If we reflect JKL across the x-axis, the new triangle that is formed has exactly that same x-coordinate by its y-coordinate has reversed sign,
Therefore,
[tex]J(0,2)\rightarrow J^{\prime}(0,-2)[/tex][tex]K(4,4)\rightarrow K(4,-4)[/tex]and
[tex]L(4,0)\rightarrow L^{\prime}(4,0)[/tex]Hence, the coordinates of the vertices of J'K'L' are
[tex]J^{\prime}(0,-2)[/tex][tex]K^{\prime}(4,-4)[/tex][tex]L^{\prime}(4,0)\text{.}[/tex]