Question

quadrilateral jklm was dilated according to the rule $d_{o, \frac{1}{2}}(x, y) \longrightarrow (\frac{1}{2}x, \frac{1}{2}y)$ to create the image quadrilateral jklm, which is shown on the graph.\
graph of quadrilateral jklm on a coordinate plane with j at (0, -2), k at (3, 2), l at (6, -2), m at (3, -6)\
what are the coordinates of vertex j of the pre-image?\
(0, 4)\
(0, -1)\
(0, -4)\
(0, 0)

Explanation:

Step1: Find coordinates of \( J' \)

From the graph, \( J' \) has coordinates \( (0, -2) \).

Step2: Apply dilation inverse rule

Dilation rule: \( D_{O,\frac{1}{2}}(x,y)\to(\frac{1}{2}x,\frac{1}{2}y) \). Let pre - image \( J=(x,y) \), then \( \frac{1}{2}x = 0 \) and \( \frac{1}{2}y=-2 \).
Solve for \( x \): \( x = 0\times2 = 0 \).
Solve for \( y \): \( y=-2\times2=-4 \)? Wait, no, wait. Wait, \( J' \) is \( (0, - 2) \)? Wait, no, looking at the graph, \( J' \) is at \( (0, - 2) \)? Wait, no, the graph: the point \( J' \) is on the y - axis, below the origin. Wait, the grid: each square is 1 unit. So \( J' \) has coordinates \( (0, - 2) \)? Wait, no, wait the dilation rule is \( (x,y)\to(\frac{1}{2}x,\frac{1}{2}y) \). So to find the pre - image, we need to reverse the dilation. If the image is \( (x',y')=(\frac{1}{2}x,\frac{1}{2}y) \), then \( x = 2x' \), \( y = 2y' \).
Wait, looking at the graph, \( J' \) is at \( (0, - 2) \)? Wait, no, the first option is \( (0,4) \), second \( (0, - 1) \), third \( (0, - 4) \), fourth \( (0,0) \). Wait, maybe I misread \( J' \) coordinates. Let's re - examine the graph. The quadrilateral \( J'K'L'M' \): \( J' \) is on the y - axis, at \( (0, - 2) \)? Wait, no, maybe \( J' \) is \( (0, - 2) \), then pre - image \( J=(2\times0,2\times(-2))=(0, - 4) \)? Wait, but let's check the dilation rule again. The dilation is \( D_{O,\frac{1}{2}} \), so scale factor \( \frac{1}{2} \), center at origin. So image coordinates are \( \frac{1}{2} \) of pre - image. So pre - image coordinates are \( 2\times \) image coordinates.
Wait, maybe \( J' \) is \( (0, - 2) \), then pre - image \( J=(0\times2, - 2\times2)=(0, - 4) \)? But wait, the options include \( (0, - 4) \). Wait, but let's check again. Wait, maybe I made a mistake in \( J' \) coordinates. Let's look at the graph: the point \( J' \) is at \( (0, - 2) \)? Wait, no, the y - axis: the origin is (0,0). Below the origin, each grid line is 1 unit. So \( J' \) is at \( (0, - 2) \)? Then pre - image \( J=(0, - 4) \)? But wait, the first option is \( (0,4) \), second \( (0, - 1) \), third \( (0, - 4) \), fourth \( (0,0) \). Wait, maybe I messed up \( J' \) coordinates. Wait, maybe \( J' \) is \( (0, - 2) \), then pre - image \( J=(0, - 4) \). But let's check the options. The third option is \( (0, - 4) \). Wait, but let's re - do the steps.

Step1: Identify \( J' \) coordinates

From the graph, \( J' \) is at \( (0, - 2) \)? Wait, no, maybe \( J' \) is at \( (0, - 2) \). Wait, the dilation rule is \( (x,y)\to(\frac{1}{2}x,\frac{1}{2}y) \). So if \( J'=(x',y')=(\frac{1}{2}x,\frac{1}{2}y) \), then \( x = 2x' \), \( y = 2y' \).

Step2: Calculate pre - image \( J \) coordinates

Given \( J'=(0, - 2) \) (from graph), then \( x = 2\times0 = 0 \), \( y = 2\times(-2)=-4 \). So \( J=(0, - 4) \).

Wait, but let's check the graph again. The quadrilateral \( J'K'L'M' \): \( J' \) is on the y - axis, at \( (0, - 2) \)? Wait, maybe I misread \( J' \) as \( (0, - 2) \), but maybe it's \( (0, - 2) \). Then pre - image is \( (0, - 4) \), which is the third option.

Wait, no, wait a second. Maybe \( J' \) is \( (0, - 2) \), then pre - image \( J=(0, - 4) \). So the coordinates of \( J \) (pre - image) are \( (0, - 4) \).

Answer:

\((0, - 4)\) (corresponding to the option \( (0, - 4) \))