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Perpendicular lines Absolute valueHere 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.

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.

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.

Use this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEUse this quiz to check your grade 6 – grade 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th to 8th grade algebra topics to identify areas of strength and support!

DOWNLOAD FREEIn 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.**

Find the distance between the points A and B.

**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)}.**

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.**

Find the distance between the points A and B.

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).

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\
&=\sqrt{(6-1)^2+(2-(-5))^2}
\end{aligned}

**Solve the equation.**

\begin{aligned}
d&=\sqrt{(6-1)^2+(2-(-5))^2} \\\\
&=\sqrt{5^2+7^2} \\\\
&=\sqrt{25+49} \\\\
&=\sqrt{74} \\\\
&=8.6 \text { (1dp)}
\end{aligned}

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).

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\
&=\sqrt{(7-1)^2+(12-4)^2}
\end{aligned}

**Solve the equation.**

\begin{aligned}
d&=\sqrt{(7-1)^2+(12-4)^2} \\\\
&=\sqrt{6^2+8^2} \\\\
&=\sqrt{36+64} \\\\
&=\sqrt{100} \\\\
&=10
\end{aligned}

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

Give your answer to 1 decimal place.

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

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\
&=\sqrt{(6-(-2))^2+(-7-5)^2}
\end{aligned}

**Solve the equation.**

\begin{aligned}
d&=\sqrt{(6-(-2))^2+(-7-5)^2}\\\\
&=\sqrt{8^2+(-12)^2} \\\\
&=\sqrt{64+144} \\\\
&=\sqrt{208} \\\\
&=14.4 \text { (1dp)}
\end{aligned}

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.

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\
17&=\sqrt{(16-1)^2+(k-5)^2}
\end{aligned}

**Solve the equation.**

\begin{aligned}
17&=\sqrt{(16-1)^{2}+(k-5)^{2}}\\\\
17&=\sqrt{15^{2}+(k-5)^{2}}\\\\
289&=225+(k-5)^{2}\\\\
64&=(k-5)^{2}\\\\
\pm8&=k-5\\\\
k&=\pm8+5\\\\
k&=13\text{ or }k=-3\\\\
\end{aligned}

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

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.

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\
15&=\sqrt{(f-2)^2+(10-9)^2}
\end{aligned}

**Solve the equation.**

\begin{aligned}
15&=\sqrt{(f-2)^2+(10-9)^2} \\\\
15&=\sqrt{(f-2)^2+1^2} \\\\
225&=1+(f-2)^2 \\\\
224&=(f-2)^2 \\\\
\pm14.9666\ldots&=f-2 \\\\
f&= \pm14.9666\ldots+2 \\\\
f&=17.0 \text { (1dp) or } f=-13.0 \text{ (1dp)}
\end{aligned}

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

- 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.

**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.

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

14

10

3.74

100

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

\begin{aligned} d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\ & =\sqrt{(6-0)^2+(8-0)^2} \\\\ & =\sqrt{6^2+8^2} \\\\ & =\sqrt{36+64} \\\\ & =\sqrt{100} \\\\ & =10 \end{aligned}

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

34

5.83

26

14

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

\begin{aligned} d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\ & =\sqrt{(24-0)^2+(0-10)^2} \\\\ & =\sqrt{24^2+(-10)^2} \\\\ & =\sqrt{576+100} \\\\ & =\sqrt{676} \\\\ & =26 \end{aligned}

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

Give your answer to 1 decimal place.

11.5

130

23.0

11.4

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

\begin{aligned} d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\ & =\sqrt{(14-5)^2+(10-3)^2} \\\\ & =\sqrt{9^2+7^2} \\\\ & =\sqrt{81+49} \\\\ & =\sqrt{130} \\\\ & =11.4 \text { (1dp)} \end{aligned}

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

Give your answer to 1 decimal place.

14.3

19

16.4

7.8

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

\begin{aligned} d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\ & =\sqrt{(-8-(-2))^2+(-9-4)^2} \\\\ & =\sqrt{(-6)^2+(-13)^2} \\\\ & =\sqrt{36+169} \\\\ & =\sqrt{205} \\\\ & =14.3 \text { (1dp)} \end{aligned}

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

Find the value of a, where a is positive.

-27

21

27

-21

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\
25&=\sqrt{(15-8)^2+(a-(-3))^2} \\\\
25&=\sqrt{7^2+(a+3)^2} \\\\
625&=49+(a+3)^2 \\\\
576&=(a+3)^2 \\\\
\pm 24&=a+3 \\\\
a&=\pm 24-3\\\\
a&=21 \text { or } a=-27
\end{aligned}

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.

3

-15

-3

15

\begin{aligned}
d&=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\\\
15&=\sqrt{(6-b)^2+(-8-4)^2} \\\\
15&=\sqrt{(6-b)^2+(-12)^2} \\\\
225&=(6-b)^2+144 \\\\
81&=(6-b)^2 \\\\
\pm 9&=6-b \\\\
-b&=\pm 9-6\\\\
-b&=3 \text { or }-b=-15 \\\\
b&=-3 \text { or } b=15
\end{aligned}

As b is negative, b=-3.

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}

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

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