assignment 6 midpoint formula

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High school geometry

Course: high school geometry   >   unit 6.

• Getting ready for analytic geometry
• Distance formula
• Distance between two points
• Midpoint formula
• Distance formula review

Midpoint formula review

What is the midpoint formula, what problems can i solve with the midpoint formula, check your understanding.

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• (Choice C)   ( 1 , 8 ) ‍   C ( 1 , 8 ) ‍  
• (Choice D)   ( 5 , 4 ) ‍   D ( 5 , 4 ) ‍  

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Midpoint Formula

Related Pages Pythagoras’ Theorem Midpoint Formula Coordinate Geometry Geometry Lessons

In this lesson, we will learn

• the midpoint formula. Click here
• how to find the midpoint given two endpoints. Click here
• how to find one endpoint given the midpoint and another endpoint. Click here
• how to proof the midpoint formula. Click here

We have included a midpoint calculator at the end of this lesson.

The Midpoint Formula

Some coordinate geometry questions may require you to find the midpoint of line segments in the coordinate plane. To find a point that is halfway between two given points, get the average of the x -values and the average of the y -values.

The following diagram shows the midpoints formula for the two points (x 1 ,y 1 ) and (x 2 ,y 2 ). Scroll down the page for more examples and solutions on how to use the midpoint formula.

For Example: The midpoint of the points A(1,4) and B(5,6) is

Find the midpoint given two endpoints

We can use the midpoint formula to find the midpoint when given two endpoints.

Example: Find the midpoint of the two points A(1, -3) and B(4, 5).

Midpoint Worksheet 1 Midpoint Worksheet 2 to calculate the midpoint.

How to use the formula for finding the midpoint of two points?

Example: Find the midpoint of the two points (5, 8) and (-5, -6).

How to use the midpoint formula given coordinates in fractions?

Example: Determine the midpoint of the two points (2/3, 1/4) and (11/6, 7/9).

Find an endpoint when given a midpoint and another endpoint

We can use the midpoint formula to find an endpoint when given a midpoint and another endpoint.

Example: M(3, 8) is the midpoint of the line AB. A has the coordinates (-2, 3), Find the coordinates of B.

Coordinates of B = (8, 13)

How to find a missing endpoint when given the midpoint and another endpoint?

How to solve problems using the Midpoint Formula?

Example: For a line segment DE, one endpoint is D(6, 5) and the midpoint M(4, 2). Find the coordinates of the other endpoint, E.

Proof of the Midpoint Formula

How to derive the midpoint formula by finding the midpoint of a line segment?

How to use the Pythagorean theorem to prove the midpoint formula? The following video gives a proof of the midpoint formula using the Pythagorean Theorem. Step 1: Use the distance formula to show the midpoint creates two congruent segments. Step 2: Use the slope formula to show that the coordinate of the midpoint is located on the line segment.

Midpoint Calculator Enter the coordinates of two points and the midpoint calculator will give the midpoint of the two points. Use this to check your answers.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

Midpoint Formula

Midpoint refers to a point that is exactly in the middle of the line segment joining two points. The two reference points are the endpoints of a line segment, and the midpoint is lying in between the two points. The midpoint divides the line joining these two points into two equal halves. Further, if a line is drawn to bisect a line segment joining these two points, the line passes through the midpoint.

The midpoint formula is used to find the midpoint between two points whose coordinates are known to us. The midpoint formula is also used to find the coordinates of the endpoint if we know the coordinates of the other endpoint and the midpoint. In the coordinate plane , if a line is drawn to connect two points (4, 2), and (8, 6), then the coordinates of the midpoint of the line joining these two points are ({4 + 8}/2, {2 + 6}/2) = (12/2, 8/2) = (6, 4). Let us learn more about the formula of the midpoint, and different midpoint methods.

What is Midpoint?

A midpoint is a point lying between two points and is in the middle of the line joining the two points. If a line is drawn joining the two points, then the midpoint is a point at the middle of the line and is equidistant from the two points. Given any two points, say A and C, the midpoint is a point B which is located halfway between points A and C. Therefore, to calculate AB or BC, we can simply measure the length of the line segment and divide by 2.

Observe that point B is equidistant from A and C. A midpoint exists only for a line segment . A line or a ray cannot have a midpoint because a line is indefinite in both directions and a ray has only one end and thus can be extended.

The midpoint formula is defined for the points in the coordinate axes. Let (x 1 , y) 1 and (x 2 , y) 2 be the endpoints of a line segment. The midpoint is equal to half of the sum of the x-coordinates of the two points, and half of the sum of the y-coordinates of the two points. The midpoint formula to calculate the midpoint of a line segment joining these points can be given as,

Midpoint Formula in Math

Given two points A (x 1 , y 1 ) and B (x 2 , y 2 ), the midpoint between A and B is given by,

M(x 3 , y 3 ) = ((x 1 + x 2 )/2, (y 1 + y 2 )/2)

where, M is the midpoint between A and B, and (x 3 , y 3 ) are its coordinates.

Derivation of Midpoint Formula

Let us look at this example and find the midpoint of two points in one-dimensional axis. Suppose, we have two points, 5 and 9, on a number line . The midpoint will be calculated as: (5 + 9)/2 = 14/2 = 7. So, 7 is the midpoint of 5 and 9.

We would apply the same logic to find the midpoint of a line segment with its endpoints, (x 1 , y 1 ) and (x 2 , y 2 ) on a coordinate plane. For any line segment, the midpoint is halfway between its two endpoints. The expression for the x-coordinate of the midpoint is [x 1 + x 2 ]/2, which is the average of the x-coordinates. Similarly, the expression for the y-coordinate is [y 1 + y 2 ]/2, which is the average of the y-coordinates.

Thus, the formula for midpoint is, ((x 1 + x 2 )/2, (y 1 + y 2 )/2)

Let us check an example to see the application of the midpoint formula.

Example: Using the midpoint formula, find the midpoint between points X(5, 3) and Y(7, 8).

Solution: Let M be the midpoint between X and Y.

M = ((5 + 7)/2, (3 + 8)/2) = (6, 11/2)

Therefore, the coordinates of the midpoint between X and Y is (6, 11/2).

How to Find Midpoint?

Further, based on the points and their coordinate values, the following two midpoint methods are used to find the midpoint of the line segment joining the two points.

Method 1: If the line segment is vertical or horizontal , then dividing the length by 2 and counting that value from any of the endpoints will get us the midpoint of the line segment. Look at the figure shown below. The coordinates of points A and B are (-3, 2) and (1, 2) respectively. The length of the horizontal line \(\overline{AB}\) is 4 units. Half of this length is 2 units. Moving 2 units from the point (-3, 2) will give (-1, 2). So, (-1, 2) is the midpoint of \(\overline{AB}\).

