Respuesta :

ANSWER TO PART A

The given triangle has vertices [tex]J(-4,1), K(-4,-2),L(-3,-1)[/tex]


The mapping for rotation through [tex]90\degree[/tex] counterclockwise has the mapping


[tex](x,y)\rightarrow (-y,x)[/tex]


Therefore

[tex]J(-4,1)\rightarrow J'(-1,-4)[/tex]


[tex]K(-4,-2)\rightarrow K'(2,-4)[/tex]


[tex]L(-3,-1)\rightarrow L'(1,-3)[/tex]


We plot all this point and connect them with straight lines.


ANSWER TO PART B


For a reflection across the y-axis we negate the x coordinates.


The mapping is



[tex](x,y)\rightarrow (-x,y)[/tex]


Therefore

[tex]J(-4,1)\rightarrow J''(4,1)[/tex]


[tex]K(-4,-2)\rightarrow K''(4,-2)[/tex]


[tex]L(-3,-1)\rightarrow L''(3,-1)[/tex]


We plot all this point and connect them with straight lines.


See graph in attachment







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