Question

Marty sketches triangle DEF on the coordinate plane. He then sketches triangle D'E'F' on the same plane. What is the scale factor going from DEF to D'E'F'?
D (0,0) D' (0,0)
E (6,-3) E' (4,-2)
F ((9,6) F' (6,4)

Answer 1

To determine the scale factor from triangle DEF to triangle D'E'F', we can compare the lengths of corresponding sides of the triangles.

We'll start by calculating the lengths of the sides of triangle DEF:

1. **Length of DE:**
\[
DE = \sqrt{(6 - 0)^2 + (-3 - 0)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}
\]

2. **Length of EF:**
\[
EF = \sqrt{(9 - 6)^2 + (6 - (-3))^2} = \sqrt{(3)^2 + (9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}
\]

3. **Length of DF:**
\[
DF = \sqrt{(9 - 0)^2 + (6 - 0)^2} = \sqrt{(9)^2 + (6)^2} = \sqrt{81 + 36} = \sqrt{117} = 3\sqrt{13}
\]

Next, let's find the lengths of the sides of triangle D'E'F':

1. **Length of D'E':**
\[
D'E' = \sqrt{(4 - 0)^2 + (-2 - 0)^2} = \sqrt{(4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}
\]

2. **Length of E'F':**
\[
E'F' = \sqrt{(6 - 4)^2 + (4 - (-2))^2} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}
\]

3. **Length of D'F':**
\[
D'F' = \sqrt{(6 - 0)^2 + (4 - 0)^2} = \sqrt{(6)^2 + (4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}
\]

Now we can calculate the scale factor for corresponding sides.

For DE to D'E':
\[
\text{Scale factor } = \frac{D'E'}{DE} = \frac{2\sqrt{5}}{3\sqrt{5}} = \frac{2}{3}
\]

For EF to E'F':
\[
\text{Scale factor } = \frac{E'F'}{EF} = \frac{2\sqrt{10}}{3\sqrt{10}} = \frac{2}{3}
\]

For DF to D'F':
\[
\text{Scale factor } = \frac{D'F'}{DF} = \frac{2\sqrt{13}}{3\sqrt{13}} = \frac{2}{3}
\]

Since all corresponding sides have the same scale factor, the scale factor going from triangle DEF to triangle D'E'F' is:

\[
\boxed{\frac{2}{3}}
\]