assignment 5 distance formula

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Distance formula

Here you will learn about the distance formula, including how to find the distance between two coordinates.

Students first learn about the distance formula in 8th grade as a part of geometry, and again in high school geometry as a part of expressing geometric properties with equations.

What is the distance formula?

The distance formula (also known as the Euclidean distance formula ) is an application of the Pythagorean theorem a^2+b^2=c^2 in coordinate geometry.

It will calculate the distance between two cartesian coordinates on a two-dimensional plane , or coordinate plane .

To do this, find the differences between the x- coordinates and the difference between the y- coordinates , square them, then find the square root of the answer.

This can be written as the distance formula,

where d is the distance between the points \left(x_1, y_1\right) and \left(x_2, y_2\right).

For example,

The line segment between the first point and the second point forms the hypotenuse of a right angled triangle.

The length of the hypotenuse of the right triangle is the distance between the two end points of the line segment.

Common Core State Standards

How does this relate to 8 th grade math?

• Grade 8 – Geometry (8.G.B.8) Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

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How to use the distance formula

In order to use the distance formula, you need to:

Identify the two points and label them \bf{\left(\textbf{x}_{1}, \textbf{y}_{1}\right)} and \bf{\left(\textbf{x}_{2}, \textbf{y}_{2}\right)}.

Substitute the values into the formula, \bf{\textbf{d}=\sqrt{\left(\textbf{x}_2-\textbf{x}_1\right)^2+\left(\textbf{y}_2-\textbf{y}_1\right)^2}}.

Solve the equation.

Distance formula examples

Example 1: distance between two points on a coordinate axes in the first quadrant.

Find the distance between the points A and B.

A=(3,1) and B=(6,5).

Let \left(x_{1}, y_{1}\right)=(3,1) and \left(x_{2}, y_{2}\right)=(6,5).

2 Substitute the values into the formula, \bf{\textbf{d}=\sqrt{\left(\textbf{x}_2-\textbf{x}_1\right)^2+\left(\textbf{y}_2-\textbf{y}_1\right)^2}}.

3 Solve the equation.

Example 2: find the distance between two points on a coordinate axes

Give your answer to 1 decimal place.

A=(1,-5) and B=(6,2).

Let \left(x_{1}, y_{1}\right)=(1,-5) and \left(x_{2}, y_{2}\right)=(6,2).

Example 3: find the distance between two given points with positive coordinates

Find the distance between the points (1,4) and (7,12).

Let \left(x_{1}, y_{1}\right)=(1,4) and \left(x_{2}, y_{2}\right)=(7,12).

Example 4: find the distance between any two given points

Find the distance between the points (-2,5) and (6,-7).

Let \left(x_{1}, y_{1}\right)=(-2,5) and \left(x_{2}, y_{2}\right)=(6,-7).

Example 5: finding a missing value given the distance

The distance between the points (1,5) and (16,k) is 17.

Find the value of k, where k is negative.

Let \left(x_{1}, y_{1}\right)=(1,5) and \left(x_{2}, y_{2}\right)=(16,k).

It is also stated that d=17.

The question states that k is negative so k=-3.

Example 6: finding a missing value given the distance

The distance between the points (2,9) and (f,10) is 15.

Find the value of f, where f is positive.

Let \left(x_{1}, y_{1}\right)=(2,9) and \left(x_{2}, y_{2}\right)=(f, 10).

It is also stated that d=15.

The question states that f is positive so f=17.0 (1dp).

Teaching tips for distance formula

• Use visual aids such as coordinate planes that highlight the x- axis and y- axis, graphs, or geometric shapes to visually represent the distance formula.
• Introduce real-world scenarios where distance calculations are essential. For example, discuss scenarios involving mapping or measuring distances between points in various contexts to allow students to see the relevance of the concept.
• Allow students to explore the distance formula through hands-on activities such as measuring distances on a coordinate plane or calculating distances between objects in the classroom.

Easy mistakes to make

• Confusing the distance formula with the midpoint formula An easy mistake to make is to find the midpoint instead of the distance. The midpoint formula is \left(\cfrac{x_{1}+x_{2}}{2}, \cfrac{y_{1}+y_{2}}{2}\right).
• Squaring negative numbers to give a negative When using the distance formula, it is common to get negative values after the subtraction step. These values will be squared, so it is important to remember that the square of a negative value is positive. For example, (-3)^2=9.
• Subtracting in the wrong order When subtracting, a common misconception is to switch the order of subtraction when plugging in the coordinates ( for example, using \left(x_{2}-x_{1}\right) and \left(y_{1}-y_{2}\right). Ensure that the subtraction is done in the same order for both coordinate values: \left(x_{2}-x_{1}\right)  and \left(y_{2}-y_{1}\right).
• Forgetting to simplify When solving, neglecting to simplify the expression inside the square root is a common mistake. After squaring each term, simplify the expression inside the square root before taking the square root.

