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9. Connecting Algebra & Geometry with Coordinates
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United States of AmericaTennessee
Geometry

9.05 Classifying polygons in the coordinate plane

Lesson
Practice
Lesson

Concept summary

If we know the coordinates of the vertices of a triangle or quadrilateral, we can classify it more precisely using the distances and slopes of line segments joining pairs of vertices.

We can use the distance formula or Pythagorean theorem to determine the lengths of sides, and we can check the slopes of sides or diagonals to see if they are parallel or perpendicular.

For triangles, we have the following classifications:

A triangle with 3 congruent sides.
Equilateral triangle (all sides congruent)
A triangle with a pair of congruent sides.
Isosceles triangle (a pair of congruent sides)
A triangle with no congruent sides.
Scalene triangle (no congruent sides)
Two right triangles: one having two congruent sides, and the other having no congruent sides.
Right triangle (pair of perpendicular sides)

Note that it is possible for a triangle to be a scalene right triangle or an isosceles right triangle.

Triangle that are not right triangles are also sometimes classified as acute triangles (if all angle measures are smaller than 90 \degree), or obtuse triangles (if one angle measure is larger than 90 \degree).

For quadrilaterals, we have the following classifications:

A quadrilateral with 4 congruent sides, and 4 right angles.
Square
A quadrilateral with 2 pairs of congruent sides, and 4 right angles.
Rectangle
A quadrilateral with 4 congruent sides.
Rhombus
A quadrilateral with 2 pairs of parallel sides.
Parallelogram
A quadrilateral with only one pair of parallel sides.
Trapezoid
A quadrilateral where two sides are parallel, and the other two sides are non parallel and congruent.
Isosceles trapezoid

We want to classify polygons as precisely as possible:

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

For example, the triangle is a right-triangle and isosceles, so we can say it is a right-isosceles triangle to be as precise as possible.

The quadrilateral could be called a rectangle or a rhombus, but the most precise classification is a square.

Worked examples

Example 1

Consider the given quadrilateral.

-6
-5
-4
-3
-2
-1
1
2
x
-6
-5
-4
-3
-2
-1
1
2
y
a

Determine the slopes of \overline{AB}, \overline{BC}, \overline{CD}, and \overline{AD}.

Approach

The slopes of all the sides will help us to determine which sides are parallel and which are perpendicular.

Since the quadrilateral is displayed on the coordinate plane, we can read the rise and run off the coordinate plane instead of using the slope formula.

Solution

Finding the slope of \overline{AB}, m_{\overline{AB}}:

\displaystyle m_{\overline{AB}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Counting squares on the diagram

Finding the slope of \overline{BC}, m_{\overline{BC}}:

\displaystyle m_{\overline{BC}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Counting squares on the diagram

Finding the slope of \overline{CD}, m_{\overline{CD}}:

\displaystyle m_{\overline{CD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Counting squares on the diagram

Finding the slope of \overline{AD}, m_{\overline{AD}}:

\displaystyle m_{\overline{AD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Counting squares on the diagram

Reflection

Notice that adjacent sides have opposite reciprocal slopes and that opposite sides have the same slopes.

b

Calculate AB and AD.

Approach

-6
-5
-4
-3
-2
-1
1
2
x
-6
-5
-4
-3
-2
-1
1
2
y

Since we have the diagram, we do not need to use the distance formula as we can use the Pythagorean theorem directly.

Solution

Calculating AB:

\displaystyle AB^2\displaystyle =\displaystyle 3^2+4^2substitute into Pythagorean theorem
\displaystyle AB^2\displaystyle =\displaystyle 25Simplify using order of operations
\displaystyle AB\displaystyle =\displaystyle \sqrt{25}Square root both sides
\displaystyle AB\displaystyle =\displaystyle 5Simplify

Calculating AD:

\displaystyle AD^2\displaystyle =\displaystyle 4^2+3^2substitute into Pythagorean theorem
\displaystyle AD^2\displaystyle =\displaystyle 25Simplify using order of operations
\displaystyle AD\displaystyle =\displaystyle \sqrt{25}Square root both sides
\displaystyle AD\displaystyle =\displaystyle 5Simplify

Reflection

We only took the positive square root because we were looking for a length, which cannot be negative.

c

Classify the quadrilateral as precisely as possible.

