Question

triangle xyz was dilated using the rule $d_{o,0.25} (x, y) \
ightarrow (0.25x, 0.25y)$. the image is shown in the diagram. what are the coordinates of z of the pre-image? $\bigcirc (-8, -4)$ $\bigcirc (-8, 2)$ $\bigcirc (-2, -1)$ $\bigcirc (-0.25, -0.5)$

Explanation:

Step1: Identify image coordinates of Z

From the diagram, the image of Z (after dilation) has coordinates \((-2, -1)\) (assuming the grid and diagram show this; let's confirm with dilation rule). Wait, no—wait, the dilation rule is \((x,y) \to (0.25x, 0.25y)\), so the image coordinates \((x', y') = (0.25x, 0.25y)\), where \((x,y)\) is pre - image. So to find pre - image, we solve \(x' = 0.25x\) and \(y' = 0.25y\) for \(x\) and \(y\).

Looking at the options, let's check the image of Z. Wait, maybe from the diagram, the image of Z is \((-2, -1)\)? Wait no, let's re - express. Let the pre - image coordinates be \((x,y)\), then image is \((0.25x, 0.25y)\). Let's assume the image of Z (from the diagram) is \((-2, -1)\)? Wait no, maybe the image of Z is \((-2, -1)\)? Wait, no—wait, let's take the options. Let's suppose the image of Z (after dilation) is \((-2, -1)\)? Wait, no, let's do the reverse. If dilation is \(D_{O,0.25}\), then pre - image \((x,y)\) maps to image \((0.25x, 0.25y)\). So to find pre - image, we have \(x=\frac{x'}{0.25}\) and \(y = \frac{y'}{0.25}\) (since \(x' = 0.25x\implies x=\frac{x'}{0.25}=4x'\), same for \(y\)).

Looking at the diagram, the image of Z (the dilated point) has coordinates, let's say from the grid, Z (image) is at \((-2, -1)\)? Wait, no, let's check the options. Let's take each option and apply dilation:

Option 1: \((-8, -4)\) dilated by 0.25: \((0.25\times(-8), 0.25\times(-4))=(-2, -1)\). Wait, but we need pre - image. Wait, no—wait, the dilation rule is \(D_{O,0.25}(x,y)\to(0.25x, 0.25y)\). So if the image of Z is, say, \((-2, -1)\), then pre - image \((x,y)\) satisfies \(0.25x=-2\) and \(0.25y = - 1\). Solving for \(x\): \(x=\frac{-2}{0.25}=-8\), \(y=\frac{-1}{0.25}=-4\)? Wait, no, that would be if image is \((-2, -1)\), pre - image is \((-8, -4)\)? Wait, no, wait: dilation with scale factor 0.25 (reduction) means pre - image is larger. Wait, maybe I got it reversed. Let's clarify: Dilation \(D_{O,k}\) where \(k = 0.25\) means each coordinate of the pre - image is multiplied by \(k\) to get the image. So image = \(k\times\) pre - image. So pre - image = image \(\div k\) (since image = \(k\times\) pre - image \(\implies\) pre - image = image / \(k\)).

From the diagram, let's find the image of Z. Looking at the grid, the image of Z (the dilated triangle) has coordinates, let's say, from the diagram, Z (image) is at \((-2, -1)\)? Wait, no, let's check the options. Let's take the image of Z as \((-2, -1)\) (maybe from the diagram). Then pre - image Z would be \((-2\div0.25, -1\div0.25)=(-8, -4)\)? Wait, but let's check the other options.

Option 2: \((-8,2)\) dilated: \((0.25\times(-8), 0.25\times2)=(-2, 0.5)\), not matching.

Option 3: \((-2,-1)\) dilated: \((-0.5, -0.25)\), not matching.

Option 4: \((-0.25, -0.5)\) dilated: \((-0.0625, -0.125)\), not matching.

Wait, maybe the image of Z is \((-2, -1)\), so pre - image is \((-8, -4)\) because \(x=-2\div0.25=-8\), \(y = - 1\div0.25=-4\).

Step1: Let image coordinates of Z be \((x', y')\)

From the diagram (assuming image of Z is \((-2, -1)\)).

Step2: Use dilation formula to find pre - image

Dilation rule: \(x' = 0.25x\), \(y' = 0.25y\), where \((x,y)\) is pre - image.

Solve for \(x\): \(x=\frac{x'}{0.25}\), solve for \(y\): \(y=\frac{y'}{0.25}\)

Substitute \(x'=-2\), \(y'=-1\):

\(x=\frac{-2}{0.25}=-8\)

\(y=\frac{-1}{0.25}=-4\)

So pre - image coordinates of Z are \((-8, -4)\)

Answer:

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