Streets in the city can be represented by the equations in the given table. Use the equations to find the type of quadrilateral that the streets form. Justify your answer. ​

It is a rhombus.

Step-by-step explanation:

We have the four equations:

3x - y = -4

3y - x = 4

3y = x + 12

y = 3*x - 4

First, we need to write all of them in slope-intercept form, so they are easier to read.

We get:

y = 3*x+ 4

y =  (1/3)*x + 4/3

y = (1/3)*x + 12/3

y = 3*x - 4

First, we can see that:

y = 3*x+ 4 and y = 3*x - 4

are parallel lines, because they have the same slope.

y =  (1/3)*x + 4/3 and y = (1/3)*x + 12/3  also are parallel lines, because they have the same slope.

Then this is a parallelogram.

Now, here we have two slopes, 3 and 1/3.

Remember that for a line y = a*x + b

A perpendicular line would be written as:

y = -(1/a)*x + c

From this we can see that there are no perpendicular lines in our set of lines.

Then this can not be a square nor a rectangle.

It can not be a trapezoid, because we would need 3 different slopes to make a trapezoid, and here we have only two.

Then this can only be a rhombus.

Below you can see the graph of the four lines to see which quadrilateral they form.

include a picture of the drawing in order for me to

step-by-step explanation:

if i put something wrong your not gonna like it so rather not say nothing.