Question

dilate △jkl by a scale factor of (\frac{1}{3}) with the center of dilation at the origin.

Explanation:

Step1: Identify coordinates of vertices

First, find the coordinates of \( J \), \( K \), and \( L \) from the graph.

• \( J \): From the grid, \( x = -9 \), \( y = 6 \), so \( J(-9, 6) \)
• \( K \): \( x = 0 \), \( y = 9 \), so \( K(0, 9) \)
• \( L \): \( x = -6 \), \( y = 0 \), so \( L(-6, 0) \)

Step2: Apply dilation formula

The rule for dilation with scale factor \( k \) centered at the origin is \( (x, y) \to (k \cdot x, k \cdot y) \). Here, \( k = \frac{1}{3} \).

• For \( J(-9, 6) \):

New \( x \)-coordinate: \( \frac{1}{3} \times (-9) = -3 \)
New \( y \)-coordinate: \( \frac{1}{3} \times 6 = 2 \)
So, \( J'(-3, 2) \)

• For \( K(0, 9) \):

New \( x \)-coordinate: \( \frac{1}{3} \times 0 = 0 \)
New \( y \)-coordinate: \( \frac{1}{3} \times 9 = 3 \)
So, \( K'(0, 3) \)

• For \( L(-6, 0) \):

New \( x \)-coordinate: \( \frac{1}{3} \times (-6) = -2 \)
New \( y \)-coordinate: \( \frac{1}{3} \times 0 = 0 \)
So, \( L'(-2, 0) \)

Step3: Plot the new points

Plot \( J'(-3, 2) \), \( K'(0, 3) \), and \( L'(-2, 0) \) on the coordinate plane and connect them to form the dilated triangle \( \triangle J'K'L' \).

Answer:

The coordinates of the dilated triangle \( \triangle J'K'L' \) are \( J'(-3, 2) \), \( K'(0, 3) \), and \( L'(-2, 0) \). (To visualize, plot these points and draw the triangle.)