
In geometry, the distance between two points refers to the length of the shortest path between them, which is always a straight line. This measurement is always non-negative, as distance cannot be less than zero.
Depending on where the points are located, we use different methods:
1D Space: Distance on a horizontal or vertical number line.
2D Space (XY Plane): Distance between points with (x, y) coordinates.
3D Space (XYZ Plane): Distance between points in a three-dimensional environment.
To find the length between coordinates, we use specific distance between two points formulas based on the dimensions involved.
If you have two points A(x_1) and B(x_2) on a single line, the distance is simply the absolute difference between them:
Formula: d = |x_2 - x_1|
For points P(x_1, y_1) and Q(x_2, y_2), we use the standard distance formula:
Formula: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
If you are measuring the distance from the point (0,0) to (x, y):
Formula: d = \sqrt{x^2 + y^2}
Below are some more examples for practice:
Example 1: Distance from the Origin
Find the distance between P(0, 0) and Q(6, 8).
Using d = \sqrt{x^2 + y^2}:
d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10\ units}.
Example 2: Points on a Horizontal Line
Find the distance between A(-2, 4) and B(5, 4).
Since the y-coordinates are the same, use the absolute difference of x:
d = |5 - (-2)| = |5 + 2| = \mathbf{7\ units}.
Example 3: Working with Negative Coordinates
Find the distance between C(-3, -1) and D(-7, -4).
d = \sqrt{(-7 - (-3))^2 + (-4 - (-1))^2}
d = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5\ units}.
Example 4: Calculating a Diagonal Path
Find the distance between X(1, 1) and Y(4, 5).
d = \sqrt{(4 - 1)^2 + (5 - 1)^2}
d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5\ units}.
Example 5: Points on a Vertical Line
Find the distance between M(3, -2) and N(3, 10).
Since the x-coordinates are the same, use the absolute difference of y:
d = |10 - (-2)| = |10 + 2| = \mathbf{12\ units}.
Example 6: Using the 3D Formula
Find the distance between R(1, 2, 3) and S(3, 4, 4).
d = \sqrt{(3 - 1)^2 + (4 - 2)^2 + (4 - 3)^2}
d = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = \mathbf{3\ units}.
Example 7: Distance with Decimal Coordinates
Find the distance between J(0.5, 2) and K(3.5, 6).
d = \sqrt{(3.5 - 0.5)^2 + (6 - 2)^2}
d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \mathbf{5\ units}.
Read More - Half Angle Formula with Examples
Understanding the distance between two points derivations helps in retaining the formula for the long term. The 2D formula is actually a direct application of the Pythagoras Theorem.
Steps for Derivation:
Assume two points P(x_1, y_1) and Q(x_2, y_2) on a coordinate plane.
Draw a horizontal line from P and a vertical line from Q until they meet at a point R, forming a right-angled triangle PQR.
The horizontal side (PR) length is (x_2 - x_1).
The vertical side (QR) length is (y_2 - y_1).
According to Pythagoras Theorem: PQ² = PR² + QR²
Substituting the values: PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.
Taking the square root gives us the final distance between two points formulas.
Let’s apply the logic to some real-world mathematical problems to see how these distance between two points examples work in practice.
Example 1: Basic 2D Distance
Find the distance between A(2, 3) and B(5, 7).
(x_1, y_1) = (2, 3)
(x_2, y_2) = (5, 7)
d = \sqrt{(5 - 2)^2 + (7 - 3)^2}
d = \sqrt{(3)^2 + (4)^2}
d = \sqrt{9 + 16} = \sqrt{25}
Distance = 5 units
Example 2: Negative Coordinates
Find the distance between P(-1, -1) and Q(2, 3).
d = \sqrt{(2 - (-1))^2 + (3 - (-1))^2}
d = \sqrt{(3)^2 + (4)^2}
d = \sqrt{9 + 16} = 5
Distance = 5 units
Read More - Perpendicular Bisector: Definition, Properties, and Practical Examples
Calculating the distance between two points is a systematic process. Whether you are working in a 2D or 3D plane, following these structured steps ensures accuracy in your results:
Identify the Coordinates: Label your first point as A(x_1, y_1) and your second point as B(x_2, y_2).
Apply the Formula: Input these values into the standard distance between two points formula: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
Simplify and Label: Solve the arithmetic within the square root and always express your final answer in "units".
Note: If you are working with a 3D plane, simply extend the calculation to include the z-axis usng the formula: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}.
Order doesn't matter: Because we square the differences (x_2 - x_1), the result is the same even if you calculate (x_1 - x_2).
Positive Result: Distance is a scalar quantity and is always positive.
Simplifying Radicals: Constantly try to simplify the square root at the end (e.g., \sqrt{8} becomes 2\sqrt{2}).