Method 2: The other way to find the midpoint is by using the midpoint formula. The coordinates of points A and B are (-3, -3) and (1, 4) respectively. Using midpoint formula, we have: ({-3 + 1}/2, {-3 + 4}/2) = (-2/2, 1/2)= (-1,1/2).

Method 3: One way to find the midpoint of a line given in a plane is by using construction. We can use a compass and straightedge construction to first construct a lens using circular arcs of equal (and large enough) radii centered at the two endpoints, then connecting the cusps of the lens (the two points where the arcs intersect). The point of intersection of the line connecting the cusps and the segment is the midpoint of the segment.

Here's an example to find the coordinate of an endpoint, given the midpoint and the coordinates of the other endpoint.

Example: Midpoint R between the points P and Q has the coordinates (4, 6). If the coordinates of Q are (8, 10), then what are the coordinates for point P? Solve it by using the midpoint formula.

Let x coordinate of P be m and y coordinate of P be n.

P = (m, n) Q = (8, 10) R = (4, 6) By using the midpoint formula,

R = ((m + 8)/2, (n + 10)/2) = (4, 6) Solving for m, (m + 8)/2 = 4 m + 8 = 8 m = 0

Solving for n, (n + 10)/2 = 6 n + 10 = 12 n = 2

Therefore, coordinates of P are (0, 2).

Formulas Related to Midpoint

The midpoint formula includes computations separately for the x-coordinate of the points, and the y-coordinate of the points. The following two formulas are closely related to the midpoint formula.

• Centroid of Triangle Formula

Section Formula

Centroid of a triangle formula.

The point of intersection of the medians of a triangle is called the centroid of the triangle. The median is a line joining the vertex to the midpoint of the opposite side of the triangle . The centroid divides the median of the triangle in the ratio 2:1. For a triangle with vertices (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) the formula to find the coordinates of the centroid of the triangle is as follows.

The section formula is helpful to find the coordinates of any point which is on the line joining the two points. Further, the ratio in which the point divided the line joining the two given points is needed to know the coordinates of the point. The point can be located between the points, or anywhere beyond the points, but on the same line. The section formula to find the coordinates of a point, which divides the line joining the points (x 1 , y 1 ), and (x 2 , y 2 ) in the ratio m:n is as follows. Here:

• The positive sign is used in the formula to find the coordinates of the point, which divides the points internally, and
• the negative sign is used if the point is dividing externally.

Important Notes on Midpoint:

The following points are the important properties of the midpoints.

• The midpoint divides a line segment in an equal ratio, that is, 1:1.
• The midpoint divides a line segment into two equal parts.
• The bisector of a line segment cuts it at its midpoint.

☛ Related Topics:

• Midpoint Calculator
• Centroid Calculator

Midpoint Formula Examples

Example 1: The diameter of a circle has endpoints, (2, -3) and (-6, 5). Find the coordinates of the center of the circle.

The center of a circle divides the diameter into 2 equal parts. So, it is a midpoint of the diameter. Let x 1 = 2, y 1 = -3, x 2 = -6, and y 2 = 5 The coordinates of the center is calculated as: ((x 1 + x 2 )/2, (y 1 + y 2 )/2) = ((2 + (-6))/2, (-3 + 5)/2) = (-4/2, 2/2) = (-2, 1).

Answer: ∴ the center of the circle is (-2, 1).

Example 2: Consider the line segment \(\overline{AB}\) shown below.

The endpoints are (1, h) and (5, 7). Find the value of h if the midpoint of \(\overline{AB}\) is (3, -2).

Let x 1 = 1, y 1 = h, x 2 = 5, and y 2 = 7. According to the definition of midpoint we have, ((x 1 + x 2 )/2, (y 1 + y 2 )/2) = ((1 + 5)/2, (h + 7)/2) = (6/2, (h + 7)/2) = (3, (h + 7)/2). Equalizing this with the midpoint value (3, -2) we have (h + 7)/2 = -2; h + 7 = -2 × 2; h + 7 = -4; h = -4 - 7; h = -11.

Answer: ∴ the value of h is -11.

Example 3: The endpoints of a line segment are (2, h) and (4, 7). Find the value of h if the midpoint is (3, -1).

Let x 1 = 2, y 1 = h, x 2 = 4, and y 2 = 7.

According to the definition of midpoint we have, ((x 1 + x 2 )/2, (y 1 + y 2 )/2) = ((2 + 4)/2, (h + 7)/2) = (6/2, (h + 7)/2) = (3, (h + 7)/2).

Equalizing this with the midpoint value (3, -1) we have (h + 7)/2 = -1;

h + 7 = -1 × 2

h = -2 - 7; h = -9.

Therefore, the value of h is -9.

Answer: ∴ The value of h is -9.

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Practice Questions on Midpoint Formula

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FAQs on Midpoint Formula

What is the midpoint formula in coordinate geometry.

The midpoint formula in coordinate geometry is defined as the formula to find the center point of a straight line, using the coordinates of its endpoints. It says the midpoint of a line segment with coordinates (x 1 y 1 ) and (x 2 , y 2 ) is given by the formula ((x 1 + x 2 )/2, (y 1 + y 2 )/2).

How to Use Midpoint Formula?

The midpoint formula is really simple when it comes to its applicability.

• Step 1: Identify the line segment or the two endpoints.
• Step 2: Find their coordinates.
• Step 3: Add the x coordinates of both the endpoints and divide by 2.
• Step 4: Add the y coordinates of both endpoints and divide by 2.
• Step 5: Write the values so obtained in steps 3 and step 4 as you mention the coordinates of any point.

☛ Also Check: You can try this midpoint calculator to verify the result obtained for the midpoint of a line segment- Midpoint Calculator

What is Meant by Midpoint?

A midpoint is defined as the point that is in the middle of the line joining two points. It is the point that is equidistant from both endpoints, thus bisecting the line segment.

What is the Midpoint Formula in Words?

For the midpoint of the line joining two points, whose coordinates are given, the midpoint formula in words can be described as the average of the x-coordinates of the two points and the average of the y-coordinates of the two points.

Can Midpoint be a Fraction?

Yes, the midpoint value can also be a fraction . It is basically dependent on the numeric value of the two points. The midpoint is the sum of the numeric value of two points, divided by 2. For points such as -4 and 5 on the number line, the midpoint is +1/2.

