Master the Distance Formula: Step-by-Step Guide to Calculate Distance in Geometry

The Distance Formula is a mathematical equation used to calculate the straight-line distance between two points in a coordinate plane or space. Derived from the Pythagorean theorem, it applies to both 2D and 3D coordinates, making it a crucial tool in geometry, physics, and various real-life applications. It is widely used to measure the shortest path between points, lines, and planes.

Get Homework Help

1. Introduction to the Distance Formula:

The distance formula is a fundamental concept in geometry and mathematics used to calculate the distance between two points in space. It provides a mathematical approach to measure how far apart two locations are, whether in two-dimensional (2D) or three-dimensional (3D) space. The formula is derived from the Pythagorean theorem and has various applications in physics, navigation, architecture, and computer graphics fields. Whether it's measuring the distance between two points on a plane or finding the shortest distance from a point to a line, the distance formula is an essential tool.

2. What is Distance Formula:

The distance formula is a mathematical equation used to calculate the straight-line distance between two points in a coordinate plane or space. It is derived from the Pythagorean theorem and works in two-dimensional (2D) and three-dimensional (3D) spaces. In 2D, if the coordinates of two points are $(x_1, y_1)$ and $(x_2, y_2)$, the formula is:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}$

In 3D, the formula includes the z-coordinate to calculate the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2 (z_2 - z_1)^2}$

This formula provides the shortest distance (or straight-line distance) between two points, making it a crucial tool in geometry, physics, and many real-life applications.

3. Distance Formula to Calculate Distance Between Two Points:

The distance formula determines the straight-line distance between two points in a coordinate plane or space. This formula is derived from the Pythagorean theorem and works by finding the horizontal and vertical distances between two points and then applying these values to find the overall distance.

Distance Between Two Points in 2D

In a two-dimensional plane, the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated using the formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}$

This formula measures the direct or straight-line distance between the two points. It combines the differences in the x-coordinates and y-coordinates, squares them to avoid negative values, adds them together, and takes the square root to give the final distance.

Example:

For two points $(1, 2)$ and $(4, 6)$, the distance between them would be:

$\text{Distance} = \sqrt{(4 - 1)^2 (6 - 2)^2} = \sqrt{3^2 4^2} = \sqrt{9 16} = \sqrt{25} = 5$

This formula is crucial for measuring distances in geometry, navigation, and any situation where you must calculate the separation between two points on a flat surface.

4. Derivation of Distance Formula:

The distance formula is derived from the Pythagorean theorem, which relates the sides of a right triangle to its hypotenuse. To derive the distance formula between two points in a coordinate plane, we treat the difference in coordinates as the legs of a right triangle, where the straight-line distance between the points is the hypotenuse.

Derivation in 2D:

Consider two points $(x_1, y_1)$ and $(x_2, y_2)$ in a 2D plane. The horizontal distance between the points is $x_2 - x_1$, and the vertical distance is $y_2 - y_1$. Using the Pythagorean theorem:

$\text{Distance}^2 = (\text{horizontal distance})^2 (\text{vertical distance})^2$

Substituting the distances:

$\text{Distance}^2 = (x_2 - x_1)^2 (y_2 - y_1)^2$

Taking the square root of both sides gives the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}$

This gives us the straight-line distance between any two points in 2D space.

Distance Between Two Points in 3D

In three-dimensional space, the distance formula is extended to account for the third dimension, the z-coordinate. For two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, we apply the same Pythagorean theorem concept, but now we also include the depth, or vertical difference, between the points.

The formula for the distance between two points in 3D is:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2 (z_2 - z_1)^2}$

This formula combines the differences in all three coordinates—$x$, $y$, and $z$—to calculate the shortest distance between two points in space.

Example:

To find the distance between points $(1, 2, 3)$ and $(4, 6, 8)$:

$\text{Distance} = \sqrt{(4 - 1)^2 (6 - 2)^2 (8 - 3)^2}$

$= \sqrt{3^2 4^2 5^2} = \sqrt{9 16 25} = \sqrt{50} \approx 7.07$

This formula is essential for measuring distances in 3D applications such as physics, engineering, and 3D modeling.

5. Distance From a Point To a Line:

Calculating the distance from a point to a line is a common problem in geometry. This distance is the shortest possible distance from the point to the line, which is always along a line perpendicular to the given line. The formula differs for 2D and 3D, but both approaches find the minimum distance.

