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

11.01 Distance and the coordinate plane

Lesson
Practice

Introduction

The Pythagorean theorem was used to find distances in the coordinate plane in 8th grade. In this lesson, we will generalize that process to help us find the distance or length of any line segment. This will give us a new formula that is useful for finding perimeter and area in the coordinate plane.

Distance and the coordinate plane

Exploration

Consider the triangle below.

4
x
-10
10
y
  1. How can we find the distance from A to C?
  2. How can we find the distance from B to C?
  3. How can we find the distance from A to B?

We can use the Pythagorean theorem to help us find distances and lengths on the coordinate plane.

x
y

The distance between two points can be calculated by creating a right triangle, where the line segment is the hypotenuse, then using the Pythagorean theorem.

Length of a: a=\left|x_2-x_1\right|

Length of b: b=\left|y_2-y_1\right|

Length of d using the Pythagorean theorem:

\begin{aligned} d^2&= a^2+b^2 \\ d^2&= (x_2-x_1)^2 + (y_2-y_1)^2 \\ d& = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{aligned}

This equation is known as the distance formula.

\displaystyle d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
\bm{d}
distance between the two points
\bm{\left(x_1,y_1\right)}
coordinates of the first point
\bm{\left(x_2,y_2\right)}
coordinates of the second point

Examples

Example 1

Find the distance between A \left(-1,9\right) and B \left(-4,1\right). Leave your answer in exact form.

Worked Solution
Create a strategy

We can either plot \overline{AB} on the coordinate plane and use the Pythagorean theorem, or we can use the distance formula. Using the distance formula will likely be more efficient.

Apply the idea

We will use A \left(-1,9\right) as \left(x_1,y_1\right) and B \left(-4,1\right) as \left(x_2,y_2\right).

\displaystyle d\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}Distance formula
\displaystyle =\displaystyle \sqrt{\left(-4--1\right)^2+\left(1-9\right)^2}Substitute \left(x_1,y_1\right)=\left(-1,9\right) and \left(x_2,y_2\right)=\left(-4,1\right)
\displaystyle =\displaystyle \sqrt{\left(-3\right)^2+\left(-8\right)^2}Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{9+64}Evaluate the squares
\displaystyle =\displaystyle \sqrt{73}Evaluate the addition

There are no perfect squares as factors of 73, so there is no further simplification to do.

The distance between A and B is \sqrt{73} units.

Reflect and check

A good final check is that our answer is positive as it is a length, so it must be a positive value.

We could also have used A \left(-1,9\right) as \left(x_2,y_2\right) and B \left(-4,1\right) as \left(x_1,y_1\right) and gotten the same answer.

Idea summary

The distance formula:

\displaystyle d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
\bm{d}
distance between the two points
\bm{\left(x_1,y_1\right)}
coordinates of the first point
\bm{\left(x_2,y_2\right)}
coordinates of the second point

Perimeter and area in the coordinate plane

To calculate the area and perimeter of polygons on the coordinate plane, we can first use the distance formula to find the relevant distances.

Perimeter

The sum of the side lengths of a polygon

Area

The measure, in square units, of the inside region of a closed two-dimensional figure

We define a square with side lengths of 1 unit to have an area of 1 square unit. With this definition we can easily find that the area of a rectangle will be the product of its length and width, and we can then use this to establish area formulas for a number of other polygons:

x
y
A=lw
x
y
A=\dfrac{1}{2}bh

Exploration

Consider the polygon shown below.

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y
  1. How could we find the perimeter of this shape?
  2. How could we find the area of this shape?

For figures which are not one of our basic shapes, we can break down the composite figure into basic shapes, then add or subtract the pieces.

Examples

Example 2

Calculate the area of the triangle shown below.

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y
Worked Solution
Create a strategy

We can draw a rectangle around the triangle, find the area of the rectangle, then subtract the areas of the triangles around the outside of the given triangle.

Apply the idea

A=A_{\text{Rectangle}}-A_{\text{Triangle 1}}-A_{\text{Triangle 2}}-A_{\text{Triangle 3}}

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

To find the area of the rectangle, we need to know the side lengths of the sides. The length of the vertical sides is l=\left|4-\left(-4\right)\right|=8, and the length of the horizontal sides is w=\left|3-\left(-4\right)\right|=7.

A_{\text{Rectangle}}=\left(8\right)\left(7\right)=56

Next, we will find the base and height of Triangle 1. Using the x-values to find the length of the base, we get b_1=\left|-4-2\right|=6. Using the y-values to find the length of the height, we get {h_1=\left|4-0\right|=4}

A_{\text{Triangle }1}=\dfrac{1}{2}\left(6\right)\left(4\right)=12

We will repeat these steps for triangles 2 and 3.

