Question

Quadrilateral ABCD has vertices A(1,0) B(5,0) C (7,2) D(3,2). Use slope to prove that ABCD is a parallelogram.

Answer 1

Answer:

$AD \parallel BC$

$AB \parallel CD$

Step-by-step explanation:

The vertices of quadrilateral ABCD are A(1,0) B(5,0) C (7,2) D(3,2).

The slope of side AB is

$M_{AB}=\frac{0-0}{5-1} =\frac{0}{4}=0$

The slope of side BC is

$M_{BC}=\frac{2-0}{7-5} =\frac{2}{2}=1$

The slope of side CD is

$M_{CD}=\frac{2-2}{3-7} =\frac{0}{-4}=0$

The slope of AD is

$M_{AD}=\frac{2-0}{3-1} =\frac{2}{2}=1$

$AD \parallel BC$

$AB \parallel CD$

We see that the opposite sides of the quadrilateral ABCD are equal.

Hence the quadrilateral is a parallelogram