Why is the Midpoint Formula Important?

Midpoint formula has varied applications in real life, such as constructions purpose, etc. It has importance in geometry, such as

• Finding the coordinates of the centroid of a triangle.
• Finding the median of a triangle .
• Finding the midpoint of a line segment .

How do you Calculate Midpoint?

The midpoint can be found with the formula ((x 1 + x 2 )/2, (y 1 + y 2 )/2). Here (x 1 , y 1 ), and (x 2 , y 2 ) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

Can the Midpoint be Zero?

The midpoint can be zero. This is dependent on the value of the two points. For two points on a number line with values -4, and 4, the midpoint is 0; And for two points such as (-2, 5), and (2, -5), the midpoint is equal to (0, 0).

What is the Midpoint of a Curve?

The midpoint of a curve is the midpoint of the largest chord which can be drawn for the curve. The midpoint of a circle is the midpoint of its largest chord, which is the diameter of the circle.

What is the Midpoint of a Line?

The midpoint of a line is a point that is equidistant from the endpoints of the line and in the middle of the line. If the endpoints of the line are (x 1 , y 1 ), and (x 2 , y 2 ), then the formula for the midpoint of the line is ((x 1 + x 2 )/2, (y 1 + y 2 )/2).

What is the Midpoint of a Circle?

The midpoint of a circle is the center of the circle. The largest chord of the circle is the diameter, and the midpoint of the diameter of the circle is the midpoint of the circle. The midpoint of the circle is equidistant from every point on the circle.

What is the Midpoint of a Triangle?

The midpoint of the triangle is the centroid of the triangle . The centroid is the point of intersection of the medians of a triangle. The center of gravity of any triangular-shaped object is at its centroid.

Choose Your Test

Sat / act prep online guides and tips, how to use the midpoint formula.

General Education

The midpoint formula lets you find the exact center between two defined points. You might encounter this formula in your economics or geometry class or while prepping for a college entrance exam like the SAT or ACT.

In this article, we’ll answer the questions what is the midpoint formula and when do you use the midpoint formula, as well as provide example problems you can try.

What Is The Midpoint Formula in Geometry?

You can probably guess what the midpoint formula does based on its name: the midpoint formula helps you find the exact halfway between two defined points. That halfway mark is the midpoint.

Here’s the actual midpoint formula:

The midpoint N of the line segment A (X 1 , Y 1 ) to B (X 2 , Y 2 ) can be found with the formula:

$$({X_1 + X_2}/2 , {Y_1+Y_2}/2)$$

Let’s take a look at that in practice.

First, you’ll want to find point A . The values for point A are X 1 = -3 and Y 1 = 2.

Next, you’ll want to find point B . The values for point B are X 2 = 4 and Y 2 = 4.

Now that we have those values, we plug them into our equation:

$$({\-3 + 4}/2, {\2 + 4}/2)$$

$$(1/2, 6/2) = (1/2, 3)$$

The exact midpoint of the line segment AB is (1/2, 3).

Midpoint Formula Geometry Examples

Here are several examples of midpoint formula geometry problems.

M = $({\4+ 2}/{\2} , {\5 + 1}/{\2})$

= $(6/2 , 6/2)$

M = $({\5+ -2}/{\2}, {\4 + 1}/{\2})$

= $(3/2, 5/2)$

CD has endpoints at C (9, 1) and D (7, 9). Find the midpoint M of CD .

M = $({\9+ 7}/{\2}, {\9 + 1}/{\2})$

= $(16/2, 10/2)$

=  (8, 5)

Final Thoughts

The midpoint formula allows you to find the exact midpoint of a line segment. The midpoint formula is $({X_1 + X_2}/2 , {Y_1+Y_2}/2)$.

What's Next?

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The Midpoint Formula

Formula Examples

Sometimes you need to find the point that is exactly midway between two other points. For instance, you might need to find a line that bisects (divides into two equal halves) a given line segment. This middle point is called the "midpoint". The concept doesn't come up often, but the Formula is quite simple and obvious, so you should easily be able to remember it for later.

Think about it this way: If you are given two numbers, you can find the number exactly between them by averaging them, by adding them together and dividing by two. For example, the number exactly halfway between 5 and 10 is:

[5 + 10] / 2 = 15 / 2 = 7.5 .

Content Continues Below

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Midpoint Formula

The Midpoint Formula works exactly the same way. If you need to find the point that is exactly halfway between two given points, just average the x -values and the y -values.

Find the midpoint P between (–1, 2) and (3, –6) .

First, I apply the Midpoint Formula; then, I'll simplify:

So the answer is P = (1, –2) .

But as long as you remember that you're averaging the two points' x - and y -values, you'll do fine. It won't matter which point you pick to be the "first" point you plug in. Just make sure that you're adding an x to an x , and a y to a y .

Find the midpoint P between (6.4, 3) and (–10.7, 4) .

I'll apply the Midpoint Formula, and simplify:

So the answer is P = (–2.15, 3.5) .

Find the value of p so that (–2, 2.5) is the midpoint between ( p , 2) and (–1, 3) .

I'll apply the Midpoint Formula:

The y -coordinates already match. This reduces the problem to needing to compare the x -coordinates, "equating" them (that is, setting them equal, because they must be the same) and solving the resulting equation to figure out what p is. This will give me the value necessary for making the x -values match. So:

So the answer is p = –3 .

Let's do some more examples ....

URL: http://www.purplemath.com/modules/midpoint.htm

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11.1 Distance and Midpoint Formulas; Circles

Learning objectives.

Use the Distance Formula

Use the Midpoint Formula

• Write the equation of a circle in standard form
• Graph a circle

Be Prepared 11.1

Before you get started, take this readiness quiz.

• Find the length of the hypotenuse of a right triangle whose legs are 12 and 16 inches. If you missed this problem, review Example 2.34 .
• Factor: x 2 − 18 x + 81 . x 2 − 18 x + 81 . If you missed this problem, review Example 6.24 .
• Solve by completing the square: x 2 − 12 x − 12 = 0 . x 2 − 12 x − 12 = 0 . If you missed this problem, review Example 9.22 .

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions . We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 4 ) ( 6 , 4 ) and ( 2 , 1 ) . ( 2 , 1 ) .

Try It 11.1

Use the rectangular coordinate system to find the distance between the points ( 6 , 1 ) ( 6 , 1 ) and ( 2 , −2 ) . ( 2 , −2 ) .

Try It 11.2

Use the rectangular coordinate system to find the distance between the points ( 5 , 3 ) ( 5 , 3 ) and ( −3 , −3 ) . ( −3 , −3 ) .