Related graphing linear equation lessons

• Graphing linear equations
• Slope intercept form
• How to find midpoint
• How to find the y intercept
• How to find the slope of a line
• Linear interpolation

Practice distance formula questions

1. Find the distance between the point (6,8) and the origin.

The origin is (0,0) so let (x_{1},y_{1})=(0,0) and (x_{2},y_{2})=(6,8).

2. Find the distance between the points (0,10) and (24,0).

Let \left(x_{1},y_{1}\right)=(0,10) and \left(x_{2},y_{2}\right)=(24,0).

3. Find the distance between the points (5,3) and (14,10).

Let \left(x_{1},y_{1}\right)=(5,3) and \left(x_{2},y_{2}\right)=(14,10).

4. Find the distance between the points (-2,4) and (-8,-9).

Let \left(x_{1},y_{1}\right)=(-2,4) and \left(x_{2},y_{2}\right)=(-8,-9).

5. The distance between the points (8,-3) and (15,a) is 25.

Find the value of a, where a is positive.

As a is positive, a=21.

6. The distance between the points (b,4) and (6,-8) is 15.

Find the value of b, where b is negative.

As b is negative, b=-3.

Distance formula FAQs

The distance formula calculates the distance between two points by treating the vertical and horizontal distances as sides of a right triangle, and then finding the length of the line (hypotenuse of a right triangle) using the Pythagorean Theorem.

The theorem is named after the ancient Greek mathematician, Pythagoras, and describes the relationships between the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two other sides.

Yes, the distance formula can be extended to three dimensions. To find the distance between the two points \left(x_{1},y_{1},z_{1}\right) and \left(x_{2},y_{2},z_{2}\right) and using the following formula: d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}

The next lessons are

• Angles in parallel lines
• Angles in polygons
• Rate of change
• Systems of equations
• Number patterns

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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

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The Distance Formula

Formula Examples

The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. The Pythagorean Theorem allows you to relate the three sides of a right triangle; in particular, it allows you to find the length of the third side of a right triangle, given the lengths of the other two sides. The Distance Formula takes two points and implicitly assigns them the role of the hypotenuse.

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Distance Formula

Here's how we get from the one to the other:

Suppose you're given the two points (−2, 1) and (1, 5) , and they want you to find out how far apart they are. The points look like this:

You can draw in the lines, parallel to the two coordinate axes, that form a right-angled triangle, using these points as two of the corners:

It's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x -values and the y -values:

Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle):

c 2 = a 2 + b 2

c 2 = (5 − 1) 2 + (1 − (−2) 2

This format always holds true. Given two points, you can always plot them, create the right triangle, and then find the length of the hypotenuse. The length of the hypotenuse is the distance between the two points. And since this format always works, it can be turned into a formula:

The Distance Formula: Given the two points ( x 1 , y 1 ) and ( x 2 , y 2 ) , the distance d between these points is given by the formula:

Don't let the subscripts scare you, by the way. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The distance between the two points will be the same, regardless of the order in which you used them.

Find the distance between the points (−2, −3) and (−4, 4) .

To find my answer, I just plug the points' coordinates into the Distance Formula:

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Distance Formula Exercises

Distance formula practice problems with answers.

Here are ten (10) practice exercises about the distance formula. As you engage with these problems, my hope is that you gain a deeper understanding of how to apply the distance formula. Good luck!

Problem 1: How far is the point [latex]\left( { – 4,6} \right)[/latex] from the origin?

[latex]\color{black}2\sqrt {13} [/latex] units

Problem 2: Find the distance between the points [latex]\left( {4,7} \right)[/latex] and [latex]\left( {1, – 6} \right)[/latex]. Round your answer to the nearest hundredth.

[latex]\color{black}13.34[/latex] units

Problem 3: Find the distance between the points on the XY-plane. Round your answer to one decimal place.

[latex]\color{black}8.6[/latex] units

Problem 4: Determine the distance between points on the coordinate plane. Round your answer to two decimal places.

[latex]\color{black}8.06[/latex] units

Problem 5: The chord of a circle has endpoints as shown below (in green dots). What is the length of the chord?

[latex]\color{black}4\sqrt 5[/latex] units

Problem 6: The diameter of a circle has endpoints as shown below (in red dots). What is the length of the diameter?

[latex]\color{black}\sqrt 13[/latex] units

Problem 7: Find the two points on the x-axis that are [latex]15[/latex] units away from the point [latex]\left( {-2, 9} \right)[/latex].