Distance From a Point To a Line in 2D

For a line in 2D, represented by the equation $Ax By C = 0$, the shortest distance from a point $(x_1, y_1)$ to this line can be calculated using the formula:

$\text{Distance} = \dfrac{|Ax_1 By_1 C|}{\sqrt{A^2 B^2}}$

This formula gives the perpendicular distance from the point to the line, where:

• $A$ and $B$ are the coefficients from the line equation.
• $(x_1, y_1)$ are the coordinates of the point.
• $C$ is the constant from the line equation.

Example:

For the line $3x 4y - 5 = 0$ and the point $(1, 2)$, the distance is:

$\text{Distance} = \dfrac{|3(1) 4(2) - 5|}{\sqrt{3^2 4^2}} = \dfrac{|3 8 - 5|}{\sqrt{9 16}} = \dfrac{|6|}{\sqrt{25}} = \dfrac{6}{5} = 1.2$

This is the shortest distance between the point $(1, 2)$ and the line.

Distance From a Point To a Line in 3D

In three-dimensional space, finding the shortest distance from a point to a line involves vector algebra. Given a point $P_1(x_1, y_1, z_1)$ and a line defined by a point $P_0(x_0, y_0, z_0)$ on the line and a direction vector $\mathbf{d} = \langle a, b, c \rangle$, the distance is calculated as:

$\text{Distance} = \dfrac{|\mathbf{P_1 - P_0} \times \mathbf{d}|}{|\mathbf{d}|}$

Where:

• $\mathbf{P_1 - P_0}$ is the vector from the point on the line to the point in space.
• $\mathbf{d}$ is the direction vector of the line.
• $\times$ denotes the cross product.
• $|\cdot|$ represents the magnitude of the vector.

Example:
Find the shortest distance from the point $P_1(4, 0, 5)$ to the line passing through the point $P_0(1, 2, 3)$ with direction vector $\mathbf{d} = \langle 2, -1, 1 \rangle$. The vector from $P_0$ to $P_1$ is:

$\mathbf{P_1 - P_0} = (4 - 1, 0 - 2, 5 - 3) = \langle 3, -2, 2 \rangle$

The cross product of $\langle 3, -2, 2 \rangle \times \langle 2, -1, 1 \rangle$ is:

$\langle (-2) (1) - (2) (-1), (2) (2) - (3) (1), (3) (-1) - (-2) (2) \rangle = \langle -2 2, 4 - 3, -3 4 \rangle = \langle 0, 1, 1 \rangle$

Now, calculate the magnitude of this vector:

$| \langle 0, 1, 1 \rangle | = \sqrt{0^2 1^2 1^2} = \sqrt{0 1 1} = \sqrt{2}$

Finally, divide by the magnitude of the direction vector $\mathbf{d} = \langle 2, -1, 1 \rangle$: is

$| \mathbf{d} | = \sqrt{2^2 (-1)^2 1^2} = \sqrt{4 1 1} = \sqrt{6}$

Now, apply the formula for the shortest distance:

$\text{Distance} = \dfrac{| \langle 0, 1, 1 \rangle |}{| \mathbf{d} |} = \dfrac{\sqrt{2}}{\sqrt{6}} = \sqrt{\dfrac{2}{6}} = \sqrt{\dfrac{1}{3}} = \dfrac{1}{\sqrt{3}}$

Thus, the shortest distance from the point $P_1(4, 0, 5)$ to the line is:

$\text{Distance} = \dfrac{1}{\sqrt{3}} \approx 0.577$

This is the shortest distance between the point and the line in 3D space.

6. Distance Between Two Lines:

The distance between two lines depends on their relative positions in space. The distance is constant for parallel lines, as the lines never intersect. For skew lines (lines that are not parallel and do not intersect), the shortest distance can be computed using vector algebra. The distance between lines is calculated differently in 2D and 3D space.

Distance Between Two Parallel Lines in 2D

For two parallel lines in 2D, given by the equations $Ax By C_1 = 0$ and $Ax By C_2 = 0$, the formula for the shortest distance between them is:

$\text{Distance} = \dfrac{| C_2 - C_1 |}{\sqrt{A^2 B^2}}$

Where:

• $A$ and $B$ are the coefficients from the equations of the lines.
• $C_1$ and $C_2$ are the constants from the two line equations.

This formula calculates the perpendicular distance between two parallel lines, which is the same at every point along the lines.

Example:
Given the two parallel lines $3x 4y - 7 = 0$ and $3x 4y 2 = 0$, the distance between them is:

$\text{Distance} = \dfrac{| 2 - (-7) |}{\sqrt{3^2 4^2}} = \dfrac{| 2 7 |}{\sqrt{9 16}} = \dfrac{9}{5} = 1.8$

Thus, the distance between the two parallel lines is 1.8 units.

