Understanding Area and Perimeter in the Coordinate Plane

Here are a few recommended readings before getting started with this lesson.

Right triangles
Pythagorean Theorem
Distance Formula
Perimeter
Area

Try your knowledge on these topics.

a State the coordinates of the points plotted on the plane.

b Determine whether a triangle with side lengths

4,

5,

and

6

is a right triangle.

c Use the Pythagorean Theorem to find the missing side length. Round the answer to one decimal place.

A right triangle with given side lengths

d Calculate

M N

by using the Distance Formula.

e Find the perimeter of the given polygon.

f Calculate the area of the rectangle.

	Length	Width
Endpoints	$(- 5, 2)$ and $(3, 5)$	$(3, 5)$ and $(5, 0)$
Substitute	$ℓ = (3 - (- 5))^{2} + (5 - 2)^{2}$	$w = (5 - 3)^{2} + (0 - 5)^{2}$
Evaluate	$ℓ \approx 8.5$	$w \approx 5.4$

Length

Width

Endpoints

(- 5, 2)

and

(3, 5)

(3, 5)

and

(5, 0)

Substitute

ℓ = (3 - (- 5))^{2} + (5 - 2)^{2}

w = (5 - 3)^{2} + (0 - 5)^{2}

Evaluate

ℓ \approx 8.5

w \approx 5.4

	Area	Perimeter
Formula	$A = w ℓ$	$P = 2 (w + ℓ)$
Substitute	$A = 5.4 (8.5)$	$P = 2 (5.4 + 8.5)$
Evaluate	$A = 45.9$	$P = 27.8$

Area

Perimeter

Formula

A = w ℓ

P = 2 (w + ℓ)

Substitute

A = 5.4 (8.5)

P = 2 (5.4 + 8.5)

Evaluate

A = 45.9

P = 27.8

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Side	Endpoints	Substitute	Evaluate
$\overline{M N}$	$M (- 2, - 2)$ and $N (3, 4)$	$M N = (3 - (- 2))^{2} + (4 - (- 2))^{2}$	$M N \approx 7.8$
$\overline{N K}$	$N (3, 4)$ and $K (4, - 2)$	$N K = (4 - 3)^{2} + (- 2 - 4)^{2}$	$N K \approx 6.1$
$\overline{M K}$	$M (- 2, - 2)$ and $K (4, - 2)$	$M K = (4 - (- 2))^{2} + (- 2 - (- 2))^{2}$	$M K = 6$

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Side

Endpoints

Substitute

Evaluate

\overline{M N}

M (- 2, - 2)

and

N (3, 4)

M N = (3 - (- 2))^{2} + (4 - (- 2))^{2}

M N \approx 7.8

\overline{N K}

N (3, 4)

and

K (4, - 2)

N K = (4 - 3)^{2} + (- 2 - 4)^{2}

N K \approx 6.1

\overline{M K}

M (- 2, - 2)

and

K (4, - 2)

M K = (4 - (- 2))^{2} + (- 2 - (- 2))^{2}

M K = 6

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Side	Endpoints	Substitute	Evaluate
$a$	$(2, 3)$ and $(3, 7.5)$	$a = (3 - 2)^{2} + (7.5 - 3)^{2}$	$a \approx 4.6$
$b$	$(3, 7.5)$ and $(7, 1)$	$b = (7 - 3)^{2} + (1 - 7.5)^{2}$	$b \approx 7.6$
$c$	$(2, 3)$ and $(7, 1)$	$c = (7 - 2)^{2} + (1 - 3)^{2}$	$c \approx 5.4$