[latex]\left( {10,0} \right)[/latex] and [latex]\left( { – 14,0} \right)[/latex]

Problem 8: Find the two points found on the y-axis which are [latex]25[/latex] units from the point [latex]\left( {7, -5} \right)[/latex].

[latex]\left( {0,19} \right)[/latex] and [latex]\left( {0, – 29} \right)[/latex]

Problem 9: Find the values of [latex]\color{red}k[/latex] such that the points [latex]\left( {{\color{red}k}, – 1} \right)[/latex] and [latex]\left( {5, 4} \right)[/latex] have a distance of [latex]{13}[/latex] units.

[latex]{k_1} = – 7[/latex] and [latex]{k_2} = 17[/latex]

Problem 10: Find the values of [latex]\color{red}m[/latex] such that the points [latex]\left( {{\color{red}m},3} \right)[/latex] and [latex]\left( {1,{\color{red}m}} \right)[/latex] are [latex]10[/latex] units apart.

[latex]{m_1} = – 5[/latex] and [latex]{m_2} = 9[/latex]

You might also like these tutorials:

• Distance Formula
• Distance between Point and Line Formula

Applications of the Distance Formula

Length between two points based on a right triangle..

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Distance Formula

The distance formula, in coordinate geometry or Euclidean geometry, is used to find the distance between the two points in an XY plane. The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y). To find the distance between any two points in a plane, we make the use of Pythagoras theorem here. This is an important topic for Class 10 students. Let us discuss more here, the formula, and its derivation with examples.

Distance Formula in Maths

In maths, the distance formula is defined for coordinate geometry. Students should not get confused here about the formula for speed and distance. Both are completely two different topics. When we speak about speed and distance, it is covered under real-life scenarios when a person or a vehicle moves at a speed to cover a distance at a particular interval of time. Then the speed is equal to the ratio of distance covered and time taken.

In case of coordinate geometry, when a person moves from one point to another point in a plane, then the distance between the start point and endpoint is calculated using the distance formula. Let us learn its formula in the next section.

What is the distance formula?

The distance formula is the formula, which is used to find the distance between any two points, only if the coordinates are known to us. These coordinates could lie on x-axis or y-axis or both. Suppose, there are two points, say P and Q in an XY plane. The coordinates of point P are (x 1 ,y 1 ) and of Q are (x 2 ,y 2 ). Then the formula to find the distance between two points PQ is given by:

Where D is the distance between the points.

Distance Formula Derivation

Let P(x 1 , y 1 ) and Q(x 2 , y 2 ) be the coordinates of two points on the coordinate plane.

Draw two lines parallel to both the x-axis and y-axis (as shown in the figure) through P and Q.

The parallel line through P will meet the perpendicular drawn to the x-axis from Q at T.

Thus, ΔPTQ  is right-angled at T.

PT = Base, QT = Perpendicular and PQ = Hypotenuse

By Pythagoras’s Theorem,

PQ 2 = PT 2  + QT 2

= (x 2  – x 1 ) 2  + (y 2  – y 1 ) 2

PQ = √[(x 2  – x 1 ) 2  + (y 2  – y 1 ) 2 ]

Hence, the distance between two points (x 1 , y 1 ) and (x 2 , y 2 ) is √[(x 2  – x 1 ) 2  + (y 2  – y 1 ) 2 ]

Similarly, the distance of a point P(x, y) from the origin O(0, 0) in the Cartesian plane is given by the formula:

OP = √(x 2  + y 2 )

Distance Formula Examples

Let us solve some problems based on the distance formula.

Example 1: Find the distance between the two points A(1, 2) and B(-2, 2).

Solution: Given, two points A and B have coordinates (1, 2) and (-2, 2) respectively.

Let A(1, 2) = (x 1 , y 1 )

B(-2, 2) = (x 2 , y 2 )

To find: the distance between A and B

By the formula of distance between two points ,

AB = √ (-3) 2 + (0) 2

AB = √ 9 = 3 unit.

Example 2: Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Solution: Let P(1, 7), Q(4, 2), R(–1, –1) and S(– 4, 4) are the coordinates of four points in an XY plane.

To prove: PQRS is a square

Solution: To prove points P, Q, R and S form a square, we have to prove that:

PQ = QR = RS = PS (Lengths of the square)

PR = QS (Diagonals of the square)

Now by the distance formula, we will find the distance between PQ, QR, RS, PS, PR and QS.

