2.1: The Rectangular Coordinate Systems and Graphs
- Using the Pythagorean Theorem
- Plot ordered pairs in a Cartesian coordinate system.
- Use the distance formula.
- Use the midpoint formula.
The Pythagorean Theorem
We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem . This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.
Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a 90° angle, marked with a small square in the corner. The side of the triangle opposite the 90°90° angle is called the hypotenuse and each of the other sides are called legs .
The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, \(a^{2}+b^{2}=c^{2}\), where a and b are the lengths of the legs and cc is the length of the hypotenuse.
Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.
In any right triangle, where \(a\) and \(b\) are the lengths of the legs, \(c\) is the length of the hypotenuse.
Then
\[a^{2}+b^{2}=c^{2} \label{Ptheorem}\]
To solve exercises that use the Pythagorean Theorem (Equation \ref{Ptheorem}), we will need to find square roots. We have used the notation \(\sqrt{m}\) and the definition:
If \(m = n^{2}\), then \(\sqrt{m} = n\), for \(n\geq 0\).
For example, we found that \(\sqrt{25}\) is 5 because \(25=5^{2}\).
Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.
Use the Pythagorean Theorem to find the length of the hypotenuse shown below.
Solution
| Label side c on the figure. |
Let
c
= the length of the hypotenuse.
|
| Write the appropriate formula. | \(a^{2} + b^{2} = c^{2}\) |
| Substitute. | \(3^{2}+4^{2}=c^{2}\) |
| Solve the equation. | \(9+16=c^{2}\) |
| Simplify. | \(25=c^{2}\) |
| Use the definition of square root. | \(\sqrt{25} = c\) |
| Simplify. | \(5=c\) |
|
Check
|
|
| Answer | The length of the hypotenuse is 5. |
Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.
- Answer
-
c=10
Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.
- Answer
-
c=13
Use the Pythagorean Theorem to find the length of the leg shown below.
Solution
| Label side \(b\). | |
| Write the appropriate formula. | \(a^{2} + b^{2} = c^{2}\) |
| Substitute. | \(5^{2}+b^{2}=13^{2}\) |
| Solve the equation. |
\(25+b^{2}=169\) |
| Isolate the variable term. | \(b^{2}=144\) |
| Use the definition of square root. | \(b = \sqrt{144}\) |
| Simplify. | \(b=12\) |
|
Check.
|
|
| Answer . | The length of the leg is 12. |
Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.
- Answer
-
8
Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.
- Answer
-
12
Find the length of the missing side of the right triangle in Figure \(\PageIndex{5}\).
Solution
As we have measurements for side \(b\) and the hypotenuse, the missing side is \(a\).
\[\begin{align*} a^2+b^2&= c^2\\ a^2+{(4)}^2&= {(12)}^2\\ a^2+16&= 144\\ a^2&= 128\\ a&= \sqrt{128}\\ &= 8\sqrt{2} \end{align*}\]
Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.
- Answer
-
\(5\) units
Kelvin is building a gazebo and wants to brace each corner by placing a 10″ piece of wood diagonally as shown above.
If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.
Solution
\(\begin{array} {ll} {\textbf{Step 1. }\text{Read the problem.}} &{} \\\\ {\textbf{Step 2. }\text{Identify what we are looking for.}} &{\text{the distance from the corner that the}} \\ {} &{\text{bracket should be attached}} \\ \\{\textbf{Step 3. }\text{Name. Choose a variable to represent it.}} &{\text{Let x = distance from the corner.}} \\ {\textbf{Step 4.} \text{Translate}} &{} \\ {\text{Write the appropriate formula and substitute.}} &{a^{2} + b^{2} = c^{2}} \\ {} &{x^{2} + x^{2} = 10^{2}} \\ \\ {\textbf{Step 5. Solve the equation.}} &{} \\ {} &{2x^{2} = 100} \\ {\text{Isolate the variable.}} &{x^{2} = 50} \\ {\text{Simplify. Approximate to the nearest tenth.}} &{x \approx 7.1}\\\\ {\textbf{Step 6. }\text{Check.}} &{}\\ {a^{2} + b^{2} = c^{2}} &{} \\ {(7.1)^{2} + (7.1)^{2} \approx 10^{2} \text{ Yes.}} &{} \\\\ {\textbf{Step 7. Answer the question.}} &{\text{Kelven should fasten each piece of}} \\ {} &{\text{wood approximately 7.1'' from the corner.}} \end{array}\)
John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?
- Answer
-
12 feet
Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?
- Answer
-
8 feet
Plotting Ordered Pairs in the Cartesian Coordinate System
An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the \(x\)-axis and the vertical axis the \(y\)-axis.
The Cartesian coordinate system, also called the rectangular coordinate system , is based on a two-dimensional plane consisting of the \(x\)-axis and the \(y\)-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure \(\PageIndex{2}\).
The center of the plane is the point at which the two axes cross. It is known as the origin, or point \((0,0)\). From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the \(x\)-axis and up the \(y\)-axis; decreasing, negative numbers to the left on the \(x\)-axis and down the \(y\)-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure \(\PageIndex{3}\).
Each point in the plane is identified by its \(x\)-coordinate, or horizontal displacement from the origin, and its \(y\)-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form \((x,y)\). An ordered pair is also known as a coordinate pair because it consists of \(x\)- and \(y\)-coordinates. For example, we can represent the point \((3,−1)\) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure \(\PageIndex{4}\).
