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11. Solving Problems in the Coordinate Plane
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United States of America
Grades 9-12

11.03 Classifying polygons in the coordinate plane

Lesson
Practice

Introduction

In previous lessons, we have explored the properties of various triangles and quadrilaterals. In this lesson, we will learn how to use the slopes of the sides of a polygon to classify them precisely based on the properties of their sides and angles in the coordinate plane.

Classifying polygons in the coordinate plane

Polygons can be classified based on their side lengths and angle measures.

For triangles, we have the following classifications:

Equilateral triangle

A triangle with three sides of equal length and three 60\degree interior angles

A triangle with all 3 sides marked congruent and all 3 angles marked congruent and labeled 60 degrees
Isosceles triangles

A triangle containing at least two sides of equal length and two equal interior angle measures

A triangle with two sides marked congruent and the angles between the congruent sides and the third side are marked congruent
Scalene triangle

A triangle containing three unequal side lengths and three unequal angle measures

A triangle where one angle has a single angle marking, one has a double angle marking and one has a triple angle marking. One side has a single marking, one hase a double marking and one hase a triple marking
Right triangle

A triangle containing an interior right angle

Two right triangles: one having two congruent sides, and the other having no congruent sides.

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

Triangles that are not right triangles can be further 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:

Square

A quadrilateral with four right angles and four equal-length sides

A quadrilateral with 4 congruent sides, and 4 right angles.
Rectangle

A quadrilateral containing four right angles

A quadrilateral with 2 pairs of congruent sides, and 4 right angles.
Rhombus

A quadrilateral containing four congruent sides.

A quadrilateral with 4 congruent sides.
Parallelogram

A quadrilateral containing two pairs of parallel sides

A quadrilateral with 2 pairs of parallel sides.

We want to classify polygons as precisely as possible:

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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.

If we know the coordinates of the vertices of a polygon, we can find the most precise classification using the lengths and slopes of line segments joining pairs of vertices.

  1. Find the slope of each side to determine if any sides are parallel or perpendicular.
  2. Find the length of each side to determine if any sides are congruent.

Examples

Example 1

Consider the given quadrilateral.

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a

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

Worked Solution
Create a strategy

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.

Apply the idea

Finding the slope of \overline{AB} :

\displaystyle \text{slope of }{\overline{AB}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Substitute \text{rise}=-3 and \text{run}=4

Finding the slope of \overline{BC}:

\displaystyle \text{slope of }{\overline{BC}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Substitute \text{rise}=4 and \text{run}=3

Finding the slope of \overline{CD}:

\displaystyle \text{slope of }{\overline{CD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle -\dfrac{3}{4}Substitute \text{rise}=-3 and \text{run}=4

Finding the slope of \overline{AD}:

\displaystyle \text{slope of }{\overline{AD}}\displaystyle =\displaystyle \dfrac{\text{rise}}{\text{run}}Definition of slope
\displaystyle =\displaystyle \dfrac{4}{3}Substitute \text{rise}=4 and \text{run}=3
Reflect and check

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

b

Calculate the lengths of the sides.

Worked Solution
Create a strategy
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Since we have the diagram, we will use the Pythagorean theorem, but we could also substitute the coordinates of the vertices into the distance formula.

Apply the idea

Calculating AB:

\displaystyle AB^2\displaystyle =\displaystyle 4^2+3^2Substitute into Pythagorean theorem
\displaystyle AB^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle AB\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle AB\displaystyle =\displaystyle 5Evaluate the square root

Calculating BC:

\displaystyle BC^2\displaystyle =\displaystyle 3^2+4^2Substitute into Pythagorean theorem
\displaystyle BC^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle BC\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle BC\displaystyle =\displaystyle 5Evaluate the square root

Calculating AD:

\displaystyle AD^2\displaystyle =\displaystyle 3^2+4^2Substitute into Pythagorean theorem
\displaystyle AD^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle AD\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle AD\displaystyle =\displaystyle 5Evaluate the square root

Calculating DC:

\displaystyle DC^2\displaystyle =\displaystyle 4^2+3^2Substitute into Pythagorean theorem
\displaystyle DC^2\displaystyle =\displaystyle 25Evaluate the squares and addition
\displaystyle DC\displaystyle =\displaystyle \sqrt{25}Take the square root of both sides
\displaystyle DC\displaystyle =\displaystyle 5Evaluate the square root
Reflect and check

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

If you notice, the legs of the triangle with hypotenuse \overline{AB} are the same length as the legs of the triangle with hypotenuse \overline{DC}. To save some time and work, we could have used this to say AB=DC. We could have used similar reasoning to explain BC=AD.

c

Classify the quadrilateral as precisely as possible.

Worked Solution
Create a strategy

We will use the slopes and lengths found in parts (a) and (b) to determine if any sides are parallel, perpendicular, or congruent.

Apply the idea

From part (a), we can determine that all adjacent sides are perpendicular because their slopes are reciprocals with opposite signs. That is,

  • \overline{AB}\perp \overline{BC}
  • \overline{BC}\perp \overline{CD}
  • \overline{CD}\perp \overline{AD}
  • \overline{AD}\perp \overline{AB}

We can also determine that opposite sides are parallel because they have the same slope. That is,

  • \overline{AB}\parallel \overline{CD}
  • \overline{AD}\parallel \overline{BC}

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

From part (b), we can determine that all 4 sides are congruent. That is,

  • \overline{AB}\cong \overline{BC} \cong \overline{CD} \cong \overline{AD}

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

Reflect and check

The shape is a quadrilateral, a parallelogram, a rectangle, and a rhombus, but the most precise classification is 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.