PQ = √[(1 – 4) 2 + (7-2) 2 ] = √(9 + 25) = √34

QR = √[(4+1) 2 + (2+1) 2 ] = √(25 + 9) = √34

RS = √[(–1+ 4) 2 + (–1 – 4) 2 ] = √(9 + 25) = √34

PS = √[(1+4) 2 + (7 – 4) 2 ] = √(25 + 9) = √34

PR =√[(1+1) 2 +(7+1) 2 ] = √(4+64) = √68

QS = √[(4+4) 2 + (2-4) 2 ] = √(64+4) = √68

Since, PQ = QR = RS = PS and PR = QS.

Therefore, PQRS is a square.

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The distance and midpoint formulas

• Midpoint II
• Midpoint III

The distance formula is used to find the distance between two points in the coordinate plane. We'll explain this using an example below

We want to calculate the distance between the two points (-2, 1) and (4, 3). We could see the line drawn between these two points is the hypotenuse of a right triangle. The legs of this triangle would be parallel to the axes which mean that we can measure the length of the legs easily.

We'll get the length of the distance d by using the Pythagorean Theorem

$$d^{2}=2^{2}+6^{2}=4+36=40$$

$$d=\sqrt{40}\approx 6.32$$

This method can be used to determine the distance between any two points in a coordinate plane and is summarized in the distance formula

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

The point that is at the same distance from two points A (x 1 , y 1 ) and B (x 2 , y 2 ) on a line is called the midpoint. You calculate the midpoint using the midpoint formula

$$m =\left ( \frac{x_{1}+x_{2}}{2} \right ),\: \: \left ( \frac{y_{1}+y_{2}}{2} \right )$$

We can use the example above to illustrate this

$$ m =\left ( \frac{4+(-2)}{2} \right ),\: \: \left ( \frac{3+1}{2} \right )=$$

$$=\left ( \frac{2}{2} \right ),\: \: \left ( \frac{4}{2} \right )=\begin{pmatrix} 1,\: 2 \end{pmatrix}$$

Video lesson

Calculate the distance between the two points

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Distance Calculator

What is distance, the distance formula for euclidean distance, distance to any continuous structure, distance to a line and between 2 lines, how to find the distance using our distance calculator, driving distance between cities: a real-world example, distance from earth to moon and sun - astronomical distances, distance beyond length.

Have you ever wanted to calculate the distance from one point to another, or the distance between cities? Have you ever wondered what the distance definition is? We have all these answers and more, including a detailed explanation of how to calculate the distance between any two objects in 2D space. As a bonus, we have a fascinating topic on how we perceive distances (for example as a percentage difference ); we're sure you'll love it!

Prefer watching over reading? Learn all you need in 90 seconds with this video we made for you :

Before we get into how to calculate distances, we should probably clarify what a distance is . The most common meaning is the 1D space between two points. This definition is one way to say what almost all of us think of distance intuitively, but it is not the only way we could talk about distance. You will see in the following sections how the concept of distance can be extended beyond length, in more than one sense that is the breakthrough behind Einstein's theory of relativity.

If we stick with the geometrical definition of distance we still have to define what kind of space we are working in . In most cases, you're probably talking about three dimensions or less, since that's all we can imagine without our brains exploding. For this calculator, we focus only on the 2D distance (with the 1D included as a special case). If you are looking for the 3D distance between 2 points we encourage you to use our 3D distance calculator made specifically for that purpose.

To find the distance between two points, the first thing you need is two points, obviously . These points are described by their coordinates in space. For each point in 2D space, we need two coordinates that are unique to that point. If you wish to find the distance between two points in 1D space you can still use this calculator by simply setting one of the coordinates to be the same for both points . Since this is a very special case, from now on we will talk only about distance in two dimensions.

The next step, if you want to be mathematical, accurate, and precise , is to define the type of space you're working in. No, wait, don't run away! It is easier than you think. If you don't know what space you're working in or if you didn't even know there is more than one type of space, you're most likely working in Euclidean space . Since this is the "default" space in which we do almost every geometrical operation, and it's the one we have set for the calculator to operate on. Let's dive a bit deeper into Euclidean space , what is it, what properties does it have and why is it so important?

The Euclidean space or Euclidean geometry is what we all usually think of 2D space is before we receive any deep mathematical training in any of these aspects. In Euclidean space, the sum of the angles of a triangle equals 180º and squares have all their angles equal to 90º; always. This is something we all take for granted, but this is not true in all spaces . Let's also not confuse Euclidean space with multidimensional spaces. Euclidean space can have as many dimensions as you want, as long as there is a finite number of them, and they still obey Euclidean rules .

We do not want to bore you with mathematical definitions of what is a space and what makes the Euclidean space unique, since that would be too complicated to explain in a simple distance calculator. However, we can try to give you some examples of other spaces that are commonly used and that might help you understand why Euclidean space is not the only space. Also, you will hopefully understand why we are not going to bother calculating distances in other spaces .