When dividing the axes into equally spaced increments, note that the \(x\)-axis may be considered separately from the \(y\)-axis. In other words, while the \(x\)-axis may be divided and labeled according to consecutive integers, the \(y\)-axis may be divided and labeled by increments of \(2\), or \(10\), or \(100\). In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
A two-dimensional plane where the
- \(x\)-axis is the horizontal axis
- \(y\)-axis is the vertical axis
A point in the plane is defined as an ordered pair, \((x,y)\), such that \(x\) is determined by its horizontal distance from the origin and \(y\) is determined by its vertical distance from the origin.
Plot the points \((−2,4)\), \((3,3)\), and \((0,−3)\) in the plane.
Solution
To plot the point \((−2,4)\), begin at the origin. The \(x\)-coordinate is \(–2\), so move two units to the left. The \(y\)-coordinate is \(4\), so then move four units up in the positive \(y\) direction.
To plot the point \((3,3)\), begin again at the origin. The \(x\)-coordinate is \(3\), so move three units to the right. The \(y\)-coordinate is also \(3\), so move three units up in the positive \(y\) direction.
To plot the point \((0,−3)\), begin again at the origin. The \(x\)-coordinate is \(0\). This tells us not to move in either direction along the \(x\)-axis. The \(y\)-coordinate is \(–3\), so move three units down in the negative \(y\) direction. See the graph in Figure \(\PageIndex{5}\).
AnalysisNote that when either coordinate is zero, the point must be on an axis. If the \(x\)-coordinate is zero, the point is on the \(y\)-axis. If the \(y\)-coordinate is zero, the point is on the \(x\)-axis.
Using the Distance Formula
Derived from the Pythagorean Theorem , the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, \(a^2+b^2=c^2\), is based on a right triangle where \(a\) and \(b\) are the lengths of the legs adjacent to the right angle, and \(c\) is the length of the hypotenuse. See Figure \(\PageIndex{15}\).
The relationship of sides \(|x_2−x_1|\) and \(|y_2−y_1|\) to side \(d\) is the same as that of sides \(a\) and \(b\) to side \(c\). We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, \(|-3|=3\). ) The symbols \(|x_2−x_1|\) and \(|y_2−y_1|\) indicate that the lengths of the sides of the triangle are positive. To find the length \(c\), take the square root of both sides of the Pythagorean Theorem.
\[c^2=a^2+b^2\rightarrow c=\sqrt{a^2+b^2}\]
It follows that the distance formula is given as
\[d^2={(x_2−x_1)}^2+{(y_2−y_1)}^2\rightarrow d=\sqrt{{(x_2−x_1)}^2+{(y_2−y_1)}^2}\]
We do not have to use the absolute value symbols in this definition because any number squared is positive.
Given endpoints \((x_1,y_1)\) and \((x_2,y_2)\), the distance between two points is given by
\[d=\sqrt{{(x_2−x_1)}^2+{(y_2−y_1)}^2}\]
Find the distance between the points \((−3,−1)\) and \((2,3)\).
Solution
Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure \(\PageIndex{16}\)
Then, calculate the length of \(d\) using the distance formula.
\[\begin{align*} d&= \sqrt{{(x_2 - x_1)}^2+{(y_2 - y_1)}^2}\\ &= \sqrt{{(2-(-3))}^2+{(3-(-1))}^2}\\ &= \sqrt{{(5)}^2+{(4)}^2}\\ &= \sqrt{25+16}\\ &= \sqrt{41} \end{align*}\]
Find the distance between two points: \((1,4)\) and \((11,9)\).
- Answer
-
\(\sqrt{125}=5\sqrt{5}\)
Using the Midpoint Formula
When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula . Given the endpoints of a line segment, \((x_1,y_1)\) and \((x_2,y_2)\), the midpoint formula states how to find the coordinates of the midpoint M.
\[M=\left (\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right )\]
A graphical view of a midpoint is shown in Figure \(\PageIndex{18}\). Notice that the line segments on either side of the midpoint are congruent.
Find the midpoint of the line segment with the endpoints \((7,−2)\) and \((9,5)\).
Solution
Use the formula to find the midpoint of the line segment.
\[\begin{align*} \left (\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right )&= \left (\dfrac{7+9}{2},\dfrac{-2+5}{2} \right )\\ &= \left (8,\dfrac{3}{2} \right ) \end{align*}\]
Find the midpoint of the line segment with endpoints \((−2,−1)\) and \((−8,6)\).
- Answer
-
\(\left (-5,\dfrac{5}{2} \right )\)
The diameter of a circle has endpoints \((−1,−4)\) and \((5,−4)\). Find the center of the circle.
Solution
The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.
\[\begin{align*} \left (\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2} \right )&= \left (\dfrac{-1+5}{2},\dfrac{-4-4}{2}) \right )\\ &= \left (\dfrac{4}{2},-\dfrac{8}{2} \right )\\ &= (2,4) \end{align*}\]
Access these online resources for additional instruction and practice with the Cartesian coordinate system.
1. Plotting points on the coordinate plane
2. Find x and y intercepts based on the graph of a line
Key Concepts
- The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See Example . We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the \(x\) - axis and displacement from the \(y\) - axis. See Example .
- The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See Example and Example .
- The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the \(x\)-coordinates and the sum of the \(y\)-coordinates of the endpoints by \(2\). See Example and Example .