The method we used in the last example leads us to the formula to find the distance between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

When we found the length of the horizontal leg we subtracted 6 − 2 6 − 2 which is x 2 − x 1 . x 2 − x 1 .

When we found the length of the vertical leg we subtracted 4 − 1 4 − 1 which is y 2 − y 1 . y 2 − y 1 .

If the triangle had been in a different position, we may have subtracted x 1 − x 2 x 1 − x 2 or y 1 − y 2 . y 1 − y 2 . The expressions x 2 − x 1 x 2 − x 1 and x 1 − x 2 x 1 − x 2 vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | . | y 2 − y 1 | .

In the Pythagorean Theorem, we substitute the general expressions | x 2 − x 1 | | x 2 − x 1 | and | y 2 − y 1 | | y 2 − y 1 | rather than the numbers.

a 2 + b 2 = c 2 Substitute in the values. ( | x 2 − x 1 | ) 2 + ( | y 2 − y 1 | ) 2 = d 2 Squaring the expressions makes them positive, so we eliminate the absolute value bars. ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = d 2 Use the Square Root Property. d = ± ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Distance is positive, so eliminate the negative value. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 a 2 + b 2 = c 2 Substitute in the values. ( | x 2 − x 1 | ) 2 + ( | y 2 − y 1 | ) 2 = d 2 Squaring the expressions makes them positive, so we eliminate the absolute value bars. ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = d 2 Use the Square Root Property. d = ± ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Distance is positive, so eliminate the negative value. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2

This is the Distance Formula we use to find the distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) .

Distance Formula

The distance d between the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

Example 11.2

Use the Distance Formula to find the distance between the points ( −5 , −3 ) ( −5 , −3 ) and ( 7 , 2 ) . ( 7 , 2 ) .

Write the Distance Formula. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Label the points, ( −5 , −3 x 1 , y 1 ) , ( 7 , 2 x 2 , y 2 ) and substitute. d = ( 7 − ( −5 ) ) 2 + ( 2 − ( −3 ) ) 2 Simplify. d = 12 2 + 5 2 d = 144 + 25 d = 169 d = 13 Write the Distance Formula. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Label the points, ( −5 , −3 x 1 , y 1 ) , ( 7 , 2 x 2 , y 2 ) and substitute. d = ( 7 − ( −5 ) ) 2 + ( 2 − ( −3 ) ) 2 Simplify. d = 12 2 + 5 2 d = 144 + 25 d = 169 d = 13

Try It 11.3

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) .

Try It 11.4

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −14 , −10 ) . ( −14 , −10 ) .

Example 11.3

Use the Distance Formula to find the distance between the points ( 10 , −4 ) ( 10 , −4 ) and ( −1 , 5 ) . ( −1 , 5 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Write the Distance Formula. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Label the points, ( 10 , −4 x 1 , y 1 ) , ( −1 , 5 x 2 , y 2 ) and substitute. d = ( −1 − 10 ) 2 + ( 5 − ( −4 ) ) 2 Simplify. d = ( −11 ) 2 + 9 2 d = 121 + 81 d = 202 Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation. d = 202 or d ≈ 14.2 Write the Distance Formula. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Label the points, ( 10 , −4 x 1 , y 1 ) , ( −1 , 5 x 2 , y 2 ) and substitute. d = ( −1 − 10 ) 2 + ( 5 − ( −4 ) ) 2 Simplify. d = ( −11 ) 2 + 9 2 d = 121 + 81 d = 202 Since 202 is not a perfect square, we can leave the answer in exact form or find a decimal approximation. d = 202 or d ≈ 14.2

Try It 11.5

Use the Distance Formula to find the distance between the points ( −4 , −5 ) ( −4 , −5 ) and ( 3 , 4 ) . ( 3 , 4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Try It 11.6

Use the Distance Formula to find the distance between the points ( −2 , −5 ) ( −2 , −5 ) and ( −3 , −4 ) . ( −3 , −4 ) . Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) ( x 2 , y 2 ) is

To find the midpoint of a line segment, we find the average of the x -coordinates and the average of the y -coordinates of the endpoints.

Example 11.4

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −5 , −4 ) ( −5 , −4 ) and ( 7 , 2 ) . ( 7 , 2 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.7

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −3 , −5 ) ( −3 , −5 ) and ( 5 , 7 ) . ( 5 , 7 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Try It 11.8

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are ( −2 , −5 ) ( −2 , −5 ) and ( 6 , −1 ) . ( 6 , −1 ) . Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) . ( x 2 , y 2 ) . It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , ( h , k ) , ( h , k ) , and the fixed distance is called the radius , r , of the circle.

This is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r .

Standard Form of the Equation a Circle

The standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r , is

Example 11.5

Write the standard form of the equation of the circle with radius 3 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.9

Write the standard form of the equation of the circle with a radius of 6 and center ( 0 , 0 ) . ( 0 , 0 ) .

Try It 11.10

Write the standard form of the equation of the circle with a radius of 8 and center ( 0 , 0 ) . ( 0 , 0 ) .

In the last example, the center was ( 0 , 0 ) . ( 0 , 0 ) . Notice what happened to the equation. Whenever the center is ( 0 , 0 ) , ( 0 , 0 ) , the standard form becomes x 2 + y 2 = r 2 . x 2 + y 2 = r 2 .

Example 11.6

Write the standard form of the equation of the circle with radius 2 and center ( −1 , 3 ) . ( −1 , 3 ) .

Try It 11.11

Write the standard form of the equation of the circle with a radius of 7 and center ( 2 , −4 ) . ( 2 , −4 ) .

Try It 11.12

Write the standard form of the equation of the circle with a radius of 9 and center ( −3 , −5 ) . ( −3 , −5 ) .

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example 11.7

Write the standard form of the equation of the circle with center ( 2 , 4 ) ( 2 , 4 ) that also contains the point ( −2 , 1 ) . ( −2 , 1 ) .