The first example we present to you is a bit obscure, but we hope you can excuse us, as we're physicists , for starting with this very important type of space: Minkowski space . The reason we've selected this is because it's very common in physics , in particular it is used in relativity theory, general relativity and even in relativistic quantum field theory. This space is very similar to Euclidean space, but differs from it in a very crucial feature: the addition of the dot product, also called the inner product (not to be confused with the cross product).

Both the Euclidean and Minkowski space are what mathematicians call flat space . This means that space itself has flat properties; for example, the shortest distance between any two points is always a straight line between them (check the linear interpolation calculator). There are, however, other types of mathematical spaces called curved spaces in which space is intrinsically curved and the shortest distance between two points is no a straight line.

This curved space is hard to imagine in 3D, but for 2D we can imagine that instead of having a flat plane area, we have a 2D space, for example, curved in the shape of the surface of a sphere. In this case, very strange things happen . The shortest distance from one point to another is not a straight line, because any line in this space is curved due to the intrinsic curvature of the space. Another very strange feature of this space is that some parallel lines do actually meet at some point . You can try to understand it by thinking of the so-called lines of longitude that divide the Earth into many time zones and cross each other at the poles.

It is important to note that this is conceptually VERY different from a change of coordinates . When we take the standard x , y , z x, y, z x , y , z coordinates and convert into polar, cylindrical, or even spherical coordinates, but we will still be in Euclidean space. When we talk about curved space, we are talking about a very different space in terms of its intrinsic properties . In spherical coordinates, you can still have a straight line and distance is still measured in a straight line, even if that would be very hard to express in numbers.

Coming back to the Euclidean space, we can now present you with the distance formula that we promised at the beginning . The distance formula is

which relates to the Pythagorean Theorem, which states that a 2 + b 2 = c 2 a^2+b^2=c^2 a 2 + b 2 = c 2 . Here, a a a and b b b are legs of a right triangle and c c c is the hypotenuse. Suppose that two points, ( x 1 , y 1 ) (x_1, y_1) ( x 1 ​ , y 1 ​ ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 ​ , y 2 ​ ) , are coordinates of the endpoints of the hypotenuse . Then ( x 2 − x 1 ) 2 (x_2 - x_1)^2 ( x 2 ​ − x 1 ​ ) 2 in the distance equation corresponds to a 2 a^2 a 2 and ( y 2 − y 1 ) 2 (y_2 - y_1)^2 ( y 2 ​ − y 1 ​ ) 2 corresponds to b 2 b^2 b 2 . Since c = a 2 + b 2 c = \sqrt{a^2 + b^2} c = a 2 + b 2 ​ , you can see why this is just an extension of the Pythagorean theorem .

The distance formula we have just seen is the standard Euclidean distance formula , but if you think about it, it can seem a bit limited. We often don't want to find just the distance between two points. Sometimes we want to calculate the distance from a point to a line or to a circle. In these cases, we first need to define what point on this line or circumference we will use for the distance calculation, and then use the distance formula that we have seen just above.

Here is when the concept of perpendicular line becomes crucial. The distance between a point and a continuous object is defined via perpendicularity. From a geometrical point of view, the first step to measure the distance from one point to another, is to create a straight line between both points, and then measure the length of that segment . When we measure the distance from a point to a line, the question becomes "Which of the many possible lines should I draw?". In this case the answer is: the line from the point that is perpendicular to the first line . This distance will be zero in the case in which the point is a part of the line. For these 1D cases, we can only consider the distance between points, since the line represents the whole 1D space .

This imposes restrictions on how to compute distances in some interesting geometrical instances. For example, we could redefine the concept of height of a triangle to be simply the distance from one vertex to the opposing side of the triangle. In this case, the triangle area is also redefined in terms of distance, since the area is a function of the height of the triangle.

Let's look at couple examples in 2D space. To calculate the distance between a point and a straight line we could go step by step (calculate the segment perpendicular to the line from the line to the point and the compute its length) or we could simply use this 'handy-dandy' equation :

where the line is given by A x + B y + C = 0 Ax+By+C = 0 A x + B y + C = 0 and the point is defined by ( x 1 , y 1 ) (x_1, y_1) ( x 1 ​ , y 1 ​ ) .

