Question

graph the image of △jkl after a translation 6 units right and 9 units up.

Explanation:

Step1: Identify coordinates of J, K, L

First, find the coordinates of the vertices of \( \triangle JKL \). From the graph:

• \( J \): Let's assume \( J \) is at \( (-5, -9) \) (since it's 5 units left of the origin on x - axis and 9 units down on y - axis)
• \( K \): \( K \) is at \( (-1, -9) \) (1 unit left of origin on x - axis, 9 units down on y - axis)
• \( L \): \( L \) is at \( (-3, -8) \) (3 units left of origin on x - axis, 8 units down on y - axis)

Step2: Apply translation (6 right, 9 up)

To translate a point \( (x,y) \) 6 units right and 9 units up, we use the transformation \( (x + 6,y + 9) \)

• For \( J(-5,-9) \):

New \( x=-5 + 6=1 \), New \( y=-9 + 9 = 0 \). So \( J'=(1,0) \)

• For \( K(-1,-9) \):

New \( x=-1+6 = 5 \), New \( y=-9 + 9=0 \). So \( K'=(5,0) \)

• For \( L(-3,-8) \):

New \( x=-3 + 6=3 \), New \( y=-8 + 9 = 1 \). So \( L'=(3,1) \)

Step3: Plot the new points

Plot the points \( J'(1,0) \), \( K'(5,0) \) and \( L'(3,1) \) and connect them to get the image of \( \triangle JKL \) after translation.

Answer:

The image of \( \triangle JKL \) has vertices at \( J'(1,0) \), \( K'(5,0) \) and \( L'(3,1) \). (To graph, plot these points and draw the triangle.)