Question

Use the image to answer the question.

Smaller solid triangle upper A upper B upper C is graphed within larger dotted triangle upper A prime upper B prime upper C prime on a coordinate plane. A(2,1) B(-1,3) C(-3,-2) A'(4,2)B'(-2,6)C'(-6,-4)

In the image, triangle ABC
is dilated with reference to the origin to create triangle A′B′C′
. What is the magnitude of dilation?

(1 point)
Triangle ABC
is dilated using a magnitude of
, which results in triangle A′B′C′
.

Answer 1

To determine the magnitude of dilation from triangle ABC to triangle A'B'C', we can use the coordinates of corresponding points.

The coordinates of triangle ABC are:
- A(2, 1)
- B(-1, 3)
- C(-3, -2)

The coordinates of triangle A'B'C' are:
- A'(4, 2)
- B'(-2, 6)
- C'(-6, -4)

The formula for dilation with reference to the origin is given by:

\[
(x', y') = k(x, y)
\]

where \( (x, y) \) are the original coordinates, \( (x', y') \) are the dilated coordinates, and \( k \) is the magnitude of dilation.

To find \( k \), we can look at the coordinates of point A:

For point A:
- Original point A = (2, 1)
- Dilated point A' = (4, 2)

Using the dilation formula:
\[
k \cdot (2, 1) = (4, 2)
\]

From this, we can set up the equations:

For x-coordinate:
\[
k \cdot 2 = 4 \implies k = \frac{4}{2} = 2
\]

For y-coordinate:
\[
k \cdot 1 = 2 \implies k = \frac{2}{1} = 2
\]

Thus, the magnitude of dilation \( k \) is 2.

Now, we can confirm this with another point, for instance point B:

For point B:
- Original point B = (-1, 3)
- Dilated point B' = (-2, 6)

Calculating \( k \):
For x-coordinate:
\[
k \cdot (-1) = -2 \implies k = \frac{-2}{-1} = 2
\]

For y-coordinate:
\[
k \cdot 3 = 6 \implies k = \frac{6}{3} = 2
\]

Both points confirm the same magnitude of dilation.

Thus, the magnitude of dilation is:
\[
\boxed{2}
\]