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center ( 2 , 4 ) ( 2 , 4 ) and point ( −2 , 1 ) ( −2 , 1 )

Use the Distance Formula to find the radius. r = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Substitute the values. ( 2 , 4 x 1 , y 1 ) , ( −2 , 1 x 2 , y 2 ) r = ( −2 − 2 ) 2 + ( 1 − 4 ) 2 Simplify. r = ( −4 ) 2 + ( −3 ) 2 r = 16 + 9 r = 25 r = 5 Use the Distance Formula to find the radius. r = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Substitute the values. ( 2 , 4 x 1 , y 1 ) , ( −2 , 1 x 2 , y 2 ) r = ( −2 − 2 ) 2 + ( 1 − 4 ) 2 Simplify. r = ( −4 ) 2 + ( −3 ) 2 r = 16 + 9 r = 25 r = 5

Now that we know the radius, r = 5 , r = 5 , and the center, ( 2 , 4 ) , ( 2 , 4 ) , we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle. ( x − h ) 2 + ( y − k ) 2 = r 2 Substitute in the values. ( x − 2 ) 2 + ( y − 4 ) 2 = 5 2 Simplify. ( x − 2 ) 2 + ( y − 4 ) 2 = 25 Use the standard form of the equation of a circle. ( x − h ) 2 + ( y − k ) 2 = r 2 Substitute in the values. ( x − 2 ) 2 + ( y − 4 ) 2 = 5 2 Simplify. ( x − 2 ) 2 + ( y − 4 ) 2 = 25

Try It 11.13

Write the standard form of the equation of the circle with center ( 2 , 1 ) ( 2 , 1 ) that also contains the point ( −2 , −2 ) . ( −2 , −2 ) .

Try It 11.14

Write the standard form of the equation of the circle with center ( 7 , 1 ) ( 7 , 1 ) that also contains the point ( −1 , −5 ) . ( −1 , −5 ) .

Graph a Circle

Any equation of the form ( x − h ) 2 + ( y − k ) 2 = r 2 ( x − h ) 2 + ( y − k ) 2 = r 2 is the standard form of the equation of a circle with center, ( h , k ) , ( h , k ) , and radius, r. We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from x and y . In the next example, the equation has x + 2 , x + 2 , so we need to rewrite the addition as subtraction of a negative.

Example 11.8

Find the center and radius, then graph the circle: ( x + 2 ) 2 + ( y − 1 ) 2 = 9 . ( x + 2 ) 2 + ( y − 1 ) 2 = 9 .

Try It 11.15

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y + 4 ) 2 = 4 . ( x − 3 ) 2 + ( y + 4 ) 2 = 4 .

Try It 11.16

ⓐ Find the center and radius, then ⓑ graph the circle: ( x − 3 ) 2 + ( y − 1 ) 2 = 16 . ( x − 3 ) 2 + ( y − 1 ) 2 = 16 .

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of x 2 , y 2 x 2 , y 2 to be one.

Example 11.9

Find the center and radius and then graph the circle, 4 x 2 + 4 y 2 = 64 . 4 x 2 + 4 y 2 = 64 .

Try It 11.17

ⓐ Find the center and radius, then ⓑ graph the circle: 3 x 2 + 3 y 2 = 27 3 x 2 + 3 y 2 = 27

Try It 11.18

ⓐ Find the center and radius, then ⓑ graph the circle: 5 x 2 + 5 y 2 = 125 5 x 2 + 5 y 2 = 125

If we expand the equation from Example 11.8 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , ( x + 2 ) 2 + ( y − 1 ) 2 = 9 , the equation of the circle looks very different.

( x + 2 ) 2 + ( y − 1 ) 2 = 9 Square the binomials. x 2 + 4 x + 4 + y 2 − 2 y + 1 = 9 Arrange the terms in descending degree order, and get zero on the right x 2 + y 2 + 4 x − 2 y − 4 = 0 ( x + 2 ) 2 + ( y − 1 ) 2 = 9 Square the binomials. x 2 + 4 x + 4 + y 2 − 2 y + 1 = 9 Arrange the terms in descending degree order, and get zero on the right x 2 + y 2 + 4 x − 2 y − 4 = 0

This form of the equation is called the general form of the equation of the circle .

General Form of the Equation of a Circle

The general form of the equation of a circle is

If we are given an equation in general form, we can change it to standard form by completing the squares in both x and y . Then we can graph the circle using its center and radius.

Example 11.10

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 4 x − 6 y + 4 = 0 . x 2 + y 2 − 4 x − 6 y + 4 = 0 .

We need to rewrite this general form into standard form in order to find the center and radius.

Try It 11.19

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 6 x − 8 y + 9 = 0 . x 2 + y 2 − 6 x − 8 y + 9 = 0 .

Try It 11.20

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 6 x − 2 y + 1 = 0 . x 2 + y 2 + 6 x − 2 y + 1 = 0 .

In the next example, there is a y -term and a y 2 y 2 -term. But notice that there is no x -term, only an x 2 x 2 -term. We have seen this before and know that it means h is 0. We will need to complete the square for the y terms, but not for the x terms.

Example 11.11

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 + 8 y = 0 . x 2 + y 2 + 8 y = 0 .

Try It 11.21

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 2 x − 3 = 0 . x 2 + y 2 − 2 x − 3 = 0 .

Try It 11.22

ⓐ Find the center and radius, then ⓑ graph the circle: x 2 + y 2 − 12 y + 11 = 0 . x 2 + y 2 − 12 y + 11 = 0 .

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

• Distance-Midpoint Formulas and Circles
• Finding the Distance and Midpoint Between Two Points
• Completing the Square to Write Equation in Standard Form of a Circle

Section 11.1 Exercises

Practice makes perfect.

In the following exercises, find the distance between the points. Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

( 2 , 0 ) ( 2 , 0 ) and ( 5 , 4 ) ( 5 , 4 )

( −4 , −3 ) ( −4 , −3 ) and ( 2 , 5 ) ( 2 , 5 )

( −4 , −3 ) ( −4 , −3 ) and ( 8 , 2 ) ( 8 , 2 )

( −7 , −3 ) ( −7 , −3 ) and ( 8 , 5 ) ( 8 , 5 )

( −1 , 4 ) ( −1 , 4 ) and ( 2 , 0 ) ( 2 , 0 )

( −1 , 3 ) ( −1 , 3 ) and ( 5 , −5 ) ( 5 , −5 )

( 1 , −4 ) ( 1 , −4 ) and ( 6 , 8 ) ( 6 , 8 )

( −8 , −2 ) ( −8 , −2 ) and ( 7 , 6 ) ( 7 , 6 )

( −3 , −5 ) ( −3 , −5 ) and ( 0 , 1 ) ( 0 , 1 )

( −1 , −2 ) ( −1 , −2 ) and ( −3 , 4 ) ( −3 , 4 )

( 3 , −1 ) ( 3 , −1 ) and ( 1 , 7 ) ( 1 , 7 )

( −4 , −5 ) ( −4 , −5 ) and ( 7 , 4 ) ( 7 , 4 )

In the following exercises, ⓐ find the midpoint of the line segments whose endpoints are given and ⓑ plot the endpoints and the midpoint on a rectangular coordinate system.

