Answer18:
The quadrilateral ABCD is not a parallelogram
Answer19:
The quadrilateral ABCD is a parallelogram
Step-by-step explanation:
For question 18:
Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)
The slope of a line is given m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
Now,
The slope of a line AB:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{6-(-1)}{(-4)-(-4)}[/tex]
m=[tex]\frac{7}{0}[/tex]
The slope is 90 degree
The slope of a line BC:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{6-6}{(-4)-(-1)}[/tex]
m=[tex]\frac{0}{(-3)}[/tex]
The slope is zero degree
The slope of a line CD:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{(-4)-6}{2-2}[/tex]
m=[tex]\frac{-10}{0}[/tex]
The slope is 90 degree
The slope of a line DA:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{(-1)-(-4)}{(-4)-(2)}[/tex]
m=[tex]\frac{3}{-6}[/tex]
m=[tex]\frac{-1}{2}[/tex]
The slope of the only line AB and CD are the same.
Thus, The quadrilateral ABCD is not a parallelogram
For question 19:
Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)
The slope of a line is given m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
Now,
The slope of a line AB:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{2-3}{3-(-2)}[/tex]
m=[tex]\frac{-1}{5}[/tex]
The slope of a line BC:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{(-1)-2}{2-3}[/tex]
m=[tex]\frac{-3}{-1}[/tex]
m=3
The slope of a line CD:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{0-(-1)}{(-3)-2}[/tex]
m=[tex]\frac{-1}{5}[/tex]
The slope of a line DA:
m=[tex]\frac{Y2-Y1}{X2-X1}[/tex]
m=[tex]\frac{3-0}{(-2)-(-3)}[/tex]
m=3
The slope of the line AB and CD are the same
The slope of the line BC and DA are the same
Thus, The quadrilateral ABCD is a parallelogram