The only problem here is that a straight line is generally given as y = m x + b y=mx+b y = m x + b , so we would need to convert this equation to the previously show form:

so we can see that A = m A=m A = m , B = − 1 B=-1 B = − 1 and C = b C=b C = b . This leaves the previous equation with the following values:

For the distance between 2 lines, we just need to compute the length of the segment that goes from one to the other and is perpendicular to both. Once again, there is a simple formula to help us :

if the lines are A 1 x + B 1 y + C 1 = 0 A_1x+B_1y+C_1=0 A 1 ​ x + B 1 ​ y + C 1 ​ = 0 and A 2 x + B 2 y + C 2 = 0 A_2x+B_2y+C_2=0 A 2 ​ x + B 2 ​ y + C 2 ​ = 0 . We can also convert to slope intercept form and obtain:

for lines y = m 1 x + b 1 y=m_1x+b_1 y = m 1 ​ x + b 1 ​ and y = m 2 x + b 2 y=m_2x+b_2 y = m 2 ​ x + b 2 ​ .

Notice that both line needs to be parallel since otherwise the would touch at some point and their distance would then be d = 0 d=0 d = 0 . That's the reason the formulas omit most of the subscripts since for parallel lines: A 1 = A 2 = A A_1=A_2=A A 1 ​ = A 2 ​ = A and B 1 = B 2 = B B_1=B_2=B B 1 ​ = B 2 ​ = B while in slope intercept form parallel lines are those for which m 1 = m 2 = m m_1=m_2=m m 1 ​ = m 2 ​ = m .

As we have mentioned before, distance can mean many things , which is why we have provided a few different options for you in this calculator. You can calculate the distance between a point and a straight line, the distance between two straight lines (they always have to be parallel), or the distance between points in space. When it comes to calculating the distances between two point, you have the option of doing so in 1, 2, 3, or 4 dimensions. We know, we know, 4 dimensions sounds scary , but you don't need to use that option. And you can always learn more about it by reading some nice resources and playing around with the calculator. We promise it won't break the Internet or the universe.

We have also added the possibility for you to define 3 different points in space, from which you will obtain the 3 pairs of distances between them, so, if you have more than two points, this will save you time. The number of dimensions you are working in will determine the number of coordinates that describe a point, which is why, as you increase the number of dimensions, the calculator will ask for more input values.

Even though using the calculator is very straightforward, we still decided to include a step-by-step solution. This way you can get acquainted with the distance formula and how to use it (as if this was the 1950's and the Internet was still not a thing). Now let's take a look at a practical example: How to find the distance between two points in 2-D.

Suppose you have two coordinates, ( 3 , 5 ) (3, 5) ( 3 , 5 ) and ( 9 , 15 ) (9, 15) ( 9 , 15 ) , and you want to calculate the distance between them. To calculate the 2-D distance between these two points, follow these steps:

• Input the values into the formula: ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} ( x 2 ​ − x 1 ​ ) 2 + ( y 2 ​ − y 1 ​ ) 2 ​ .
• In the formula, subtract the values in the parentheses.
• Square both quantities in the parentheses.
• Add the results.
• Take the square root.
• Use the distance calculator to check your results.

Working out the example by hand, you get:

which is equal to approximately 11.66 11.66 11.66 . Note, that when you take the square root, you will get a positive and negative result, but since you are dealing with distance, you are only concerned with the positive result . The calculator will go through this calculations step by step to give you the result in exact and approximate formats.

Let's take a look of one of the applications of the distance calculator. Suppose you are traveling between cities A and B, and the only stop is in city C, with a route A to B perpendicular to route B to C. We can determine the distance from A to B, and then, knowing the gas price, determine fuel cost , fuel used and cost per person while traveling. The gas calculator, that you can find on our site, can ease that for you.

The difficulty here is to calculate the distances between cities accurately . A straight line (like what we use in this calculator) can be a good approximation, but it can be quite off if the route you're taking is not direct but takes some detour, maybe to avoid mountains or to pass by another city. In that case, just use Google maps or any other tool that calculates the distance along a path not just the distance from one point to another as the crow flies.

Our calculator can give proper measurements and predictions for distances between objects, not the length of a path . With this in mind, there are still multiple scenarios in which you might actually be interested in the distance between objects, regardless of the path you would have to take. One such example is the distance between astronomical objects .

When we look at a distance within our Earth, it is hard to go far without bumping into some problems , from the intrinsic curvature of this space (due to the Earth curvature being non-zero) to the limited maximum distance between two points on the Earth. It is because of this, and also because there is a whole universe beyond our Earth , that distances in the universe are of big interest for many people. Since we have no proper means of interplanetary traveling, let alone interstellar travels, let's focus for now on the actual Euclidean distance to some celestial objects. For example the distance from the Earth to the Sun, or the distance from the Earth to the Moon.

