction and is reflected across the x -axis.

ΔABC
is dilated by a scale factor of 2 and is reflected across the y-axis.
triangle upper A upper B upper C is dilated by a scale factor of 2 and is reflected across the y -axis.

ΔABC
is dilated by a scale factor of 14
and is reflected across the y-axis.

Answer 1

To determine which transformations could show that triangles \( \Delta ABC \) and \( \Delta DEF \) are similar, we need to analyze the vertices of both triangles.

The coordinates of the vertices are:
- For \( \Delta ABC \):
- \( A(0, 3) \)
- \( B(0, 1) \)
- \( C(-4, 2) \)

- For \( \Delta DEF \):
- \( D(0, 6) \)
- \( E(0, 2) \)
- \( F(8, 4) \)

First, we need to find the side lengths of both triangles using the distance formula.

For \( \Delta ABC \):
1. Length \( AB \):
\[
AB = \sqrt{(0 - 0)^2 + (3 - 1)^2} = \sqrt{0 + 4} = 2
\]

2. Length \( BC \):
\[
BC = \sqrt{(0 - (-4))^2 + (1 - 2)^2} = \sqrt{(4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]

3. Length \( AC \):
\[
AC = \sqrt{(0 - (-4))^2 + (3 - 2)^2} = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17}
\]

By calculating, we see that:
- \( AB = 2 \)
- \( BC = \sqrt{17} \)
- \( AC = \sqrt{17} \)

For \( \Delta DEF \):
1. Length \( DE \):
\[
DE = \sqrt{(0 - 0)^2 + (6 - 2)^2} = \sqrt{0 + 16} = 4
\]

2. Length \( EF \):
\[
EF = \sqrt{(0 - 8)^2 + (2 - 4)^2} = \sqrt{(8)^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]

3. Length \( DF \):
\[
DF = \sqrt{(0 - 8)^2 + (6 - 4)^2} = \sqrt{(8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}
\]

By calculating, we see that:
- \( DE = 4 \)
- \( EF = \sqrt{68} \)
- \( DF = \sqrt{68} \)

Now, to determine similarity we should verify the ratios of corresponding sides:
- Ratio of \( AB \) to \( DE \):
\[
\frac{AB}{DE} = \frac{2}{4} = \frac{1}{2}
\]

- Ratio of \( BC \) to \( EF \):
\[
\frac{BC}{EF} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{1}{2}
\]

- Ratio of \( AC \) to \( DF \):
\[
\frac{AC}{DF} = \frac{\sqrt{17}}{\sqrt{68}} = \frac{1}{2}
\]

All three ratios are equal to \( \frac{1}{2} \), indicating that triangles \( \Delta ABC \) and \( \Delta DEF \) are indeed similar.

### Conclusion:
From the choices provided, the most suitable transformations that map \( \Delta ABC \) to \( \Delta DEF \) are:
- \( \Delta ABC \) is dilated by a scale factor of \( \frac{1}{2} \) and is reflected across the x-axis.

So, the answer to the question is:
- \( \Delta ABC \) is dilated by a scale factor of \( \frac{1}{2} \) and is reflected across the x-axis.