( 0 , −5 ) ( 0 , −5 ) and ( 4 , −3 ) ( 4 , −3 )

( −2 , −6 ) ( −2 , −6 ) and ( 6 , −2 ) ( 6 , −2 )

( 3 , −1 ) ( 3 , −1 ) and ( 4 , −2 ) ( 4 , −2 )

( −3 , −3 ) ( −3 , −3 ) and ( 6 , −1 ) ( 6 , −1 )

In the following exercises, write the standard form of the equation of the circle with the given radius and center ( 0 , 0 ) . ( 0 , 0 ) .

Radius: 2 2

Radius: 5 5

In the following exercises, write the standard form of the equation of the circle with the given radius and center

Radius: 1, center: ( 3 , 5 ) ( 3 , 5 )

Radius: 10, center: ( −2 , 6 ) ( −2 , 6 )

Radius: 2.5 , 2.5 , center: ( 1.5 , −3.5 ) ( 1.5 , −3.5 )

Radius: 1.5 , 1.5 , center: ( −5.5 , −6.5 ) ( −5.5 , −6.5 )

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

Center ( 3 , −2 ) ( 3 , −2 ) with point ( 3 , 6 ) ( 3 , 6 )

Center ( 6 , −6 ) ( 6 , −6 ) with point ( 2 , −3 ) ( 2 , −3 )

Center ( 4 , 4 ) ( 4 , 4 ) with point ( 2 , 2 ) ( 2 , 2 )

Center ( −5 , 6 ) ( −5 , 6 ) with point ( −2 , 3 ) ( −2 , 3 )

In the following exercises, ⓐ find the center and radius, then ⓑ graph each circle.

( x + 5 ) 2 + ( y + 3 ) 2 = 1 ( x + 5 ) 2 + ( y + 3 ) 2 = 1

( x − 2 ) 2 + ( y − 3 ) 2 = 9 ( x − 2 ) 2 + ( y − 3 ) 2 = 9

( x − 4 ) 2 + ( y + 2 ) 2 = 16 ( x − 4 ) 2 + ( y + 2 ) 2 = 16

( x + 2 ) 2 + ( y − 5 ) 2 = 4 ( x + 2 ) 2 + ( y − 5 ) 2 = 4

x 2 + ( y + 2 ) 2 = 25 x 2 + ( y + 2 ) 2 = 25

( x − 1 ) 2 + y 2 = 36 ( x − 1 ) 2 + y 2 = 36

( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25 ( x − 1.5 ) 2 + ( y + 2.5 ) 2 = 0.25

( x − 1 ) 2 + ( y − 3 ) 2 = 9 4 ( x − 1 ) 2 + ( y − 3 ) 2 = 9 4

x 2 + y 2 = 64 x 2 + y 2 = 64

x 2 + y 2 = 49 x 2 + y 2 = 49

2 x 2 + 2 y 2 = 8 2 x 2 + 2 y 2 = 8

6 x 2 + 6 y 2 = 216 6 x 2 + 6 y 2 = 216

In the following exercises, ⓐ identify the center and radius and ⓑ graph.

x 2 + y 2 + 2 x + 6 y + 9 = 0 x 2 + y 2 + 2 x + 6 y + 9 = 0

x 2 + y 2 − 6 x − 8 y = 0 x 2 + y 2 − 6 x − 8 y = 0

x 2 + y 2 − 4 x + 10 y − 7 = 0 x 2 + y 2 − 4 x + 10 y − 7 = 0

x 2 + y 2 + 12 x − 14 y + 21 = 0 x 2 + y 2 + 12 x − 14 y + 21 = 0

x 2 + y 2 + 6 y + 5 = 0 x 2 + y 2 + 6 y + 5 = 0

x 2 + y 2 − 10 y = 0 x 2 + y 2 − 10 y = 0

x 2 + y 2 + 4 x = 0 x 2 + y 2 + 4 x = 0

x 2 + y 2 − 14 x + 13 = 0 x 2 + y 2 − 14 x + 13 = 0

Writing Exercises

Explain the relationship between the distance formula and the equation of a circle.

Is a circle a function? Explain why or why not.

In your own words, state the definition of a circle.

In your own words, explain the steps you would take to change the general form of the equation of a circle to the standard form.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Midpoint formula

The midpoint formula is a formula used to find the halfway point between two coordinates on a graph. Given a line segment with endpoints A and B, the midpoint is the point located exactly between A and B, meaning that it is the same distance from A and B, as in the figure below.

The midpoint formula can be used when two points on a graph in the coordinate plane are known. Given two points ( x 1 , y 1 ) and ( x 2 , y 2 ), their midpoint M is:

Notice that the midpoint formula involves taking an average of the x- and y-values of two coordinates. This can make it easier to remember the midpoint formula. The average of two numbers is just the sum of the two numbers divided by 2. The midpoint of a line segment in the coordinate plane is located at the vertical and horizontal halfway points between two points, which is equivalent to the average between the x -coordinates, and the average between the y -coordinates of the line segment. Remembering this should make it relatively easy to recall the midpoint formula whenever needed.

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11.2: Distance and Midpoint Formulas and Circles

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Learning Objectives

By the end of this section, you will be able to:

Use the Distance Formula

Use the midpoint formula.

• Write the equation of a circle in standard form
• Graph a circle

Before you get started, take this readiness quiz.

• Find the length of the hypotenuse of a right triangle whose legs are \(12\) and \(16\) inches. If you missed this problem, review Example 2.34.
• Factor: \(x^{2}-18 x+81\). If you missed this problem, review Example 6.24.
• Solve by completing the square: \(x^{2}-12 x-12=0\). If you missed this problem, review Example 9.22.

In this chapter we will be looking at the conic sections, usually called the conics, and their properties. The conics are curves that result from a plane intersecting a double cone—two cones placed point-to-point. Each half of a double cone is called a nappe.

There are four conics—the circle , parabola , ellipse , and hyperbola . The next figure shows how the plane intersecting the double cone results in each curve.

Each of the curves has many applications that affect your daily life, from your cell phone to acoustics and navigation systems. In this section we will look at the properties of a circle.

We have used the Pythagorean Theorem to find the lengths of the sides of a right triangle. Here we will use this theorem again to find distances on the rectangular coordinate system. By finding distance on the rectangular coordinate system, we can make a connection between the geometry of a conic and algebra—which opens up a world of opportunities for application.