Worked Solution
Create a strategy

Using technology or drawing a rough sketch can help us visualize the triangle, and that will help us come up with a strategy.

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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.

However, we must calculate the slopes and lengths of the sides to draw a correct conclusion.

Apply the idea

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 the coordinates
\displaystyle =\displaystyle \sqrt{\left(-2\right)^2+\left(-6\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{4+36} Evaluate the squares
\displaystyle =\displaystyle \sqrt{40} Evaluate the addition

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-\left(-2\right)\right)^2} Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-18\right)^2+\left(6\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{324+36} Evaluate the squares
\displaystyle =\displaystyle \sqrt{360} Evaluate the addition

Find AC:

\displaystyle AC\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-\left(-2\right)\right)^2} Substitute the coordinates
\displaystyle =\displaystyle \sqrt{\left(-20\right)^2+\left(0\right)^2} Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{400} Evaluate the squares
\displaystyle =\displaystyle 20 Evaluate the square root

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

Find the slope of \overline{AB}:

\displaystyle \text{slope of }{\overline{AB}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle =\displaystyle \dfrac{4-\left(-2\right)}{-4-\left(-6\right)}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{6}{2}Evaluate the subtraction
\displaystyle =\displaystyle 3Evaluate the division

Find the slope of \overline{BC}:

\displaystyle \text{slope of }{\overline{BC}}\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Slope formula
\displaystyle =\displaystyle \dfrac{-2-4}{14-\left(-4\right)}Substitute the coordinates
\displaystyle =\displaystyle \dfrac{-6}{18}Evaluate the subtraction
\displaystyle =\displaystyle -\dfrac{1}{3}Simplify the fraction

These two sides are perpendicular because their slopes are opposite reciprocals, so they form a right angle.

\triangle ABC is a right-scalene triangle.

Reflect and check

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.

Example 3

Show that the quadrilateral with vertices at A\left(2,1\right),\,B\left(1,4\right),\,C\left(6,5\right),\, D\left(7,2\right) is not a rectangle.

Worked Solution
Create a strategy

By definition, a rectangle has 4 right angles, meaning the adjacent sides are all perpendicular to each other. If we find any two sides that are not perpendicular, then we have shown that the quadrilateral is not a rectangle.

Apply the idea

Slope of \overline{AB}:

\displaystyle m_{\overline{AB}}\displaystyle =\displaystyle \dfrac{4-1}{1-2}
\displaystyle =\displaystyle -3

Slope of \overline{BC}:

\displaystyle m_{\overline{BC}}\displaystyle =\displaystyle \dfrac{5-4}{6-1}
\displaystyle =\displaystyle \dfrac{1}{5}

These slopes are not opposite reciprocals, which means \overline{AB} is not perpendicular to \overline{BC}. This also means those sides do not form a 90\degree angle, so the quadrilateral is not a rectangle.

Reflect and check
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This sketch makes it easy to see that this figure does not have 4 right angles.

However, we could show that this figure has 2 sets of parallel sides, making it a parallelogram.

Example 4

Let ABCD be a rhombus with vertices at A\left(-2,-1\right),\, B\left(x,y\right),\, C\left(-2,-7\right) and D\left(-6,-4\right).

a

Find the coordinates of vertex B.

Worked Solution
Create a strategy

The sides of a rhombus are all the same length, so we need to find the length of one side first. We can do this by counting the rise (length of the vertical distance) and the run (length of the horizontal distance) from point C to point D. The rise and run from point C to B will have to be the same.

Apply the idea
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The horizontal distance from C to D is 4. The vertical distance from C to D is 3.

To find the coordinates of B, we will count 4 units right and 3 units up.

Vertex B is located at \left(2,-4\right).

Reflect and check

Completing the polygon, we can see that this does appear to be a rhombus.

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b

Verify ABCD is a rhombus.

Worked Solution
Create a strategy

To verify this is a rhombus, we need to prove all sides are the same length. We can do this using the distance formula d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}.

Apply the idea
\displaystyle AB\displaystyle =\displaystyle \sqrt{\left(-2-2\right)^2+\left(-1-\left(-4\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(-4\right)^2+\left(3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle BC\displaystyle =\displaystyle \sqrt{\left(2-\left(-2\right)\right)^2+\left(-4-\left(-7\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(4\right)^2+\left(3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle CD\displaystyle =\displaystyle \sqrt{\left(-2-\left(-6\right)\right)^2+\left(-7-\left(-4\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(4\right)^2+\left(-3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5
\displaystyle DA\displaystyle =\displaystyle \sqrt{\left(-6-\left(-2\right)\right)^2+\left(-4-\left(-1\right)\right)^2}
\displaystyle =\displaystyle \sqrt{\left(-4\right)^2+\left(-3\right)^2}
\displaystyle =\displaystyle \sqrt{16+9}
\displaystyle =\displaystyle \sqrt{25}
\displaystyle =\displaystyle 5

Since \overline{AB}\cong \overline{BC} \cong \overline{CD} \cong \overline{DA}, the quadrilateral is a rhombus.

Reflect and check

If the question had asked us to verify that it was a parallelogram instead, we would have needed to find the slopes of each side to prove there are 2 sets of parallel sides.

Idea summary

If we know the coordinates of the vertices of a polygon, we can find the most precise classification using the lengths and slopes of line segments joining pairs of vertices.

  1. Find the slope of each side to determine if any sides are parallel or perpendicular.
  2. Find the length of each side to determine if any sides are congruent.

Outcomes

G.CO.D.13

Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

G.GPE.B.4

Use coordinates to prove simple geometric theorems algebraically.

G.GPE.B.5

Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems.

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