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Side

Endpoints

Substitute

Evaluate

a

(2, 3)

and

(3, 7.5)

a = (3 - 2)^{2} + (7.5 - 3)^{2}

a \approx 4.6

b

(3, 7.5)

and

(7, 1)

b = (7 - 3)^{2} + (1 - 7.5)^{2}

b \approx 7.6

c

(2, 3)

and

(7, 1)

c = (7 - 2)^{2} + (1 - 3)^{2}

c \approx 5.4

$d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Side	Endpoints	Substitute	Evaluate
$\overline{A B}$	$A (0, 0)$ and $B (4, 7)$	$A B = (4 - (0))^{2} + (7 - 0)^{2}$	$A B \approx 8.1$
$\overline{B C}$	$B (4, 7)$ and $C (8, 0)$	$B C = (8 - 4)^{2} + (0 - 7)^{2}$	$B C \approx 8.1$

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Side

Endpoints

Substitute

Evaluate

\overline{A B}

A (0, 0)

and

B (4, 7)

A B = (4 - (0))^{2} + (7 - 0)^{2}

A B \approx 8.1

\overline{B C}

B (4, 7)

and

C (8, 0)

B C = (8 - 4)^{2} + (0 - 7)^{2}

B C \approx 8.1

A linear function

f (x)

passes through the origin. Find the function's slope if the point

(8, f (8))

17

units from the origin.

We have been told that f(x) is a linear function that passes though the origin. This means that its y-intercept is 0. With this information, we can write a partial equation in slope-intercept form for f(x).

We know that the point (8 ,f(8)) is on the function's graph. Let's write this point using our equation.

The distance from this point to the origin ( 0, 0) is 17 units. By substituting all of these values into the Distance Formula, we get an equation, which we can solve for m.

The slope of the line is either 158 or - 158.

Tearrik's school is located

6

miles northeast of his home. His favorite gym is

4

miles southeast of his home. How far is it from the school to the gym? Answer in exact form.

We will first find the position of the school and the position of the gym in relation to Tearrik's home. After that we will find the distance between them.

Position of the School

To model this situation, we can use a coordinate plane where the y-axis points north and south and the x-axis points east and west. We can let Tearrik's home be at the origin and let one unit represent a distance of one mile. Let's mark the position of Tearrik's school, 6 miles to the northeast.

The coordinates of a position to the northeast of the origin ( 0, 0) can be written on the form ( a, a). To find a, we can use the Distance Formula and the information that the distance to the school is 6 miles.

Position of the Gym

The gym is 4 miles from Tearrik's home. Since it is located to the southeast, coordinates on the form ( b, - b) represent its position in our coordinate plane.

Let's find b using the information that the distance between the home and the gym is 4 miles.

Distance Between the School and the Gym

We have now found the position of both the school and the gym in our coordinate plane.

Let's use the Distance Formula to find the distance between the school and the gym.

The distance between the school and the gym is sqrt(52) miles.

Calculate the distance between the points $A$ and $B .$ Answer in exact form.

To find the distance between A and B, we will use the Distance Formula. We know that the x-coordinate of A is -2 and that the y-coordinate of B is - 13. In order to use the Distance Formula, we first need to find the missing coordinate for each of these points.

Point A

Let's substitute x= - 2 into the function rule and solve for y.

Point A has the coordinates ( - 2, 3).

Point B

Here we will substitute -13 for y and solve for x.

From the diagram we know that the x-coordinate must be nonnegative, which is why we only kept the principal root when solving the equation. Therefore, the coordinates of point B are ( 6, -13).

Distance of AB

We will now substitute the coordinates of the points into the Distance Formula.

The distance between the points is 8sqrt(5) units.

Perimeter and Area in a Coordinate Plane

Catch-Up and Review

Investigate Shapes by Using Coordinates

Solving Problems in the Real World by Using Coordinates

Hint

Solution

Investigate the Properties of a Triangle

Hint

Solution

Investigate the Properties of a Compound Shape

Hint

Solution

Square Room

Rectangular Room

Triangular Room

Total Area

Investigate the Properties of a Triangle

Hint

Solution

Classifying Triangles by Using Coordinates

Hint

Solution

Position of the School

Position of the Gym

Distance Between the School and the Gym

Point A

Point B

Distance of AB

Perimeter and Area in a Coordinate Plane

Recommended exercises

	7 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions