To finish the demonstration that the quadrilateral JKLM is a rhombus we need to prove that side JK is congruent with side LM.
The length of a segment with endpoints (x1, y1) and (x2, y2) is calculated as follows:
![\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://img.qammunity.org/2023/formulas/mathematics/college/s91s233vun4sfi433g8exahmp5kohnyvcn.png)
Substituting with points L(1,6) and M(4,2) we get:
![\begin{gathered} LM=\sqrt[]{(4-1)^2+(2-6)^2} \\ LM=\sqrt[]{3^2+(-4)^2} \\ LM=\sqrt[]{9+16^{}} \\ LM=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8jtbdepile21rpcaoou7zt5jeau0kspxb6.png)
Given that opposite sides are parallel, all sides have the same length, and, from the diagram, the quadrilateral is not a square, we conclude that it is a rhombus.