Question

identifying vertices of the pre-image of a dilation
triangle xyz was dilated using the rule $d_{0,0.25} (x, y) \to (0.25x, 0.25y)$.
the image is shown in the diagram. what are the coordinates of z of the pre-image?
$\circ$ $(-8, -4)$
$\circ$ $(-8, 2)$
$\circ$ $(-2, -1)$
$\circ$ $(-0.25, -0.5)$

Explanation:

Step1: Find coordinates of \( Z' \)

From the graph, \( Z' \) is at \( (-2, -1) \) (since it's 2 units left of origin on x - axis and 1 unit down on y - axis).

Step2: Use dilation rule to find pre - image \( Z \)

Dilation rule is \( D_{O,0.25}(x,y)\to(0.25x,0.25y) \). Let pre - image \( Z=(x,y) \), then image \( Z'=(0.25x,0.25y) \). We know \( Z' = (-2,-1) \), so we set up equations:
\( 0.25x=-2 \) and \( 0.25y = - 1 \)
For \( x \): \( x=\frac{-2}{0.25}=-8 \)
For \( y \): \( y=\frac{-1}{0.25}=-4 \)
So pre - image \( Z=(-8,-4) \)
Wait, no, wait. Wait, the dilation rule is \( (x,y)\to(0.25x,0.25y) \), so if \( Z'=(0.25x,0.25y) \), then to find \( (x,y) \) we need to solve \( x=\frac{x'}{0.25} \), \( y = \frac{y'}{0.25} \)
Wait, let's re - check the coordinates of \( Z' \). Looking at the graph, \( Z' \) is at \( (-2,-1) \)? Wait, no, looking at the grid, the x - coordinate of \( Z' \): from the origin (0,0), moving left 2 units? Wait, no, the grid lines: each square is 1 unit. Let's see, the triangle \( X'Y'Z' \): \( Z' \) is at \( (-2,-1) \)? Wait, no, maybe I misread. Wait, the x - axis: the origin is (0,0), the negative x - axis is to the left. Let's look at the coordinates of \( Z' \): from the graph, \( Z' \) is at \( (-2,-1) \)? Wait, no, let's count the grid. Let's see, the point \( Z' \): x - coordinate: let's see the vertical line through \( Z' \): it's at x=-2, y=-1? Wait, no, maybe I made a mistake. Wait, the dilation factor is 0.25, so pre - image is 4 times the image (since 0.25 is \( \frac{1}{4} \), so inverse is 4). Wait, let's check the options. The options are (-8,-4), (-8,2), (-2,-1), (-0.25,-0.5). Wait, if \( Z' \) is at (-2,-1), then pre - image would be (-2/0.25, - 1/0.25)=(-8,-4). But wait, maybe I misread \( Z' \) coordinates. Wait, maybe \( Z' \) is at (-2,-1)? Wait, no, let's look again. The graph: the triangle \( X'Y'Z' \), \( Z' \) is at ( - 2, - 1)? Wait, no, maybe \( Z' \) is at ( - 2, - 1)? Wait, let's check the y - coordinate. The y - axis: above the x - axis is positive, below is negative. So \( Z' \) is below the x - axis, so y is negative. x is negative. So \( Z'=(-2,-1) \). Then pre - image: \( x=\frac{-2}{0.25}=-8 \), \( y=\frac{-1}{0.25}=-4 \). So the pre - image \( Z \) is (-8,-4)
Wait, but let's check the other way. If the dilation is \( D_{O,0.25} \), which means center at origin, scale factor 0.25. So image = scale factor * pre - image. So pre - image = image / scale factor. So if image \( Z'=(x',y') \), pre - image \( Z=(x',y')/0.25=(4x',4y') \)
Now, let's find \( Z' \) coordinates correctly. Looking at the graph, \( Z' \) is at (-2,-1)? Wait, no, maybe \( Z' \) is at (-2,-1)? Wait, let's count the grid. Let's see, the point \( Z' \): x - coordinate: from origin, left 2 units, y - coordinate: down 1 unit. So \( Z'=(-2,-1) \). Then pre - image \( Z=(4(-2),4(-1))=(-8,-4) \), which is one of the options.
Wait, but let's check again. Maybe I misread \( Z' \) coordinates. Let's look at the graph again. The triangle \( X'Y'Z' \): \( Z' \) is at ( - 2, - 1)? Wait, no, maybe \( Z' \) is at ( - 2, - 1)? Wait, the options: (-8,-4) is an option. Let's verify: if pre - image \( Z=(-8,-4) \), then dilation by 0.25: (0.25(-8),0.25(-4))=(-2,-1), which matches \( Z' \) (if \( Z'=(-2,-1) \)). So that's correct.

Answer:

(-8, -4) (the first option: \(\boldsymbol{(-8, -4)}\))