Our first step is to develop a formula to find distances between points on the rectangular coordinate system. We will plot the points and create a right triangle much as we did when we found slope in Graphs and Functions. We then take it one step further and use the Pythagorean Theorem to find the length of the hypotenuse of the triangle—which is the distance between the points.

Example \(\PageIndex{1}\)

Use the rectangular coordinate system to find the distance between the points \((6,4)\) and \((2,1)\).

Exercise \(\PageIndex{1}\)

Use the rectangular coordinate system to find the distance between the points \((6,1)\) and \((2,-2)\).

Exercise \(\PageIndex{2}\)

Use the rectangular coordinate system to find the distance between the points \((5,3)\) and \((-3,-3)\).

The method we used in the last example leads us to the formula to find the distance between the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\).

When we found the length of the horizontal leg we subtracted \(6−2\) which is \(x_{2}-x_{1}\).

When we found the length of the vertical leg we subtracted \(4−1\) which is \(y_{2}-y_{1}\).

If the triangle had been in a different position, we may have subtracted \(x_{1}-x_{2}\) or \(y_{1}-y_{2}\). The expressions \(x_{2}-x_{1}\) and \(x_{1}-x_{2}\) vary only in the sign of the resulting number. To get the positive value-since distance is positive- we can use absolute value. So to generalize we will say \(\left|x_{2}-x_{1}\right|\) and \(\left|y_{2}-y_{1}\right|\).

In the Pythagorean Theorem, we substitute the general expressions \(\left|x_{2}-x_{1}\right|\) and \(\left|y_{2}-y_{1}\right|\) rather than the numbers.

\(\begin{array}{l c}{} & {a^{2}+b^{2}=c^{2}} \\ {\text {Substitute in the values. }}&{(|x_{2}-x_{1}|)^{2}+(|y_{2}-y_{1}|)^{2}=d^{2}} \\ {\text{Squaring the expressions makes}}&{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=d^{2}} \\ \text{them positive, so we eliminate} \\\text{the absolute value bars.}\\ {\text{Use the Square Root Property.}}&{d=\pm\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\ {\text{Distance is positive, so eliminate}}&{d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\\\text{the negative value.}\end{array}\)

This is the Distance Formula we use to find the distance \(d\) between the two points \((x_{1},y_{1})\) and \((x_{2}, y_{2})\).

Definition \(\PageIndex{1}\)

Distance Formula

The distance \(d\) between the two points \((x_{1},y_{1})\) and \((x_{2}, y_{2})\) is

\(d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Example \(\PageIndex{2}\)

Use the Distance Formula to find the distance between the points \((-5,-3)\) and \((7,2)\).

Write the Distance Formula.

Label the points, \(\left( \begin{array}{c}{x_{1}, y_{1}} \\ {-5,-3}\end{array}\right)\), \(\left( \begin{array}{l}{x_{2}, y_{2}} \\ {7,2}\end{array}\right)\) and substitute.

\(d=\sqrt{(7-(-5))^{2}+(2-(-3))^{2}}\)

\(d=\sqrt{12^{2}+5^{2}}\) \(d=\sqrt{144+25}\) \(d=\sqrt{169}\) \(d=13\)

Exercise \(\PageIndex{3}\)

Use the Distance Formula to find the distance between the points \((-4,-5)\) and \((5,7)\).

Exercise \(\PageIndex{4}\)

Use the Distance Formula to find the distance between the points \((-2,-5)\) and \((-14,-10)\).

Example \(\PageIndex{3}\)

Use the Distance Formula to find the distance between the points \((10,−4)\) and \((−1,5)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

Label the points, \(\left( \begin{array}{c}{x_{1}, y_{1}} \\ {10,-4}\end{array}\right)\), \(\left( \begin{array}{c}{x_{2}, y_{2}} \\ {-1,5}\end{array}\right)\) and substitute.

\(d=\sqrt{(-1-10)^{2}+(5-(-4))^{2}}\)

\(d=\sqrt{(-11)^{2}+9^{2}}\) \(d=\sqrt{121+81}\) \(d=\sqrt{202}\)

Since \(202\) is not a perfect square, we can leave the answer in exact form or find a decimal approximation.

\(d=\sqrt{202}\) or \(d \approx 14.2\)

Exercise \(\PageIndex{5}\)

Use the Distance Formula to find the distance between the points \((−4,−5)\) and \((3,4)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

\(d=\sqrt{130}, d \approx 11.4\)

Exercise \(\PageIndex{6}\)

Use the Distance Formula to find the distance between the points \((−2,−5)\) and \((−3,−4)\). Write the answer in exact form and then find the decimal approximation, rounded to the nearest tenth if needed.

\(d=\sqrt{2}, d \approx 1.4\)

It is often useful to be able to find the midpoint of a segment. For example, if you have the endpoints of the diameter of a circle, you may want to find the center of the circle which is the midpoint of the diameter. To find the midpoint of a line segment, we find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates of the endpoints.

Definition \(\PageIndex{2}\)

Midpoint Formula

The midpoint of the line segment whose endpoints are the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is

\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

To find the midpoint of a line segment, we find the average of the \(x\)-coordinates and the average of the \(y\)-coordinates of the endpoints.

Example \(\PageIndex{4}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−5,−4)\) and \((7,2)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Exercise \(\PageIndex{7}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−3,−5)\) and \((5,7)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Exercise \(\PageIndex{8}\)

Use the Midpoint Formula to find the midpoint of the line segments whose endpoints are \((−2,−5)\) and \((6,−1)\). Plot the endpoints and the midpoint on a rectangular coordinate system.

Both the Distance Formula and the Midpoint Formula depend on two points, \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\). It is easy to confuse which formula requires addition and which subtraction of the coordinates. If we remember where the formulas come from, is may be easier to remember the formulas.

Write the Equation of a Circle in Standard Form

As we mentioned, our goal is to connect the geometry of a conic with algebra. By using the coordinate plane, we are able to do this easily.

We define a circle as all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, \((h,k)\), and the fixed distance is called the radius , \(r\), of the circle.

Definition \(\PageIndex{3}\)

A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center , \((h,k)\), and the fixed distance is called the radius , \(r\), of the circle.

This is the standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\).

Definition \(\PageIndex{4}\)

The standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\), is

Example \(\PageIndex{5}\)

Write the standard form of the equation of the circle with radius \(3\) and center \((0,0)\).

Exercise \(\PageIndex{9}\)

Write the standard form of the equation of the circle with a radius of \(6\) and center \((0,0)\).

\(x^{2}+y^{2}=36\)

Exercise \(\PageIndex{10}\)

Write the standard form of the equation of the circle with a radius of \(8\) and center \((0,0)\).

