If we graph the given points, we have:
One property of the parallelograms is that their opposite sides are equal.
Then, we have to verify if the segments QT and RS are equal.
[tex]QT=RS[/tex]
To find the measure of segments QT and RS, we can use the distance formula.
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}\Rightarrow\text{ Distance formula} \\ \text{ Where} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are the coordinates of the points} \end{gathered}[/tex]
• Measure of segment QT
[tex]\begin{gathered} (x_1,y_1)=Q(-10,-2) \\ (x_2,y_2)=T(-11,-8) \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt[]{(-11-(-10))^2+(-8-(-2))^2} \\ d=\sqrt[]{(-11+10)^2+(-8+2)^2} \\ d=\sqrt[]{(-1)^2+(-6)^2}\rbrack \\ d=\sqrt[]{1+36} \\ d=\sqrt[]{37} \\ d\approx6.08\Rightarrow\text{ The symbol }\approx\text{ is read 'approximately'} \end{gathered}[/tex]
• Measure of segment RS
[tex]\begin{gathered} (x_1,y_1)=R(1,-1) \\ (x_2,y_2)=S(1,-7) \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt[]{(1-1)^2+(-7-(-1))^2} \\ d=\sqrt[]{(0)^2+(-7+1)^2} \\ d=\sqrt[]{0+(-6)^2} \\ d=\sqrt[]{(-6)^2} \\ d=\sqrt[]{36} \\ d=6 \end{gathered}[/tex]
As we can see, the segments QT and RS are different.
[tex]\begin{gathered} QT\ne RS \\ 6.08\ne6 \end{gathered}[/tex]
Then, the figure does not satisfy the mentioned property of parallelograms.
Therefore, the figure is not a parallelogram.
