Question

$\triangle jkp$ is the image of $\triangle jkp$. describe the dilation. then write a coordinate rule to represent the dilation.
the dilation of $\triangle jkp$ is centered at $p(3,2)$ and has a scale factor of
$(x, y) \to ( \quad (x- \quad )+ \quad, \quad (y- \quad )+ \quad )$

Explanation:

Step1: Identify the center and scale factor

The dilation is centered at \( P(3,2) \). To find the scale factor, we can compare the distances from the center to a point and its image. Let's take point \( J \) and \( J' \). From the graph, we can see the transformation from \( J \) to \( J' \) involves a scale factor. By analyzing the coordinates (assuming \( J \) has coordinates relative to \( P \)), the scale factor \( k \) is 2 (since the image is larger, and the distance from \( P \) to \( J' \) is twice the distance from \( P \) to \( J \)).

Step2: Derive the coordinate rule

The general rule for a dilation centered at \( (h,k) \) with scale factor \( k \) is \( (x,y) \to (h + k(x - h), k + k(y - k)) \). Here, \( h = 3 \), \( k = 2 \), and the scale factor \( k_{scale}=2 \). So substituting, we get:
For the \( x \)-coordinate: \( 3 + 2(x - 3) \)
For the \( y \)-coordinate: \( 2 + 2(y - 2) \)

Simplifying the \( x \)-coordinate: \( 3 + 2x - 6 = 2x - 3 \)? Wait, no, wait. Wait, the general formula for dilation centered at \( (h,k) \) is \( (x,y) \to (h + s(x - h), k + s(y - k)) \), where \( s \) is the scale factor. So with \( h = 3 \), \( k = 2 \), \( s = 2 \), we have:
\( x' = 3 + 2(x - 3) = 3 + 2x - 6 = 2x - 3 \)
\( y' = 2 + 2(y - 2) = 2 + 2y - 4 = 2y - 2 \)
But let's check the form given. The given form is \( (x,y) \to ( \square(x - \square) + \square, \square(y - \square) + \square ) \). So substituting \( h = 3 \), \( s = 2 \), \( k = 2 \):
For the \( x \)-part: \( 2(x - 3) + 3 \)
For the \( y \)-part: \( 2(y - 2) + 2 \)

Answer:

The dilation of \( \triangle JKP \) is centered at \( P(3,2) \) and has a scale factor of \( \boldsymbol{2} \).

The coordinate rule is \( (x, y) \to ( \boldsymbol{2}(x - \boldsymbol{3}) + \boldsymbol{3}, \boldsymbol{2}(y - \boldsymbol{2}) + \boldsymbol{2} ) \)