These distances are beyond imaginable for our ape-like brains. We struggle to comprehend the size of our planet, never mind the vast, infinite universe. This is so difficult that we need to use either scientific notation or light years, as a unit of distance for such long lengths . The longest trips you can do on Earth are barely a couple thousand kilometers, while the distance from Earth to the Moon, the closest astronomical object to us, is 384,000 km . On top of that, the distance to our closest star, that is the distance from Earth to the Sun , is 150,000,000 km or a little over 8 light minutes .

When you compare these distances with the distance to our second nearest star (Alpha Centauri) , which is 4 light years , suddenly they start to look much smaller. If we want to go even more ridiculous in comparison we can always think about a flight from New York to Sydney, which typically takes more than 20 h and it's merely over 16,000 km , and compare it with the size of the observable universe, which is about 46,600,000,000 light years !

Here, we have inadvertently risen a fascinating point , which is that we measure distances not in length but in time. Thus, we extend the notion of distance beyond its geometrical sense. We will explore this possibility in the next section as we speak about the importance and usefulness of distance beyond the purely geometrical sense . This is a very interesting path to take and is mostly inspired by the philosophical need to extend every concept to have a universal meaning, as well as from the obvious physical theory to mention, when talking about permutations of the space and time, or any other variable that can be measured .

Typically, the concept of distance refers to the geometric Euclidean distance and is linked to length. However, you can extend the definition of distance to mean just the difference between two things , and then a world of possibilities opens up. Suddenly one can decide what is the best way to measure the distance between two things and put it in terms of the most useful quantity. A very simple step to take is to think about the distance between two numbers, which is nothing more than the 1D difference between these numbers . To obtain it, we simply subtract one from the other and the result would be the difference, a.k.a. the distance.

We could jump from this numerical distance to, for example, difference or distance in terms of the percentage difference, which in some cases might provide a better way of comparison . This is still just one level of abstraction in which we simply remove the units of measurement. But what if we were to use different units altogether?

By extending the concept of distance to mean something closer to difference , we can calculate the difference between two temperatures, or other related quantity like pressure. But we don't need to get really extreme, let's see how two points can be separated by a different distance, depending on the assumptions made . Coming back to the driving distance example, we could measure the distance of the journey in time, instead of length. In this case, we need an assumption to allow such translation; namely the way of transport.

There is a big difference in the time taken to travel 10 km by plane versus the time it takes by car . Sometimes, however, the assumption is clear and implicitly agreed on, like when we measure the lightning distance in time which we then convert to length. This brings up an interesting point, that the conversion factor between distances in time and length is what we call "speed" or "velocity" (remember they are not exactly the same thing). Truth be told, this speed doesn't have to be constant as exemplified by accelerated motions such as that of a free fall under gravitational force, or the one that links stopping time and stopping distance via the breaking force and drag or, in very extreme cases, via the force of a car crash.

Another place where you can find weird units of distance are in solid state physics, where the distance a particle travels inside of a material is often expressed as an average of interactions or collisions. This distance is linked to length by using the mean free path, which is the mean distance (in length) a particle travels between interactions. If we want to get even more exotic we can think about the distance from the present value to the future value of something like a car.

We don't want to, however, make anyone's brain explode, so please don't think too hard about this . Just take this calculator and use it for length-based distance in 2D space. You can always return to this philosophical view on distances if you ever find yourself bored!

How to find the distance between two points?

To find the distance between two points we will use the distance formula: √[(x₂ - x₁)² + (y₂ - y₁)²]:

• Get the coordinates of both points in space.
• Subtract the x-coordinates of one point from the other, same for the y components.
• Square both results separately.
• Sum the values you got in the previous step.
• Find the square root of the result above.

If you think this is too much effort, you can simply use the Distance Calculator from Omni

Is distance a vector?

Distance is not a vector . The distance between points is a scalar quantity, meaning it is only defined by its value. However, the displacement is a vector with value and direction. So the distance between A and B is the same as the distance from B to A, but the displacement is different depending on their order.

What is a klick in distance?

Click is slang for a kilometer which is 0.62 miles. It is actually written with "k" (Klick) as it is derived from the word kilometer. It is commonly used in the military and motorcyclists.

What is the distance formula?

The distance formula is: √[(x₂ - x₁)² + (y₂ - y₁)²] . This works for any two points in 2D space with coordinates (x₁, y₁) for the first point and (x₂, y₂) for the second point. You can memorize it easily if you notice that it is Pythagoras theorem and the distance is the hypothenuse, and the lengths of the catheti are the difference between the x and y components of the points.

How to find the distance of a vector?

The distance of a vector is its magnitude . If you know its components:

• Take each of the components of the vector and square them.
• Sum them up.
• Find the square root of the previous result.
• Enjoy the good work!

If you know its polar representation , it will be a number and an angle. That number is the magnitude of the vector, which is its distance.

