Distance Formula

The distance formula in maths, especially in coordinate geometry, is used to find the distance between two points in an XY plane. It is a fundamental topic that is covered in the Distance Formula class 9.

1.0What is the Distance Formula?

To calculate distance formula between two points (x₁,y₁) and (x₂,y₂) in a two-dimensional plane, you have to:

$D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Here, 

• D is the distance between two points.
• (x₁,y₁) and (x₂,y₂) are the coordinates of the points. 

2.0Derivation of the Distance Formula

The distance formula geometry is based on the Pythagorean theorem, which states:

a² + b² = c²

Here, c represents the length of the hypotenuse in a right triangle; a and b are the other sides. 

Considering the two points A (x₁,y₁) and B(x₂,y₂), we can form a right triangle by drawing a horizontal and a vertical line from these two points. The horizontal distance (x₂ - x₁) represents one leg of the triangle, and the vertical difference (y₂-y₁) represents the other leg. So, applying the Pythagorean theorem, we get: 

$\begin{array}{rl}& D^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\\ & D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\end{array}$

3.0Distance Formula Example

Let’s take distance formula examples and solve them so you can grasp the concept better. 

Example: Find the distance between the points A (2,3) and B (6,7)

Solution: Using the distance formula in Maths, we can see, 

$\begin{array}{rl}& D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & D=\sqrt{(6-2{)}^2+(7-3{)}^2}\\ & D=\sqrt{(4{)}^2+(4{)}^2}\\ & D=\sqrt{32}\\ & D\approx 5.66\end{array}$

Thus, the distance between the two points is 5.66 units.

4.0Distance Formula in Different Coordinate Systems

Refer to the table below for the distance formula geometry in different coordinate systems: 

5.0Distance Formula Problems

To master the concept of distance formula, one needs to look closely at distance formula examples with answers. Let’s look at some distance formula problems to understand how to apply the concepts mathematically and in real-life situations. 

Problem 1: Find the distance between the points (4, –2) and (–3, 5).

Solution: 

$\begin{array}{rl}& D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & D=\sqrt{(-3-4{)}^2+(5-(-2){)}^2}\\ & D=\sqrt{(-7{)}^2+(7{)}^2}\\ & D=\sqrt{49+49}=\sqrt{98}\\ & D\approx 9.90\end{array}$

Problem 2: The distance between the points A(4, 5) and B(x, 9) is 5 units. Find the possible values of x.

Solution: Let A(4, 5) be A(x₁,y₁), and B(x, 9) be B(x₂,y₂). Now according to the question: 

$\begin{array}{rl}& D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & 5=\sqrt{(x-4{)}^2+(9-5{)}^2}\\ & 5=\sqrt{(x-4{)}^2+(4{)}^2}\end{array}$

Squaring both sides: 

25 = x²+16–8x +16

x² – 8x + 7 = 0

x² – 7x – x +7 =0

x(x-7) – 1(x–7)=0

(x – 1)(x–7) = 0

x = 1, 7 

Problem 3: Three vertices of a triangle are given as A(0, 0), B(6, 0), and C(6, 8). Prove whether it is a right-angled triangle.

Solution: Let’s use distance formula $\left(D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right)$ to prove if the vertices belong to a right-angled triangle or not. 

$\begin{array}{rl}& \mathrm{AB}=\sqrt{(6-0{)}^2+(0-0{)}^2}=6\\ & \mathrm{BC}=\sqrt{(6-6{)}^2+(8-0{)}^2}=8\\ & \mathrm{AC}=\sqrt{(6-0{)}^2+(8-0{)}^2}=\sqrt{100}=10\end{array}$

Now, applying Pythagoras Theorem: Check if AC² = AB² + BC²

AC² = (10)² = 100

AB² + BC² = 6^{2 .}+ 8² = 36 + 64 = 100

Here, AC² = AB² + BC², hence ABC is a right-angled triangle. 

6.0Distance Formula Problem-Solving Tips

While doing distance formula problem-solving, you should keep in mind the following tips and tricks to efficiently master the exercises. 

• Plot the Points: Visual aids work better than just numbers. You can always plot the points in a graph to understand the problem better. 
• Check Coordinates Carefully: Check the coordinates carefully before making the calculations. Small mistakes can lead to an incorrect result in the end.
• Square Each Difference First: Always compute the square difference first before adding.
• Use Approximation When Needed: Some square roots may not be exact, so you should round off accordingly.

7.0Applications of the Distance Formula

The distance formula in math is an important concept that has crucial real-life applications. 

• Navigation Systems: It is used in GPS technology to calculate the distance between two locations.
• Physics: Helps in motion analysis and speed calculations.
• Computer Graphics: It is essential for rendering objects in a three-dimensional space. 
• Sports Analytics: It is used in various sports to understand player movements and ball trajectories.

8.0Practice Questions

Below are a few distance formula problems the students can use to strengthen their understanding of the distance formula.

• Find the distance between the points (5, 8) and (2, 3).
• Calculate the distance between (-7, 4) and (1, -2).
• A triangle has vertices at (0, 0), (3, 4), and (6, 0). Find the length of each side.

9.0Sample Questions

1. How do you calculate the distance formula?

The distance between two points (x₁,y₂) and (x₂,y₂) in a Cartesian plane would be

$D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\cdot \sqrt{}\left({\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2\right)$

10.0Conclusion

The distance formula is a powerful tool in mathematics that has great real-life applications. Understanding the distance formula with real-life examples and problems would improve the understanding of students on this particular subject matter. By mastering this concept, students can confidently tackle real-world distance-related problems.