Question

is the quadrilateral a parallelogram? justify your answer.

Explanation:

Step1: Find the coordinates of the vertices

From the graph, \(A(-1,4)\), \(B(2,4)\), \(C(-3,-4)\), \(D(-6,-4)\)

Step2: Calculate the slopes of \(AB\) and \(CD\)

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
For \(AB\): \(m_{AB}=\frac{4 - 4}{2-(-1)}=\frac{0}{3} = 0\)
For \(CD\): \(m_{CD}=\frac{-4-(-4)}{-3-(-6)}=\frac{0}{3}=0\)
Since \(m_{AB}=m_{CD}\), \(AB\parallel CD\)

Step3: Calculate the lengths of \(AB\) and \(CD\)

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(AB\): \(d_{AB}=\sqrt{(2 - (-1))^2+(4 - 4)^2}=\sqrt{3^2+0^2}=3\)
For \(CD\): \(d_{CD}=\sqrt{(-3 - (-6))^2+(-4 - (-4))^2}=\sqrt{3^2+0^2}=3\)
Since \(AB = CD\) and \(AB\parallel CD\)

Answer:

Yes, the quadrilateral is a parallelogram because one pair of opposite sides (\(AB\) and \(CD\)) are both parallel (have the same slope) and congruent (have the same length)