Question

a dilation centered at the origin with a scale factor of 2 is applied to △xyz. the result is △xyz, as shown below. (a) the arrows below show that the coordinates on the left are mapped to the coordinates on the right. fill in the blanks to give the coordinates after the dilation. original coordinates → final coordinates x(-2, 5) → x blank y(4, 3) → y blank z(-3, -5) → z blank (b) choose the general rule below that describes the dilation mapping △xyz to △xyz. options: (x,y)→(2x,y), (x,y)→(2x,2y), (x,y)→(½x,2y), (x,y)→(x,2y), (x,y)→(½x,½y), (x,y)→(2y,2x), (x,y)→(½y,½x), (x,y)→(2x,½y)

Explanation:

Part (a)

Step 1: Dilate \( X(-2, 5) \)

A dilation with scale factor \( 2 \) centered at the origin multiplies each coordinate by \( 2 \). So for \( X(-2, 5) \), we calculate \( x' = -2\times2 = -4 \)? Wait, no, wait. Wait, the problem says the dilation is centered at the origin with scale factor \( 2 \), so the rule is \( (x, y) \to (2x, 2y) \). Wait, but let's check the figure. Wait, maybe I misread. Wait, the original triangle is \( \triangle XYZ \) and the dilated one is \( \triangle X'Y'Z' \) with scale factor \( 2 \), so we multiply each coordinate by \( 2 \).

For \( X(-2, 5) \): \( x' = -2\times2 = -4 \)? Wait, no, maybe the scale factor is \( 2 \), so \( X(-2,5) \to X'(-2\times2, 5\times2) = (-4, 10) \)? Wait, but let's check the other points.

For \( Y(4, 3) \): \( Y' = (4\times2, 3\times2) = (8, 6) \)

For \( Z(-3, -5) \): \( Z' = (-3\times2, -5\times2) = (-6, -10) \)

Wait, but maybe the scale factor is \( 2 \), so we multiply each coordinate by \( 2 \). So:

• \( X(-2, 5) \to X'(-2\times2, 5\times2) = (-4, 10) \)
• \( Y(4, 3) \to Y'(4\times2, 3\times2) = (8, 6) \)
• \( Z(-3, -5) \to Z'(-3\times2, -5\times2) = (-6, -10) \)

Step 2: Verify with the figure

Looking at the figure, the dilated triangle is larger, so multiplying by \( 2 \) makes sense. So the coordinates after dilation are \( X'(-4, 10) \), \( Y'(8, 6) \), \( Z'(-6, -10) \).

Part (b)

Step 1: Recall dilation rule

A dilation centered at the origin with scale factor \( k \) has the rule \( (x, y) \to (kx, ky) \). Here, the scale factor is \( 2 \), so \( k = 2 \).

Step 2: Match the rule

So the rule should be \( (x, y) \to (2x, 2y) \), which is the second option in the bottom row (since the options are:

Top row: \( (x,y)\to(2x,y) \), \( (x,y)\to(\frac{1}{2}x, 2y) \), \( (x,y)\to(\frac{1}{2}x, \frac{1}{2}y) \), \( (x,y)\to(\frac{1}{2}y, \frac{1}{2}x) \)

Bottom row: \( (x,y)\to(2x, 2y) \), \( (x,y)\to(x, 2y) \), \( (x,y)\to(2y, 2x) \), \( (x,y)\to(2x, \frac{1}{2}y) \)

So the correct rule is \( (x, y) \to (2x, 2y) \), which is the first option in the bottom row (the second column? Wait, the options are:

First column top: \( (x,y)\to(2x,y) \), bottom: \( (x,y)\to(2x, 2y) \)

Second column top: \( (x,y)\to(\frac{1}{2}x, 2y) \), bottom: \( (x,y)\to(x, 2y) \)

Third column top: \( (x,y)\to(\frac{1}{2}x, \frac{1}{2}y) \), bottom: \( (x,y)\to(2y, 2x) \)

Fourth column top: \( (x,y)\to(\frac{1}{2}y, \frac{1}{2}x) \), bottom: \( (x,y)\to(2x, \frac{1}{2}y) \)

So the correct rule is \( (x, y) \to (2x, 2y) \), which is the bottom row, first column option.

Answer:

Part (a)

• \( X(-2, 5) \to X'(-4, 10) \)
• \( Y(4, 3) \to Y'(8, 6) \)
• \( Z(-3, -5) \to Z'(-6, -10) \)

Part (b)

The correct rule is \( (x, y) \to (2x, 2y) \) (the option in the bottom row, first column: \( (x,y)\to(2x, 2y) \))