Question

the parallelogram ghij is a dilation of the parallelogram ghij. what is the scale factor of the dilation? simplify your answer and write it as a proper fraction, an improper fraction, or a whole number.

Explanation:

Step1: Identify coordinates of a point and its image

Let's take point \( G \) and \( G' \). From the graph, \( G \) is at \( (-8, -8) \) and \( G' \) is at \( (-10, -10) \)? Wait, no, maybe better to take \( J \) and \( J' \). \( J \) is at \( (-8, 0) \) and \( J' \) is at \( (-10, 0) \)? Wait, no, let's check \( I \) and \( I' \). \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \)? Wait, no, looking at the grid: \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \)? Wait, no, let's check the vertical distance or horizontal. Wait, maybe \( G \) is at \( (-8, -8) \), \( G' \) is at \( (-10, -10) \)? Wait, no, the blue parallelogram \( GHIJ \): \( G(-8, -8) \), \( H(4, -4) \), \( I(4, 4) \), \( J(-8, 0) \). The black parallelogram \( G'H'I'J' \): \( G'(-10, -10) \), \( H'(5, -5) \), \( I'(5, 5) \), \( J'(-10, 0) \). Wait, no, maybe the scale factor is the ratio of the length of a side of the image to the original. Let's take the vector from \( J \) to \( I \): \( J(-8, 0) \) to \( I(4, 4) \), the length? Wait, no, dilation scale factor is the ratio of the coordinates of the image to the original. Let's take point \( I \) at \( (4, 4) \) and \( I' \) at \( (5, 5) \)? Wait, no, looking at the grid, \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \)? Wait, no, the x-coordinate of \( I \) is 4, \( I' \) is 5? Wait, no, the grid lines: each square is 1 unit. \( J \) is at \( (-8, 0) \), \( J' \) is at \( (-10, 0) \). So the x-coordinate of \( J \) is -8, \( J' \) is -10. So the scale factor \( k \) is \( \frac{x_{J'}}{x_J} = \frac{-10}{-8} = \frac{5}{4} \)? Wait, no, maybe \( J \) is at \( (-8, 0) \), \( J' \) is at \( (-10, 0) \)? Wait, no, the blue \( J \) is at \( (-8, 0) \), black \( J' \) is at \( (-10, 0) \). So the distance from \( J \) to \( H \): \( J(-8, 0) \) to \( H(4, -4) \)? Wait, no, better to take a point and its image. Let's take \( G \): \( G(-8, -8) \), \( G'(-10, -10) \). So the scale factor \( k \) is \( \frac{x_{G'}}{x_G} = \frac{-10}{-8} = \frac{5}{4} \)? Wait, no, wait \( G \) is at \( (-8, -8) \), \( G' \) is at \( (-10, -10) \). So the ratio of \( G' \) coordinates to \( G \) coordinates: \( \frac{-10}{-8} = \frac{5}{4} \), \( \frac{-10}{-8} = \frac{5}{4} \). Alternatively, take \( I(4, 4) \) and \( I'(5, 5) \): \( \frac{5}{4} \). Wait, but maybe I made a mistake. Wait, the original parallelogram \( GHIJ \): let's check the length of \( JI \). \( J(-8, 0) \) to \( I(4, 4) \): the vector is \( (12, 4) \)? No, wait, \( J(-8, 0) \) to \( I(4, 4) \): x difference 12, y difference 4? No, that's not right. Wait, no, \( G(-8, -8) \), \( H(4, -4) \), \( I(4, 4) \), \( J(-8, 0) \). So \( GH \) is from \( (-8, -8) \) to \( (4, -4) \): x difference 12, y difference 4. \( G'H' \) is from \( (-10, -10) \) to \( (5, -5) \): x difference 15, y difference 5. So the scale factor is \( \frac{15}{12} = \frac{5}{4} \) or \( \frac{5}{4} \). Wait, 15/12 simplifies to 5/4. Alternatively, take the length of \( GI \): \( G(-8, -8) \) to \( I(4, 4) \): distance is \( \sqrt{(12)^2 + (12)^2} = 12\sqrt{2} \). \( G'(-10, -10) \) to \( I'(5, 5) \): distance is \( \sqrt{(15)^2 + (15)^2} = 15\sqrt{2} \). So scale factor is \( \frac{15\sqrt{2}}{12\sqrt{2}} = \frac{15}{12} = \frac{5}{4} \). Wait, but maybe I messed up the coordinates. Wait, looking at the graph, \( J \) is at \( (-8, 0) \), \( J' \) is at \( (-10, 0) \). So the x-coordinate of \( J \) is -8, \( J' \) is -10. So the scale factor is \( \frac{-10}{-8} = \frac{5}{4} \). Wait, but let's check another point: \( H \) is at \( (4, -4) \), \( H' \) is at \( (5, -5) \). So \( \frac{5}{4} \) for x and y. So the scale factor is \( \frac{5}{4} \)? Wait, no, wait the original \( G \) is at \( (-8, -8) \), \( G' \) is at \( (-10, -10) \). So \( -10 = k \times (-8) \), so \( k = \frac{-10}{-8} = \frac{5}{4} \). Yes, that makes sense. So the scale factor is \( \frac{5}{4} \)? Wait, but let's check the vertical distance. \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \). So \( 5 = k \times 4 \), so \( k = \frac{5}{4} \). Yes, that's correct.

