The vertices of the given quadrilateral GHJK are:
G = (3, 2)
H = (-3, 3)
J = (-5, -2)
K = (1, -3)
a) To find the length of JK and GH we will use the rule of the distance
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]For JK, let J = (x1, y1) and K = (x2, y2)
[tex]\begin{gathered} JK=\sqrt[]{(1--5)^2+(-3--2)^2} \\ JK=\sqrt[]{(1+5)^2+(-3+2)^2} \\ JK=\sqrt[]{(6)^2+(-1)^2} \\ JK=\sqrt[]{36+1} \\ JK=\sqrt[]{37} \end{gathered}[/tex]For GH, let G = (x1, y1) and H = (x2, y2)
[tex]\begin{gathered} GH=\sqrt[]{(-3-3)^2+(3-2)^2} \\ GH=\sqrt[]{(-6)^2+(1)^2} \\ GH=\sqrt[]{36+1} \\ GH=\sqrt[]{37} \end{gathered}[/tex]b) We will use the rule of the slope to find the slopes of JK and GH
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]For JK
[tex]\begin{gathered} m_{JK}=\frac{1--5}{-3--2} \\ m_{JK}=\frac{1+5}{-3+2} \\ m_{JK}=\frac{6}{-1} \\ m_{JK}=-6 \end{gathered}[/tex]For GH
[tex]\begin{gathered} m_{GH}=\frac{-3-3}{3-2} \\ m_{GH}=\frac{-6}{1} \\ m_{GH}=-6 \end{gathered}[/tex]c) From parts a and b
JK = GH
JK // GH because they have the same slopes
Then GHJK is a parallelogram because it has a pair of opposite sides that are both congruent and parallel
The answer is A
