# the type of quadrilateral

I am being asked to determine the type of quadrilateral defined by the point of intersection (using geometry & algebra) of the lines:

x - 4y + 3 = 0

5x - y - 8 = 0

x - 4y -16 = 0

6x - 5y - 1 = 0

https://brainmass.com/math/basic-algebra/type-quadrilateral-286683

## SOLUTION This solution is **FREE** courtesy of BrainMass!

x - 4y + 3 = 0 --equation 1

5x - y - 8 = 0 --equation 2

x - 4y -16 = 0 --equation 3

6x - 5y - 1 = 0 --equation 4

Express all equation in terms of y. That would mean only y is found in the left side of the equation.

---------------------equation 1

x - 4y + 3 = 0

- 4y = -x -3

y = (x+3)/4

Assign values for x then solve for y as shown in the table

x -3 1 5

y 0 1 2

Graph these points and connect with a straight line.

---------------------equation 2

5x - y - 8 = 0

- y = -5x + 8

y = 5x - 8

Assign values for x then solve for y as shown in the table

x 0 1 2 3

y -8 -3 2 7

Graph these points and connect with a straight line.

---------------------equation 3

x - 4y -16 = 0

-4y = -x +16

y = (x - 16)/4

Assign values for x then solve for y as shown in the table

x -4 0 4

y -5 4 3

Graph these points and connect with a straight line.

---------------------equation 4

6x - 5y - 1 = 0

- 5y = -6x + 1

y = (6x-1)/5

Assign values for x then solve for y as shown in the table

x -4 -2 1 3

y -5 -2.6 1 3.4

Graph these points and connect with a straight line.

The graph of the four lines is shown below. The bounded area formed by the intersection of the lines is a TRAPEZOID !

Take note that the two sides are parallel.

https://brainmass.com/math/basic-algebra/type-quadrilateral-286683