Wait, maybe I made a mistake in coordinates. Let's re-express:

Original parallelogram \( GHIJ \):

- \( G \): (-8, -8)
- \( H \): (4, -4)
- \( I \): (4, 4)
- \( J \): (-8, 0)

Image parallelogram \( G'H'I'J' \):

- \( G' \): (-10, -10)
- \( H' \): (5, -5)
- \( I' \): (5, 5)
- \( J' \): (-10, 0)

Now, take point \( G(-8, -8) \) and \( G'(-10, -10) \). The scale factor \( k \) is given by \( G' = k \times G \), so:

\( -10 = k \times (-8) \)

Solving for \( k \):

\( k = \frac{-10}{-8} = \frac{5}{4} \)

Similarly, for \( H(4, -4) \) and \( H'(5, -5) \):

\( 5 = k \times 4 \)

\( k = \frac{5}{4} \)

And for \( I(4, 4) \) and \( I'(5, 5) \):

\( 5 = k \times 4 \)

\( k = \frac{5}{4} \)

And for \( J(-8, 0) \) and \( J'(-10, 0) \):

\( -10 = k \times (-8) \)

\( k = \frac{5}{4} \)

So the scale factor is \( \frac{5}{4} \)? Wait, but wait, the original length from \( J \) to \( I \): \( J(-8, 0) \) to \( I(4, 4) \): the vector is \( (12, 4) \), length \( \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \). The image length from \( J'(-10, 0) \) to \( I'(5, 5) \): vector \( (15, 5) \), length \( \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \). Then the scale factor is \( \frac{5\sqrt{10}}{4\sqrt{10}} = \frac{5}{4} \). Yes, that confirms it.

Step2: Calculate the scale factor

The scale factor \( k \) is the ratio of the coordinates (or lengths) of the image to the original. Using point \( I(4, 4) \) and \( I'(5, 5) \), the scale factor is \( \frac{5}{4} \). Wait, but wait, maybe I had the coordinates wrong. Wait, looking at the graph again: \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \)? Wait, no, the x-coordinate of \( I \) is 4, \( I' \) is 5? Wait, the grid lines: each square is 1 unit. So from \( x=4 \) to \( x=5 \) is 1 unit, \( y=4 \) to \( y=5 \) is 1 unit. So \( I \) is at \( (4, 4) \), \( I' \) is at \( (5, 5) \). So the scale factor is \( \frac{5}{4} \). Wait, but let's check \( J \): \( J \) is at \( (-8, 0) \), \( J' \) is at \( (-10, 0) \). So \( -10 = k \times (-8) \), so \( k = \frac{10}{8} = \frac{5}{4} \). Yes, that's correct.

Answer:

\( \frac{5}{4} \)