Question

in the graph below, kite abcd is the image of abcd after a dilation. what are the scale factor and center of the dilation? simplify your answers and write them as fractions or whole numbers. scale factor: center of the dilation: (, )

Explanation:

Step1: Find coordinates of corresponding points

First, identify the coordinates of points \( A \), \( A' \), \( B \), \( B' \) (or other corresponding points).
- Coordinates of \( A \): \( (-9, 1) \)
- Coordinates of \( A' \): \( (-1, 5) \)
- Coordinates of \( B \): \( (6, 1) \)
- Coordinates of \( B' \): \( (4, 5) \)

Step2: Determine the center of dilation

The center of dilation \( (h, k) \) satisfies the property that for any point \( (x, y) \) and its image \( (x', y') \) after dilation, the center lies on the line connecting \( (x, y) \) and \( (x', y') \), and the ratio of distances from the center to the image and to the original point is the scale factor.

Let's assume the center is \( (h, k) \). For point \( A \) and \( A' \):
The vector from \( A \) to \( A' \) is \( (-1 - (-9), 5 - 1) = (8, 4) \)
For point \( B \) and \( B' \):
The vector from \( B \) to \( B' \) is \( (4 - 6, 5 - 1) = (-2, 4) \) Wait, maybe better to use the property that the center lies at the intersection of lines \( AA' \) and \( BB' \).

Equation of line \( AA' \):
Slope \( m_{AA'} = \frac{5 - 1}{-1 - (-9)} = \frac{4}{8} = \frac{1}{2} \)
Equation: \( y - 1 = \frac{1}{2}(x + 9) \) → \( y = \frac{1}{2}x + \frac{9}{2} + 1 = \frac{1}{2}x + \frac{11}{2} \)

Equation of line \( BB' \):
Slope \( m_{BB'} = \frac{5 - 1}{4 - 6} = \frac{4}{-2} = -2 \)
Equation: \( y - 1 = -2(x - 6) \) → \( y = -2x + 12 + 1 = -2x + 13 \)

Find intersection of \( y = \frac{1}{2}x + \frac{11}{2} \) and \( y = -2x + 13 \):
Set \( \frac{1}{2}x + \frac{11}{2} = -2x + 13 \)
Multiply both sides by 2: \( x + 11 = -4x + 26 \)
\( 5x = 15 \) → \( x = 3 \)
Substitute \( x = 3 \) into \( y = -2x + 13 \): \( y = -6 + 13 = 7 \)? Wait, that can't be right. Wait, maybe I made a mistake in coordinates. Wait, looking at the graph again, maybe I misread the coordinates. Let's recheck the graph:

Wait, the original kite \( ABCD \): \( A \) is at \( (-9, 1) \), \( B \) at \( (6, 1) \), \( D \) at \( (3, -8) \), \( C \) at \( (8, -5) \)? Wait no, the image \( A'B'C'D' \): \( A'(-1,5) \), \( B'(4,5) \), \( D'(3,2) \), \( C'(5,3) \). Wait, maybe better to use the midpoint or check the center. Wait, another approach: the center of dilation is the point that is the same when we reverse the dilation (from \( A'B'C'D' \) to \( ABCD \) would be a dilation with scale factor \( 1/2 \)? Wait, let's check the distance between \( A \) and \( B \): \( AB \) length: \( 6 - (-9) = 15 \) (since y-coordinates are same, horizontal distance). \( A'B' \) length: \( 4 - (-1) = 5 \) (horizontal distance, y-coordinates same). So the ratio of \( A'B' \) to \( AB \) is \( 5/15 = 1/3 \)? Wait no, \( AB \) is from \( x=-9 \) to \( x=6 \), so length \( 6 - (-9) = 15 \). \( A'B' \) is from \( x=-1 \) to \( x=4 \), length \( 4 - (-1) = 5 \). So \( 5/15 = 1/3 \)? Wait, but let's check the vertical distance. For \( A \), y=1; \( A' \), y=5. The vertical change from \( A \) to \( A' \) is 4, from center to \( A \) and center to \( A' \). Wait, maybe my coordinate reading was wrong. Let's re-express the coordinates correctly:

Looking at the graph:
- \( A \): (-9, 1) (x=-9, y=1)
- \( A' \): (-1, 5) (x=-1, y=5)
- \( B \): (6, 1) (x=6, y=1)
- \( B' \): (4, 5) (x=4, y=5)

Now, let's find the center by solving the system of equations for lines \( AA' \) and \( BB' \).

Line \( AA' \): passes through (-9,1) and (-1,5). Slope \( m = (5-1)/(-1 - (-9)) = 4/8 = 1/2 \). Equation: \( y - 1 = (1/2)(x + 9) \) → \( y = (1/2)x + 9/2 + 1 = (1/2)x + 11/2 \).

Line \( BB' \): passes through (6,1) and (4,5). Slope \( m = (5-1)/(4 - 6) = 4/(-2) = -2 \). Equation: \( y - 1 = -2(x - 6) \) → \( y = -2x + 12 + 1 = -2x + 13 \).

Find intersection:
Set \( (1/2)x + 11/2 = -2x + 13 \)
Multiply both sides by 2: \( x + 11 = -4x + 26 \)
\( 5x = 15 \) → \( x = 3 \)
Then \( y = -2(3) + 13 = 7 \). Wait, but (3,7) – let's check with another point, say \( D \) and \( D' \). \( D \) is (3, -8), \( D' \) is (3, 2). The line \( DD' \) is vertical (x=3), which passes through (3,7). Perfect! So the center is (3,7).

Step3: Calculate the scale factor

Now, take a point, say \( A(-9,1) \), center \( (3,7) \). Distance from center to \( A \): \( \sqrt{(-9 - 3)^2 + (1 - 7)^2} = \sqrt{(-12)^2 + (-6)^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5} \)

Distance from center to \( A'(-1,5) \): \( \sqrt{(-1 - 3)^2 + (5 - 7)^2} = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)

Scale factor \( k = \frac{\text{distance from center to } A'}{\text{distance from center to } A} = \frac{2\sqrt{5}}{6\sqrt{5}} = \frac{1}{3} \)? Wait, no, wait: Wait, \( A' \) is the image of \( A \) after dilation, so the scale factor is \( \frac{\text{length of } A'B'}{\text{length of } AB} \). Wait, \( AB \) length: \( 6 - (-9) = 15 \) (horizontal distance, since y=1 for both). \( A'B' \) length: \( 4 - (-1) = 5 \) (horizontal distance, y=5 for both). So \( 5/15 = 1/3 \). Wait, but when we check the vertical distance for \( A \): from \( A(-9,1) \) to center (3,7): vertical change is \( 7 - 1 = 6 \), horizontal change \( 3 - (-9) = 12 \). From center (3,7) to \( A'(-1,5) \): vertical change \( 5 - 7 = -2 \), horizontal change \( -1 - 3 = -4 \). The ratio of vertical changes: \( -2/6 = -1/3 \), ratio of horizontal changes: \( -4/12 = -1/3 \). So the scale factor is \( \frac{1}{3} \) (since dilation is a reduction, scale factor less than 1).

Wait, let's verify with point \( D(3, -8) \) and \( D'(3, 2) \). Distance from center (3,7) to \( D \): \( |-8 - 7| = 15 \) (vertical distance, since x=3). Distance from center to \( D' \): \( |2 - 7| = 5 \). So ratio \( 5/15 = 1/3 \). Perfect, that matches.

Answer:

scale factor: \(\frac{1}{3}\)
center of the dilation: \((3, 7)\)