Answer :
Answer: [tex](\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .[/tex]
Step-by-step explanation:
Transformation rule for dilation:
[tex](x,y)\to(kx,ky)[/tex] , where k = scale factor
Given : Scale factor = [tex]-\dfrac13[/tex]
Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3)
Vertices after dilation:
[tex](-1,6)\to(-1\times\dfrac{-1}{3},6\times\dfrac{-1}{3})=(\dfrac13,-2)[/tex]
[tex](0,9)\times(0\times-\dfrac13,9\times-\dfrac13)=(0,-3)[/tex]
[tex](6,-3)\to(6\times-\dfrac13,\ -3\times-\dfrac13)=(-2,1)[/tex]
[tex](3,-3)\to (3\times-\dfrac13,\ -3\times-\dfrac13)=(-1,1)[/tex]
Hence, the coordinates of the image if the parallelogram = [tex](\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .[/tex]