The missing figure is attached down
Answer:
The most specific name for quadrilateral ABCD is parallelogram
Step-by-step explanation:
The quadrilateral is a parallelogram if
From the attached figure
∵ The diagonals of the quadrilateral are AC and BD
∵ A = (-2 , 3) , C = (0 , -3)
- Find the slope of AC and its length using the rule of the slope
and the rule of the distance
∵ [tex]m_{AC}=\frac{-3-3}{0--2}[/tex]
∴ [tex]m_{AC}=\frac{-6}{2}=-3[/tex]
∵ [tex]d_{AC}=\sqrt{(0--2)^{2}+(-3-3)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}[/tex]
∴ [tex]d_{AC}=\sqrt{4+36}=\sqrt{40}[/tex]
- Find the mid-point of AC
∵ [tex]M_{AC}=(\frac{-2+0}{2},\frac{3+-3}{2})[/tex]
∴ [tex]M_{AC}=(-1,0)[/tex]
∵ B = (2 , 2) , C = (-4 , -2)
- Find the slope of AC and its length using the rule of the slope
and the rule of the distance
∵ [tex]m_{BD}=\frac{-2-2}{-4-2}[/tex]
∴ [tex]m_{BD}=\frac{-4}{-6}=\frac{2}{3}[/tex]
∵ [tex]d_{BD}=\sqrt{(-4-2)^{2}+(-2-2)^{2}}=\sqrt{(-6)^{2}+(-4)^{2}}[/tex]
∴ [tex]d_{BD}=\sqrt{36+16}=\sqrt{52}[/tex]
- Find the mid-point of AC
∵ [tex]M_{BD}=(\frac{2+-4}{2},\frac{2+-2}{2})[/tex]
∴ [tex]M_{BD}=(-1,0)[/tex]
∵ [tex]M_{AC}=M_{BD}[/tex] ⇒ diagonals bisect each other
∵ [tex]d_{AC}[/tex] ≠ [tex]d_{BD}[/tex] ⇒ diagonals not equal in length
∵ The product of their slopes = -3 × [tex]\frac{2}{3}[/tex] = -2
∵ The product of the slopes of the perpendicular lines is -1
∴ AC and BD are not perpendicular
∴ ABCD is a parallelogram
The most specific name for quadrilateral ABCD is parallelogram