What is the SI unit of distance?

The SI unit of distance is the meter , abbreviated to "m". A meter is approximately 3.28 feet. Other common units in the International System of units are the centimeter (one one-hundredth of a meter, or 0.39 inches) and the kilometer (one thousand meters or 0.62 miles), among others.

What is the distance from A to B?

The distance from A to B is the length of the straight line going from A to B. The distance from B to A is the same as the distance from A to B because distance is a scalar

What is the dimension of distance?

Distance is a measure of one-dimensional space. The distance between two points is the shortest length of 1D space between them. If you divide distance over time you will get speed, which has dimensions of space over time.

Is light-year time or distance?

A light-year is a measurement of distance. It is 9.461×10 12 kilometers or 5.879×10 12 miles, which is the distance traveled by a ray of light in a perfect vacuum over the span of a year.

How to solve for distance with velocity and time?

The velocity and the moving time of an object you can calculate the distance:

• Make sure the speed and time have compatible units (miles per hour and hours, meter per second, and seconds…).
• If they aren't, convert them to the necessary units.
• Multiply the velocity by the time .
• The result should be the distance traveled in whichever length units your speed was using!

Diameter of a cone

Grams to cups, humans vs vampires, right square pyramid.

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Distance Calculator

This calculator computes the distance between two points in two or three dimensions. It also finds the distance between two places on the world map, which are determined by their longitude and latitude. The calculator shows formulas and all steps.

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About this calculator

Definition:

The distance between two points in the coordinate plane or space is the line segment length that connects these two points.

Distance in the Coordinate Plane

To find the distance between points A (X1, y1) and B (x2, y2) in a plane, we usually use the Distance formula :

Find distance between points A(3, -4) and B(-1, 3)

First we need to identify constant x1, y1, x2 and y2: x1 = 3, y1 = -4, x2 = -1 y2 = 3. Now we can apply above formula:

d(A,B) = √[(x2 - x1)^2 + (y2-y1)^2] d(A,B) = √[(-1-3)^2 + (3-(-4))^2 d(A,B) = √[(-4)^2 + 7^2] d(A,B) = √[16+49] d(A,B) = √65

Distance in the Euclidean Space

The distance between points A (X1, y1, z1) and B (x2, y2, z2) in spcace is given by the formula:

Find distance between A(2, -1, 5) and B(3, 5, 2)

In this example the constants are x1 = 2, y1 = -1, z1 = 5, x2 = 3, y2 = 5, z2 = 2. Now we can apply above formula:

d(A,B) = √[(x2 - x1)^2 + (y2-y1)^2 + (z2-z1)^2] d(A,B) = √[(3-2)^2 + (5-(-1))^2 + (2-5)^2 d(A,B) = √[(1)^2 + 6^2 + (-3)^2] d(A,B) = √[1 + 36 + 9] d(A,B) = √46

Distance between points on sphere

The distance formula for two points on the Earth’s surface is:

d(A,B) = arccos[ sin(lat1) * sin(lat2) +cos(lat1) * cos(lat2) * cos(lon2 - lon1)] ∗6371

where lat1, lon1, lat2, and lon2 are their latitude and longitude coordinates.

Find distance between A(14.213, -38.481) and B(-2.13, 0.829)

In this example the latitudes and longitudes: lat1 = 14.213, lon1 = -38.481, lat2 = -2.13, lon2 = 0.829. After substituting into the formula, we get:

d(A,B) = arccos[ sin(lat1) * sin(lat2) +cos(lat1) * cos(lat2) * cos(lon2 - lon1)] ∗6371 d(A,B) = arccos[ sin(14.213) * sin(-2.13) + cos(14.213) * cos(-2.13) * cos(0.829 + 38.481 )] ∗6371 d(A,B) = arccos[ - 0.2455 * 0.0371 + 0.9696 * 0.9993 * 0.7737] ∗6371 d(A,B) = arccos[ - 0.00910 + 0.7496] d(A,B) = arccos[ 0.7405 ] ∗6371 d(A,B) = 5867.4

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Number Line

• distance\:(-3\sqrt{7},\:6),\:(3\sqrt{7},\:4)
• distance\:(-5,\:8d),\:(0,\:4)
• distance\:(-2,\:-3),\:(-1,\:-2)
• distance\:(p,\:1),\:(0,\:q)
• distance\:(3\sqrt{2},7\sqrt{5})(\sqrt{2},-\sqrt{5})
• distance\:(-2,-3),(-1,-2)

distance-calculator

• High School Math Solutions – Perpendicular & Parallel Lines Calculator Parallel lines have the same slope, to find the parallel line at a given point you should simply calculate the...

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