A_{\text{Triangle }2}=\dfrac{1}{2}\left(1\right)\left(8\right)=4

A_{\text{Triangle }3}=\dfrac{1}{2}\left(7\right)\left(4\right)=14

Now substituting these values into the formula we created before, we find the area of the given triangle to be:

A=56-12-4-14=26

Example 3

Consider a quadrilateral with vertices A \left(-4,3\right), B \left(-2,-4\right), C \left(4,-4\right), and D \left(5,3\right).

a

Determine the perimeter of the quadrilateral, rounding your answer to two decimal places.

Worked Solution
Create a strategy
-4
-3
-2
-1
1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
4
y

It can be very helpful to make a sketch to see what type of quadrilateral we have.

From this sketch, we see we have a trapezoid. We know it's a trapezoid because it has one pair of parallel sides.

Apply the idea

To find the perimeter, we want to find the length of each side and then find their sum by adding them all together.

From the sketch, the two parallel sides are horizontal line segments. This means that we can find their lengths by calculating the difference in the x-coordinates of their end points.

\displaystyle AD\displaystyle =\displaystyle \left| -4-5\right|
\displaystyle =\displaystyle 9
\displaystyle BC\displaystyle =\displaystyle \left| -2-4\right|
\displaystyle =\displaystyle 6

For the other two sides we will need to use the distance formula or the Pythagorean theorem, substituting in the coordinates of the end points for each side.

\displaystyle AB\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State distance formula
\displaystyle =\displaystyle \sqrt{\left(-4-\left(-2\right)\right)^2+\left(3-\left(-4\right)\right)^2}Substitute \left(x_1,y_1\right)=\left(-2,-4\right) and \left(x_2,y_2\right)=\left(-4,3\right)
\displaystyle =\displaystyle \sqrt{\left(2\right)^2+\left(7\right)^2}Evaluate the parentheses
\displaystyle =\displaystyle \sqrt{4+49}Evaluate the squares
\displaystyle =\displaystyle \sqrt{53}Evaluate the addition
\displaystyle CD\displaystyle =\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}State distance formula
\displaystyle =\displaystyle \sqrt{\left(5-1\right)^2+\left(3-\left(-4\right)\right)^2}Substitute \left(x_1,y_1\right)=\left(4,-4\right) and \left(x_2,y_2\right)=\left(5,3\right)
\displaystyle =\displaystyle \sqrt{\left(1\right)^2+\left(7\right)^2}Evaluate the parenthesis
\displaystyle =\displaystyle \sqrt{1+49}Evaluate the squares
\displaystyle =\displaystyle \sqrt{50}Evaluate the addition

Now that we have all of the side lengths we can find their sum on the calculator and round to two decimal places.

\displaystyle P_{ABCD}\displaystyle =\displaystyle 9+6+\sqrt{53}+\sqrt{50}
\displaystyle =\displaystyle 29.35

The perimeter of ABCD is 29.35 units.

Reflect and check

It is generally best to wait until the end to round for the most precise answer.

b

Determine the area of the quadrilateral.

Worked Solution
Create a strategy

This quadrilateral can be broken into 2 triangles and a rectangle.

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

After finding the areas of each shape, we can add the areas together.

Apply the idea

The vertical length of the square is l=\left|3-\left(-4\right)\right|=7, and the horizontal length is {w=\left|4-\left(-2\right)\right|=6}.

A_{\text{Square}}=\left(7\right)\left(6\right)=42

The base of the left triangle is 2 and the height is 7, so

A_{\text{Triangle }1}=\dfrac{1}{2}\left(2\right)\left(7\right)=7

The base of the right triangle is 1 and the height is 7, so

A_{\text{Triangle }2}=\dfrac{1}{2}\left(1\right)\left(7\right)=3.5

The toatl area of the quadrilateral is

A=42+7+3.5=52.5 units^2.

Reflect and check

This is a trapezoid, so we could have used the formula:

A_{ABCD}=\dfrac{1}{2}\left(a + b\right)h where a and b are the parallel sides.

From the previous part, we have a=6 and b=9.

In the calculation of AB and CD, we found the vertical distance as a part of the calculation (the difference between the y-coordinates).

\displaystyle h\displaystyle =\displaystyle \left| 3--4 \right|
\displaystyle =\displaystyle 7

Now we can substitute these values into the formula for the are of a trapezoid.

\displaystyle A_{ABCD}\displaystyle =\displaystyle \dfrac{1}{2}\left(a+b\right)h
\displaystyle =\displaystyle \dfrac{1}{2}\left(6+9\right)\left(7\right)
\displaystyle =\displaystyle \dfrac{1}{2}\left(15\right)\left(7\right)
\displaystyle =\displaystyle 52.5

The area of the quadrilateral is 52.5 units^2.

Idea summary

To find the perimeter of a shape, we can use the distance formula to identify the side lengths. Then we find the sum of all of the side lengths.

To find the area of a shape, we can break the shape into familiar polygons. We can then find the areas of those polygons and add them together.

Outcomes

G.GPE.B.7

Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g. Using the distance formula.

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