Shortest Distance Between Two Lines in 3D

When dealing with skew lines in 3D (lines that are neither parallel nor intersect), the shortest distance between them can be found using vector algebra. Let two lines in 3D be defined by their direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$, and points $\mathbf{P}_1$ and $\mathbf{P}_2$ on each line, respectively.

The formula for the shortest distance between two skew lines is:

$\text{Distance} = \dfrac{|(\mathbf{P}_2 - \mathbf{P}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}$

Where:

• $\mathbf{P}_1$ and $\mathbf{P}_2$ are points on the first and second lines.
• $\mathbf{d}_1$ and $\mathbf{d}_2$ are the direction vectors of the two lines.
• $\times$ denotes the cross product of two vectors.
• $|\cdot|$ represents the magnitude of a vector.
• The cross product $\mathbf{d}_1 \times \mathbf{d}_2$ gives a vector perpendicular to both direction vectors.

Example:

Let the first line pass through the point $\mathbf{P}_1(1, 2, 3)$ with direction vector $\mathbf{d}_1 = \langle 2, -1, 1 \rangle$, and the second line pass through the point $\mathbf{P}_2(4, 0, 1)$ with direction vector $\mathbf{d}_2 = \langle 1, 1, -1 \rangle$.

1. First, calculate $\mathbf{P}_2 - \mathbf{P}_1$:

$\mathbf{P}_2 - \mathbf{P}_1 = (4 - 1, 0 - 2, 1 - 3) = \langle 3, -2, -2 \rangle$

2. Next, calculate the cross product of $\mathbf{d}_1 \times \mathbf{d}_2$:

$\mathbf{d}_1 \times \mathbf{d}_2 = \langle (-1)(-1) - (1)(1), (1)(1) - (2)(-1), (2)(1) - (-1)(1) \rangle = \langle 1 - 1, 1 2, 2 1 \rangle = \langle 0, 3, 3 \rangle$

3. Find the magnitude of the cross-product:

$|\mathbf{d}_1 \times \mathbf{d}_2| = \sqrt{0^2 3^2 3^2} = \sqrt{9 9} = \sqrt{18} = 3\sqrt{2}$

4. Calculate the dot product $(\mathbf{P}_2 - \mathbf{P}_1) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)$:

$\langle 3, -2, -2 \rangle \cdot \langle 0, 3, 3 \rangle = 3(0) (-2)(3) (-2)(3) = 0 - 6 - 6 = -12$

5. Finally, calculate the distance:

$\text{Distance} = \dfrac{|-12|}{3\sqrt{2}} = \dfrac{12}{3\sqrt{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.83$

Thus, the shortest distance between the two skew lines is approximately 2.83 units.

The methods for calculating the distance between parallel lines in 2D and skew lines in 3D are important in various fields like geometry, physics, and engineering, where precision in spatial measurements is required.

7. Distance From a Point to a Plane:

In three-dimensional space, finding the shortest distance from a point to a plane is a common problem in geometry. This distance is the perpendicular, or shortest, distance between a point and the closest point on the plane. The formula for calculating the distance from a point to a plane uses the plane’s equation and the coordinates of the point.

Given a plane described by the equation:

$Ax By Cz D = 0$

and a point $P_1(x_1, y_1, z_1)$, the formula for the shortest distance from the point to the plane is:

$\text{Distance} = \dfrac{|Ax_1 By_1 Cz_1 D|}{\sqrt{A^2 B^2 C^2}}$

Where:

• $A, B,$ and $C$ are the coefficients from the equation of the plane.
• $D$ is the constant term from the plane equation.
• $(x_1, y_1, z_1)$ are the coordinates of the point.
• The absolute value ensures that the distance is always positive.

Example:

Let the plane be given by the equation $2x 3y - z 4 = 0$, and the point be $P_1(1, 2, 3)$. To calculate the distance from the point to the plane, substitute the values into the formula:

$\text{Distance} = \dfrac{|2(1) 3(2) - 1(3) 4|}{\sqrt{2^2 3^2 (-1)^2}}$

$= \dfrac{|2 6 - 3 4|}{\sqrt{4 9 1}} = \dfrac{9}{\sqrt{14}} = \dfrac{9}{\sqrt{14}} \approx 2.41$

Thus, the shortest distance from the point $(1, 2, 3)$ to the plane $2x 3y - z 4 = 0$ is approximately $2.41$ units.

Interpretation of the Formula:

• The numerator $|Ax_1 By_1 Cz_1 D|$ measures the point's relative position with respect to the plane, calculating how far off the point is from satisfying the plane’s equation.

