Question

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.

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Answer 1

It is a rhombus.

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.