Answer:
See graph in attachment.
Step-by-step explanation:
The given quadrilateral has vertices
K(-5,-2) A(-4,1) I(0,-1) J(-2,-4)
A reflection in the line [tex]y=x[/tex] has the mapping.
[tex](x,y)\to (y,x)[/tex]
This implies that;
[tex]K(-5,-2)\to K'(-2,-5)[/tex]
[tex]A(-4,1)\to A'(1,-4)[/tex]
[tex]I(0,-1)\to I'(-1,0)[/tex]
[tex]J(-2,-4)\to J'(-4,-2)[/tex]
We plot the points and join them with a straight edge to obtain, the graph in the attachment.
