Question

Plzzz help me with this I’ll give brainliest

Answer 1

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=$\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{6-(-1)}{(-4)-(-4)}$

m=$\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{6-6}{(-4)-(-1)}$

m=$\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{(-4)-6}{2-2}$

m=$\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{(-1)-(-4)}{(-4)-(2)}$

m=$\frac{3}{-6}$

m=$\frac{-1}{2}$

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=$\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{2-3}{3-(-2)}$

m=$\frac{-1}{5}$

The slope of a line BC:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{(-1)-2}{2-3}$

m=$\frac{-3}{-1}$

m=3

The slope of a line CD:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{0-(-1)}{(-3)-2}$

m=$\frac{-1}{5}$

The slope of a line DA:

m=$\frac{Y2-Y1}{X2-X1}$

m=$\frac{3-0}{(-2)-(-3)}$

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