We can find the slopes using the following formula:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
And the lengths using the following formulas:
[tex]d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]
Therefore:
[tex]m_{QR}=\frac{5-2}{-1-(-9)}=\frac{3}{8}[/tex][tex]m_{RS}=\frac{9-5}{1-(-1)}=\frac{4}{2}=2[/tex][tex]m_{ST}=\frac{6-3}{-7-1}=\frac{3}{8}[/tex][tex]m_{TQ}=\frac{6-2}{-7-(-9)}=\frac{4}{2}=2[/tex][tex]\begin{gathered} L_{QR}=\sqrt[]{(-1-(-9))^2+(5-2)^2} \\ L_{QR}=\sqrt[]{73} \end{gathered}[/tex][tex]\begin{gathered} L_{RS}=\sqrt[]{(1-(-1))^2+(9-5)^2} \\ L_{RS}=2\sqrt[]{5} \end{gathered}[/tex][tex]\begin{gathered} L_{ST}=\sqrt[]{(6-9)^2+(-7-1)^2} \\ L_{ST}=\sqrt[]{73} \end{gathered}[/tex][tex]\begin{gathered} L_{TQ}=\sqrt[]{(6-2)^2+(-7-(-9))^2}_{} \\ L_{TQ}=2\sqrt[]{5} \end{gathered}[/tex]
Since:
[tex]\begin{gathered} m_{RS}=m_{TQ}\to m_{RS}\parallel m_{TQ} \\ m_{QR}=m_{ST}\to m_{QR}\parallel m_{ST} \end{gathered}[/tex]
And:
[tex]\begin{gathered} L_{QR}=L_{ST} \\ L_{RS}=L_{QT} \end{gathered}[/tex]
According to this, we can conclude it is a parallelogram