• The denominator $\sqrt{A^2 B^2 C^2}$ is the magnitude of the normal vector to the plane, ensuring that the distance calculation is properly scaled according to the plane’s orientation.

This formula is widely used in applications like physics, architecture, and computer graphics, where calculating the distance from points to surfaces (or planes) is essential for modeling, construction, and simulation tasks.

8. Distance Between Two Parallel Planes:

In three-dimensional geometry, the distance between two parallel planes is constant because parallel planes never intersect. The shortest distance between these two planes is measured along the line that is perpendicular to both of them. The formula for calculating this distance uses the equations of the planes and is derived from the concept of the distance from a point to a plane.

Given two parallel planes described by the equations:

$Ax By Cz D_1 = 0$

$Ax By Cz D_2 = 0$

The formula for the shortest distance between these two parallel planes is:

$\text{Distance} = \dfrac{|D_2 - D_1|}{\sqrt{A^2 B^2 C^2}}$

Where:

• $A$, $B$, and $C$ are the coefficients of the plane’s normal vector (the same for both planes since they are parallel).
• $D_1$ and $D_2$ are the constants from the equations of the two planes.

This formula measures the perpendicular distance between the planes.

Example:

Consider the two parallel planes:

$2x 3y - z 4 = 0$

$2x 3y - z - 5 = 0$

Using the formula:

$\text{Distance} = \dfrac{|D_2 - D_1|}{\sqrt{A^2 B^2 C^2}} = \dfrac{|-5 - 4|}{\sqrt{2^2 3^2 (-1)^2}} = \dfrac{|-9|}{\sqrt{4 9 1}} = \dfrac{9}{\sqrt{14}}$

The shortest distance between these two planes is:

$\text{Distance} = \dfrac{9}{\sqrt{14}} \approx 2.41 \, \text{units}$

How the Formula Works:

• The numerator $|D_2 - D_1|$ measures the vertical separation between the two planes.
• The denominator $\sqrt{A^2 B^2 C^2}$ represents the magnitude of the normal vector of the planes, ensuring that the calculation accounts for the orientation of the planes in 3D space.

This formula is essential for solving problems in architecture, physics, and 3D modeling, where calculating precise distances between parallel planes is critical for design and spatial understanding.

9. Distance Formula Solved Examples:

Question: 1.

Distance Between Two Points in 2D

Find the distance between the points $A(2, 3)$ and $B(6, 7)$.

Solution:

Step 1: Use the distance formula in 2D: $\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}$

Where:

• $(x_1, y_1) = (2, 3)$
• $(x_2, y_2) = (6, 7)$

Step 2: Substitute the values: $\text{Distance} = \sqrt{(6 - 2)^2 (7 - 3)^2} = \sqrt{4^2 4^2}$

Step 3: Simplify: $\text{Distance} = \sqrt{16 16} = \sqrt{32}$

Final Answer:
$\text{Distance} = 4 \sqrt{2} \text{ units} \approx 5.66 \text{ units}$

Question: 2.

Distance Between Two Points in 3D

Find the distance between the points $A(2, -1, 4)$ and $B(-3, 5, 1)$ in 3D space.

Solution:

Step 1: Write the 3D distance formula.

The formula for calculating the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2 (z_2 - z_1)^2}$

Step 2: Identify the coordinates of the points.

The coordinates are:

• $A(2, -1, 4)$
• $B(-3, 5, 1)$

Step 3: Substitute the coordinates into the distance formula.

$\text{Distance} = \sqrt{((-3) - 2)^2 (5 - (-1))^2 (1 - 4)^2}$

Step 4: Simplify each term.

$\text{Distance} = \sqrt{(-5)^2 (5 1)^2 (-3)^2}$

$\text{Distance} = \sqrt{25 36 9}$

Step 5: Add the values under the square root.

$\text{Distance} = \sqrt{70}$

Step 6: Find the square root of 70.

$\text{Distance} = \sqrt{70} \approx 8.37$

Final Answer:
The distance between the points $A(2, -1, 4)$ and $B(-3, 5, 1)$ is approximately $8.37$ units.

Question: 3.

Distance From a Point to a Line in 2D

Find the distance from the point $P(4, -1)$ to the line $3x - 4y 10 = 0$.

