What if you were given the coordinates of four points that form a quadrilateral? How could you determine if that quadrilateral qualifies as one of the special four-sided figures you learned about in the previous Concepts? After completing this Concept, you'll be able to make such a determination.

### Watch This

CK-12 Classifying Quadrilaterals in the Coordinate Plane

### Guidance

*In order to be successful in this concept you need to already be familiar with the definitions and properties of the following quadrilaterals: parallelograms, rhombuses, rectangles, squares, kites and trapezoids. The definitions for each are provided in the vocabulary section as a resource, and further information can be found by searching on those topic words.*

When working in the coordinate plane, you will sometimes want to know what type of shape a given shape is. You should easily be able to tell that it is a quadrilateral if it has four sides. But how can you classify it beyond that?

First you should graph the shape if it has not already been graphed. Look at it and see if it looks like any special quadrilateral. Do the sides appear to be congruent? Do they meet at right angles? This will give you a place to start.

Once you have a guess for what type of quadrilateral it is, your job is to prove your guess. To prove that a quadrilateral is a parallelogram, rectangle, rhombus, square, kite or trapezoid, you must show that it meets the definition of that shape OR that it has properties that only that shape has.

If it turns out that your guess was wrong because the shape does not fulfill the necessary properties, you can guess again. If it appears to be no type of special quadrilateral then it is simply a *quadrilateral*.

The examples below will help you to see what this process might look like.

#### Example A

Determine what type of parallelogram \begin{align*}TUNE\end{align*} is: \begin{align*}T(0, 10), U(4, 2), N(-2, -1)\end{align*}, and \begin{align*}E(-6, 7)\end{align*}.

This *looks* like a rectangle. Let’s see if the diagonals are equal. If they are, then \begin{align*}TUNE\end{align*} is a rectangle.

\begin{align*}EU & = \sqrt{(-6 -4)^2 + (7-2)^2} && TN = \sqrt{(0 + 2)^2 +(10 + 1)^2}\\ & = \sqrt{(-10)^2 + 5^2} && \quad \ \ = \sqrt{2^2 + 11^2}\\ & = \sqrt{100 + 25} && \quad \ \ = \sqrt{4 + 121}\\ & = \sqrt{125} && \quad \ \ = \sqrt{125}\end{align*}

If the diagonals are also perpendicular, then \begin{align*}TUNE\end{align*} is a square.

\begin{align*}\text{Slope of}\ EU = \frac{7 - 2}{-6 - 4} = -\frac{5}{10} = -\frac{1}{2} \quad \text{Slope of}\ TN = \frac{10 - (-1)}{0-(-2)} = \frac{11}{2}\end{align*}

The slope of \begin{align*}EU \neq\end{align*} slope of \begin{align*}TN\end{align*}, so \begin{align*}TUNE\end{align*} is a rectangle.

#### Example B

A quadrilateral is defined by the four lines \begin{align*}y=2x+1\end{align*}, \begin{align*}y=-x+5\end{align*}, \begin{align*}y=2x-4\end{align*}, and \begin{align*}y=-x-5\end{align*}. Is this quadrilateral a parallelogram?

To check if its a parallelogram we have to check that it has two pairs of parallel sides. From the equations we can see that the slopes of the lines are \begin{align*}2\end{align*}, \begin{align*}-1\end{align*}, \begin{align*}2\end{align*} and \begin{align*}-1\end{align*}. Because two pairs of slopes match, this shape has two pairs of parallel sides and is a parallelogram.

#### Example C

Determine what type of quadrilateral \begin{align*}RSTV\end{align*} is.

This *looks* like a kite. Find the lengths of all the sides to check if the adjacent sides are congruent.

\begin{align*}RS & = \sqrt{(-5 - 2)^2 + (7 - 6)^2} && ST = \sqrt{(2 - 5)^2 + (6-(-3))^2}\\ & = \sqrt{(-7)^2 + 1^2} && \quad \ = \sqrt{(-3)^2 + 9^2}\\ & = \sqrt{50} && \quad \ = \sqrt{90}\end{align*}

\begin{align*}RV & = \sqrt{(-5-(-4))^2 + (7-0)^2} && VT = \sqrt{(-4-5)^2 + (0-(-3))^2}\\ & = \sqrt{(-1)^2 + 7^2} && \quad \ \ = \sqrt{(-9)^2 + 3^2}\\ & = \sqrt{50} && \quad \ \ = \sqrt{90}\end{align*}

From this we see that the adjacent sides are congruent. Therefore, \begin{align*}RSTV\end{align*} is a kite.

CK-12 Classifying Quadrilaterals in the Coordinate Plane

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### Guided Practice

1. A quadrilateral is defined by the four lines \begin{align*}y=2x+1\end{align*}, \begin{align*}y=-2x+5\end{align*}, \begin{align*}y=2x-4\end{align*}, and \begin{align*}y=-2x-5\end{align*}. Is this quadrilateral a rectangle?

2. Determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. \begin{align*}A(-3, 3), B(1, 5), C(4, -1), D(1, -5)\end{align*}.

