Distance Formula

The Distance Formula is a fundamental concept in coordinate geometry that allows us to calculate the distance between two points in a two-dimensional coordinate plane (the x-y plane). It’s derived directly from the Pythagorean theorem and provides a straightforward way to find the length of the straight line segment connecting two points.

Formulae

The Distance Formula is expressed as:

The distance, *d*, between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

Examples

Let’s explore how to use the Distance Formula with some examples:

Example-1:

Find the distance between the points A(1, 2) and B(4, 6).

Solution:

1. Identify the coordinates: $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (4, 6)$.
2. Substitute the values into the formula: $d = \sqrt{(4 – 1)^2 + (6 – 2)^2}$.
3. Simplify: $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25}$.
4. Calculate the final distance: $d = 5$.

Therefore, the distance between points A and B is 5 units.

Example-2:

Find the distance between the points C(-2, 3) and D(1, -1).

Solution:

1. Identify the coordinates: $(x_1, y_1) = (-2, 3)$ and $(x_2, y_2) = (1, -1)$.
2. Substitute the values into the formula: $d = \sqrt{(1 – (-2))^2 + (-1 – 3)^2}$.
3. Simplify: $d = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25}$.
4. Calculate the final distance: $d = 5$.

Therefore, the distance between points C and D is 5 units.

Theorem with Proof

Theorem: The Distance Formula is a direct consequence of the Pythagorean theorem.

Proof:

1. Consider two points, A$(x_1, y_1)$ and B$(x_2, y_2)$, on a coordinate plane.
2. Draw a right triangle with AB as the hypotenuse. The legs of the triangle are parallel to the x and y axes.
3. The horizontal leg has length $|x_2 – x_1|$ and the vertical leg has length $|y_2 – y_1|$.
4. By the Pythagorean theorem, the square of the hypotenuse (the distance *d* between A and B) is equal to the sum of the squares of the legs: $d^2 = (x_2 – x_1)^2 + (y_2 – y_1)^2$.
5. Taking the square root of both sides, we get the Distance Formula: $d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$.

Common mistakes by students

Students often make the following mistakes:

• Incorrectly substituting values: Make sure to correctly identify which x and y values correspond to which points.
• Arithmetic errors: Be careful with negative signs and order of operations when squaring and subtracting.
• Forgetting the square root: The final step is always to take the square root of the sum of the squares.

Real Life Application

The Distance Formula has numerous real-life applications, including:

• Navigation: GPS systems use the distance formula to calculate distances between locations and determine routes.
• Mapping and Surveying: Surveyors use the distance formula to accurately measure distances on land.
• Computer Graphics: Calculating distances between objects in 2D or 3D space is essential for rendering realistic images.
• Game Development: Games frequently use the distance formula for collision detection and other calculations.

Fun Fact

The Distance Formula is essentially a generalization of the Pythagorean theorem to two dimensions. You can extend this concept to three dimensions (and beyond!) to calculate distances in higher-dimensional spaces.

Recommended YouTube Videos for Deeper Understanding

Practice Questions

Q.1 What is the distance between the points A(2, 3) and B(5, 7)?

Check Solution

Ans: A

Using the distance formula, $\sqrt{(5-2)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$

Q.2 Find the value of k if the distance between the points P(k, 2) and Q(4, 5) is 5 units.

Check Solution

Ans: A

$\sqrt{(4-k)^2 + (5-2)^2} = 5$. Squaring both sides: $(4-k)^2 + 9 = 25$. $(4-k)^2 = 16$. $4-k = \pm 4$. $k = 0$ or $k = 8$.

Q.3 The vertices of a triangle are A(1, 1), B(4, 5), and C(7, 2). What type of triangle is it?

Check Solution

Ans: C

AB = $\sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9+16} = 5$. BC = $\sqrt{(7-4)^2 + (2-5)^2} = \sqrt{9+9} = 3\sqrt{2}$. CA = $\sqrt{(1-7)^2 + (1-2)^2} = \sqrt{36+1} = \sqrt{37}$. Since $AB^2 + BC^2 \ne CA^2$, $AC^2 + BC^2 \ne AB^2$, and $AB^2 + AC^2 \ne BC^2$, it is a scalene triangle.

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