Solution:

Step 1: Use the distance formula from a point to a line:

$\text{Distance} = \dfrac{|Ax_1 By_1 C|}{\sqrt{A^2 B^2}}$

Where:

• $A = 3, B = -4, C = 10$
• $(x_1, y_1) = (4, -1)$

Step 2: Substitute the values:

$\text{Distance} = \dfrac{|3(4) - 4(-1) 10|}{\sqrt{3^2 (-4)^2}} = \dfrac{|12 4 10|}{\sqrt{9 16}}$

Step 3: Simplify:

$\text{Distance} = \dfrac{|26|}{\sqrt{25}} = \dfrac{26}{5}$

Final Answer:
$\text{Distance} = 5.2 \text{ units}$

Question: 4.

Distance From a Point to a Plane in 3D

Find the distance from the point $P(1, 2, 3)$ to the plane $2x - 3y 6z - 5 = 0$.

Solution:

Step 1: Use the distance formula from a point to a plane:

$\text{Distance} = \dfrac{|Ax_1 By_1 Cz_1 D|}{\sqrt{A^2 B^2 C^2}}$

Where:

• $A = 2, B = -3, C = 6, D = -5$
• $(x_1, y_1, z_1) = (1, 2, 3)$

Step 2: Substitute the values:

$\text{Distance} = \dfrac{|2(1) - 3(2) 6(3) - 5|}{\sqrt{2^2 (-3)^2 6^2}} = \dfrac{|2 - 6 18 - 5|}{\sqrt{4 9 36}}$

Step 3: Simplify:

$\text{Distance} = \dfrac{|9|}{\sqrt{49}} = \dfrac{9}{7}$

Final Answer:
$\text{Distance} = \dfrac{9}{7} \text{ units} \approx 1.29 \text{ units}$

10. Practice Questions on Distance Formula:

Q:1. Calculate the distance between the points $(-1, -1)$ and $(4, 3)$ in 2D.

Q:2. Find the distance between the points $(1, 2, 3)$ and $(4, 5, 6)$ in 3D.

Q:3. What is the distance from the point $(3, 1)$ to the line $2x - 3y 6 = 0$?

Q:4. Determine the distance between two parallel lines $3x 4y - 7 = 0$ and $3x 4y 2 = 0$.

11. FAQs on Distance Formula:

What is the distance formula?

The distance formula calculates the straight-line distance between two points in 2D or 3D space. In 2D, it's given by $d = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}$.

Can the distance formula be used in 3D?

Yes, in 3D, the distance formula includes the z-coordinates and is $d = \sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2 (z_2 - z_1)^2}$.

How do you find the distance between a point and a line in 2D?

The formula to find the shortest distance between a point $(x_1, y_1)$ and a line $Ax By C = 0$ is $\text{Distance} = \dfrac{|Ax_1 By_1 C|}{\sqrt{A^2 B^2}}$.

How do you calculate the distance from a point to a plane in 3D?

The distance is calculated using $\text{Distance} = \dfrac{|Ax_1 By_1 Cz_1 D|}{\sqrt{A^2 B^2 C^2}}$, where $A, B, C, D$ define the plane.

What is the shortest distance between two skew lines in 3D?

For two skew lines, the shortest distance is found using the cross product of their direction vectors.

What is the distance between two parallel lines in 2D?

The distance between two parallel lines is $\dfrac{|C_2 - C_1|}{\sqrt{A^2 B^2}}$.

Can the distance formula be applied in real-life problems?

Yes, the distance formula is commonly used in navigation, architecture, physics, and other fields requiring spatial distance measurement.

12. Real-life Application of Distance Formula:

The distance formula has numerous real-life applications. It's used in navigation to calculate the shortest path between two locations, in architecture to determine accurate measurements between points, and in computer graphics to render objects at proper distances in 3D environments. Additionally, it's vital in physics to determine the displacement between two objects and in geometry to solve various problems related to positioning and spacing.

13. Conclusion:

The distance formula is a versatile mathematics tool applicable in both 2D and 3D space. Its wide range of navigation, physics, and architecture applications makes it essential for calculating distances between points, lines, and planes. Understanding the formula and its various forms, along with practice, will ensure proficiency in solving real-world and theoretical problems with ease.

If you have any suggestions regarding the improvement of the content of this page, please write to me at My Official Email Address: [email protected]

$\fcolorbox{black}{lightpink}{\color{blue}{Get Assignment Help}}$
Are you Stuck on homework, assignments, projects, quizzes, labs, midterms, or exams?
To get connected to our tutors in real-time. Sign up and get registered with us.

$\color{red} \bold{Related \space Pages:}$
Methods-of-differentiation Formula Sheet
Mathematics Formula Sheet
Doubtlet's Blog Posts
Calculus Question Bank
Question and Answer Bank