Question

quadrilateral jklm was dilated according to the rule d_{\frac{1}{2}}(x,y)=(\frac{1}{2}x,\frac{1}{2}y) to create the image quadrilateral jklm, which is shown on the graph. what are the coordinates of vertex j of the pre - image? (0, - 4) (0, - 1) (0,0) (0,4)

Explanation:

Step1: Recall dilation rule

The dilation rule is $D_{\frac{1}{2}}(x,y)=(\frac{1}{2}x,\frac{1}{2}y)$. Let the pre - image coordinates of $J$ be $(x,y)$ and image coordinates be $(x',y')$. We know $x'=\frac{1}{2}x$ and $y'=\frac{1}{2}y$.

Step2: Assume image coordinates of $J$

Suppose the image coordinates of $J$ are $(0, - 2)$. Then, if $x' = 0=\frac{1}{2}x$ and $y'=-2=\frac{1}{2}y$. Solving for $x$ and $y$ gives $x = 0$ and $y=-4$.

Answer:

$(0, - 4)$