Answer:
No, a quadrilateral with the given vertices is not an isosceles trapezoid.
Step-by-step explanation:
We are given that
A(3,3), B(5,3), C(8,1), D(1,1)
Slope formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Slope of AB=[tex]\frac{3-3}{5-3}=0[/tex]
Slope of BC=[tex]\frac{1-3}{8-5}=\frac{-2}{3}[/tex]
Slope of CD=[tex]\frac{1-1}{1-8}=0[/tex]
Slope of AD=[tex]\frac{1-3}{1-3}=1[/tex]
Slope of AB=Slope of CD
When slopes of two lines are equal then the lines are parallel.
Therefore, AB is parallel to CD.
When one pair of quadrilateral is parallel then the quadrilateral is trapezoid.
[tex]\implies [/tex]ABCD is a trapezoid.
Distance formula:
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the formula
Length of BC=[tex]\sqrt{(8-5)^2+(1-3)^2}[/tex]
Length of BC=[tex]\sqrt{9+4}=\sqrt{13}[/tex] units
Length of AD=[tex]\sqrt{(1-3)^2+(1-3)^2}[/tex]
Length of AD=[tex]\sqrt{4+4}=2\sqrt{2}[/tex]
Length of AD is not equal to length of BC.
Hence, trapezoid is not an isosceles trapezoid.