\(x^{2}+y^{2}=64\)

In the last example, the center was \((0,0)\). Notice what happened to the equation. Whenever the center is \((0,0)\), the standard form becomes \(x^{2}+y^{2}=r^{2}\).

Example \(\PageIndex{6}\)

Write the standard form of the equation of the circle with radius \(2\) and center \((−1,3)\).

Exercise \(\PageIndex{11}\)

Write the standard form of the equation of the circle with a radius of \(7\) and center \((2,−4)\).

\((x-2)^{2}+(y+4)^{2}=49\)

Exercise \(\PageIndex{12}\)

Write the standard form of the equation of the circle with a radius of \(9\) and center \((−3,−5)\).

\((x+3)^{2}+(y+5)^{2}=81\)

In the next example, the radius is not given. To calculate the radius, we use the Distance Formula with the two given points.

Example \(\PageIndex{7}\)

Write the standard form of the equation of the circle with center \((2,4)\) that also contains the point \((−2,1)\).

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it. We will use the center \((2,4)\) and point \((−2,1)\)

Use the Distance Formula to find the radius.

\(r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

Substitute the values. \(\left( \begin{array}{l}{x_{1}, y_{1}} \\ {2,4}\end{array}\right), \left( \begin{array}{c}{x_{2}, y_{2}} \\ {-2,1}\end{array}\right)\)

\(r=\sqrt{(-2-2)^{2}+(1-4)^{2}}\)

\(r=\sqrt{(-4)^{2}+(-3)^{2}}\) \(r=\sqrt{16+9}\) \(r=\sqrt{25}\) \(r=5\)

Now that we know the radius, \(r=5\), and the center, \((2,4)\), we can use the standard form of the equation of a circle to find the equation.

Use the standard form of the equation of a circle.

\((x-h)^{2}+(y-k)^{2}=r^{2}\)

Substitute in the values.

\((x-2)^{2}+(y-4)^{2}=5^{2}\)

\((x-2)^{2}+(y-4)^{2}=25\)

Exercise \(\PageIndex{13}\)

Write the standard form of the equation of the circle with center \((2,1)\) that also contains the point \((−2,−2)\).

\((x-2)^{2}+(y-1)^{2}=25\)

Exercise \(\PageIndex{14}\)

Write the standard form of the equation of the circle with center \((7,1)\) that also contains the point \((−1,−5)\).

\((x-7)^{2}+(y-1)^{2}=100\)

Graph a Circle

Any equation of the form \((x-h)^{2}+(y-k)^{2}=r^{2}\) is the standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\) . We can then graph the circle on a rectangular coordinate system.

Note that the standard form calls for subtraction from \(x\) and \(y\). In the next example, the equation has \(x+2\), so we need to rewrite the addition as subtraction of a negative.

Example \(\PageIndex{8}\)

Find the center and radius, then graph the circle: \((x+2)^{2}+(y-1)^{2}=9\).

Exercise \(\PageIndex{15}\)

• Find the center and radius, then
• Graph the circle: \((x-3)^{2}+(y+4)^{2}=4\).
• The circle is centered at \((3,-4)\) with a radius of \(2\).

Exercise \(\PageIndex{16}\)

• Graph the circle: \((x-3)^{2}+(y-1)^{2}=16\).
• The circle is centered at \((3,1)\) with a radius of \(4\).

To find the center and radius, we must write the equation in standard form. In the next example, we must first get the coefficient of \(x^{2}, y^{2}\) to be one.

Example \(\PageIndex{9}\)

Find the center and radius and then graph the circle, \(4 x^{2}+4 y^{2}=64\).

Exercise \(\PageIndex{17}\)

• Graph the circle: \(3 x^{2}+3 y^{2}=27\)
• The circle is centered at \((0,0)\) with a radius of \(3\).

Exercise \(\PageIndex{18}\)

• Graph the circle: \(5 x^{2}+5 y^{2}=125\)
• The circle is centered at \((0,0)\) with a radius of \(5\).

If we expand the equation from Example 11.1.8, \((x+2)^{2}+(y-1)^{2}=9\), the equation of the circle looks very different.

\((x+2)^{2}+(y-1)^{2}=9\)

Square the binomials.

\(x^{2}+4 x+4+y^{2}-2 y+1=9\)

Arrange the terms in descending degree order, and get zero on the right

\(x^{2}+y^{2}+4 x-2 y-4=0\)

This form of the equation is called the general form of the equation of the circle .

Definition \(\PageIndex{5}\)

The general form of the equation of a circle is

\(x^{2}+y^{2}+a x+b y+c=0\)

If we are given an equation in general form, we can change it to standard form by completing the squares in both \(x\) and \(y\). Then we can graph the circle using its center and radius.

Example \(\PageIndex{10}\)

• Graph the circle: \(x^{2}+y^{2}-4 x-6 y+4=0\)

We need to rewrite this general form into standard form in order to find the center and radius.

Exercise \(\PageIndex{19}\)

• Graph the circle: \(x^{2}+y^{2}-6 x-8 y+9=0\).
• The circle is centered at \((3,4)\) with a radius of \(4\).

Exercise \(\PageIndex{20}\)

• Graph the circle: \(x^{2}+y^{2}+6 x-2 y+1=0\)
• The circle is centered at \((-3,1)\) with a radius of \(3\).

In the next example, there is a \(y\)-term and a \(y^{2}\)-term. But notice that there is no \(x\)-term, only an \(x^{2}\)-term. We have seen this before and know that it means \(h\) is \(0\). We will need to complete the square for the \(y\) terms, but not for the \(x\) terms.

Example \(\PageIndex{11}\)

• Graph the circle: \(x^{2}+y^{2}+8 y=0\)

Exercise \(\PageIndex{21}\)

• Graph the circle: \(x^{2}+y^{2}-2 x-3=0\).
• The circle is centered at \((-1,0)\) with a radius of \(2\).

Exercise \(\PageIndex{22}\)

• Graph the circle: \(x^{2}+y^{2}-12 y+11=0\).
• The circle is centered at \((0,6)\) with a radius of \(5\).

Access these online resources for additional instructions and practice with using the distance and midpoint formulas, and graphing circles.

• Distance-Midpoint Formulas and Circles
• Finding the Distance and Midpoint Between Two Points
• Completing the Square to Write Equation in Standard Form of a Circle

Key Concepts

• Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, \((h,k)\), and the fixed distance is called the radius, \(r\), of the circle.
• Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, \((h,k)\), and radius, \(r\) , is