3. Determine what type of quadrilateral \begin{align*}EFGH\end{align*} is. \begin{align*}E(5, -1), F(11, -3), G(5, -5), H(-1, -3)\end{align*}

**Answers:**

1. To be a rectangle a shape must have four right angles. This means that the sides must be perpendicular to each other. From the given equations we see that the slopes are \begin{align*}2\end{align*}, \begin{align*}-2\end{align*}, \begin{align*}2\end{align*} and \begin{align*}-2\end{align*}. Because the slopes are not opposite reciprocals of each other, the sides are not perpendicular, and the shape is not a rectangle.

2. First, graph \begin{align*}ABCD\end{align*}. This will make it easier to figure out what type of quadrilateral it is. From the graph, we can tell this is not a parallelogram. Find the slopes of \begin{align*}\overline{BC}\end{align*} and \begin{align*}\overline{AD}\end{align*} to see if they are parallel.

Slope of \begin{align*}\overline{BC} = \frac{5-(-1)}{1-4} = \frac{6}{-3} = -2\end{align*}

Slope of \begin{align*}\overline{AD} = \frac{3-(-5)}{-3-1} = \frac{8}{-4} = -2\end{align*}

\begin{align*}\overline{BC} \| \overline{AD}\end{align*}, so \begin{align*}ABCD\end{align*} is a trapezoid. To determine if it is an isosceles trapezoid, find \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.

\begin{align*}AB & = \sqrt{(-3-1)^2 + (3 - 5)^2} && ST = \sqrt{(4 - 1)^2 + (-1-(-5))^2}\\ & = \sqrt{(-4)^2 + (-2)^2} && \quad \ = \sqrt{3^2 + 4^2}\\ & = \sqrt{20} = 2 \sqrt{5} && \quad \ = \sqrt{25} = 5\end{align*}

\begin{align*}AB \neq CD\end{align*}, therefore this is not an isosceles trapezoid.

3. We will not graph this example. Let’s find the length of all four sides.

\begin{align*}EF & = \sqrt{(5 - 11)^2 + (-1-(-3))^2} && \ \sqrt{FG} = \sqrt{(11-5)^2 + (-3-(-5))^2}\\ & = \sqrt{(-6)^2 + 2^2} = \sqrt{40} && \qquad \ \ = \sqrt{6^2 + 2^2} = \sqrt{40}\\ GH & = \sqrt{(5-(-1))^2 + (-5-(-3))^2} && \quad HE = \sqrt{(-1-5)^2 +(-3-(-1))^2}\\ & = \sqrt{6^2 + (-2)^2} = \sqrt{40} && \qquad \ \ = \sqrt{(-6)^2 + (-2)^2} = \sqrt{40}\end{align*}

All four sides are equal. This quadrilateral is either a ** rhombus** or a

**. Let’s find the length of the diagonals.**

*square*\begin{align*}EG & = \sqrt{(5 - 5)^2 + (-1-(-5))^2} && \ FH = \sqrt{(11-(-1))^2 + (-3-(-3))^2}\\ & = \sqrt{0^2 + 4^2} && \qquad = \sqrt{12^2 + 0^2}\\ & = \sqrt{16} = 4 && \qquad = \sqrt{144} = 12\end{align*}

The diagonals are not congruent, so \begin{align*}EFGH\end{align*} is a rhombus and not a square.

### Explore More

Determine what type of quadrilateral \begin{align*}ABCD\end{align*} is.

- \begin{align*}A(-2, 4), B(-1, 2), C(-3, 1), D(-4, 3)\end{align*}
- \begin{align*}A(-2, 3), B(3, 4), C(2, -1), D(-3, -2)\end{align*}
- \begin{align*}A(1, -1), B(7, 1), C(8, -2), D(2, -4)\end{align*}
- \begin{align*}A(10, 4), B(8, -2), C(2, 2), D(4, 8)\end{align*}
- \begin{align*}A(0, 0), B(5, 0), C(0, 4), D(5, 4)\end{align*}
- \begin{align*}A(-1, 0), B(0, 1), C(1, 0), D(0, -1)\end{align*}
- \begin{align*}A(2, 0), B(3, 5), C(5, 0), D(6, 5)\end{align*}

\begin{align*}SRUE\end{align*} is a rectangle and \begin{align*}PRUC\end{align*} is a square.

- What type of quadrilateral is \begin{align*}SPCE\end{align*}?
- If \begin{align*}SR = 20\end{align*} and \begin{align*}RU = 12\end{align*}, find \begin{align*}CE\end{align*}.
- Find \begin{align*}SC\end{align*} and \begin{align*}RC\end{align*} based on the information from part b. Round your answers to the nearest hundredth.

For questions 11-14, determine what type of quadrilateral \begin{align*}ABCD\end{align*} is. If it is no type of special quadrilateral, just write *quadrilateral*.

- \begin{align*}A(1, -2), B(7, -5), C(4, -8), D(-2, -5)\end{align*}
- \begin{align*}A(6, 6), B(10, 8), C(12, 4), D(8, 2)\end{align*}
- \begin{align*}A(-1, 8), B(1, 4), C(-5, -4), D(-5, 6)\end{align*}
- \begin{align*}A(5, -1), B(9, -4), C(6, -10), D(3, -5)\end{align*}

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 6.8.