Solution

From part (a), we can determine that:

  • Adjacent sides are perpendicular.
  • Opposite sides are parallel.

From this, the quadrilateral is a rectangle or a square.

From part (b), we can determine that:

  • Two adjacent sides are congruent.

From this, we can now classify the quadrilateral as a square.

Example 2

Classify the triangle with vertices at A \left(-6,-2\right), B \left(-4,4\right), and C \left(14,-2\right) as precisely as possible.

Approach

Drawing a rough sketch can help us to come up with a strategy.

x
y

From the sketch we can see that \overline{AB} and \overline{BC} appear to be perpendicular. It also appears that all the side lengths are different.

We can now calculate the slopes and lengths of the sides to draw a correct conclusion.

Solution

Find AB:

\displaystyle AB\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-6-(-4)\right)^2+\left(-2-4\right)^2} Substitute
\displaystyle =\displaystyle \sqrt{\left(-2\right)^2+\left(-6\right)^2} Simplify the parentheses
\displaystyle =\displaystyle \sqrt{4+36} Square terms
\displaystyle =\displaystyle \sqrt{40} Simplify

Find BC:

\displaystyle BC\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-4-14\right)^2+\left(4-(-2)\right)^2} Substitute
\displaystyle =\displaystyle \sqrt{\left(-18\right)^2+\left(6\right)^2} Simplify parentheses
\displaystyle =\displaystyle \sqrt{324+36} Square terms
\displaystyle =\displaystyle \sqrt{360} Simplify

Find AC:

\displaystyle BC\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} Distance formula
\displaystyle =\displaystyle \sqrt{\left(-6-14\right)^2+\left(-2-(-2)\right)^2} Substitute
\displaystyle =\displaystyle \sqrt{\left(-20\right)^2+\left(0\right)^2} Simplify parentheses
\displaystyle =\displaystyle \sqrt{400} Square terms
\displaystyle =\displaystyle 20 Simplify

None of the sides are congruent, so it is a scalene triangle.

Find the slope of \overline{AB},m_{\overline{AB}}:

\displaystyle m_{\overline{AB}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Formula for slope
\displaystyle =\displaystyle \dfrac{4-(-2)}{-4-(-6)}Substitute
\displaystyle =\displaystyle \dfrac{6}{2}Simplify numerator
\displaystyle =\displaystyle 3Simplify fraction

Find the slope of \overline{BC},m_{\overline{BC}}:

\displaystyle m_{\overline{BC}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Formula for slope
\displaystyle =\displaystyle \dfrac{-2-4}{14-(-4)}Substitute
\displaystyle =\displaystyle \dfrac{-6}{18}Simplify numerator
\displaystyle =\displaystyle -\dfrac{1}{3}Simplify fraction

These two sides are perpendicular, so they form a right angle.

\triangle ABC is a right-scalene triangle.

Reflection

If \overline{AB} and \overline{BC} were not perpendicular, then we would also need to find the slope of \overline{AC} to check whether or not any two sides are perpendicular.

Outcomes

G.GPE.A.1

Use coordinates to justify geometric relationships algebraically and to solve problems.

G.MP1

Make sense of problems and persevere in solving them.

G.MP2

Reason abstractly and quantitatively.

G.MP3

Construct viable arguments and critique the reasoning of others.

G.MP4

Model with mathematics.

G.MP5

Use appropriate tools strategically.

G.MP6

Attend to precision.

G.MP7

Look for and make use of structure.

G.MP8

Look for and express regularity in repeated reasoning.

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