Distance of a point from a line
Distance of a Point from a Line
The distance of a point from a line is the shortest distance between the point and any point on the line. This distance is always measured along the perpendicular from the point to the line.
Formula for Distance of a Point from a Line
Given a line in the Cartesian plane with the equation ( Ax + By + C = 0 ), and a point ( P(x_1, y_1) ), the distance ( d ) from the point to the line is given by the formula:
[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} ]
This formula is derived from the general equation of a line and the concept of the dot product in vector algebra.
Steps to Calculate the Distance
- Identify the equation of the line in the form ( Ax + By + C = 0 ).
- Identify the coordinates of the point ( (x_1, y_1) ).
- Substitute the coordinates of the point into the distance formula.
- Calculate the numerator and denominator separately.
- Divide the absolute value of the numerator by the square root of the sum of the squares of ( A ) and ( B ).
- The result is the distance from the point to the line.
Table of Differences and Important Points
| Aspect | Description |
|---|---|
| Equation of Line | Standard form: ( Ax + By + C = 0 ) |
| Point Coordinates | Given point ( P(x_1, y_1) ) |
| Distance Formula | ( d = \frac{ |
| Perpendicularity | The distance is measured along the line perpendicular to the given line from the point. |
| Units | The distance is measured in the same units as the coordinates of the point and the line. |
| Significance of ( A ) and ( B ) | Coefficients ( A ) and ( B ) determine the slope of the line. |
| Absolute Value | The numerator is taken as an absolute value to ensure the distance is non-negative. |
| Application | Used in geometry, physics, engineering, and computer graphics for various calculations. |
Examples
Example 1: Horizontal Line
Consider the horizontal line ( y = 3 ) and the point ( P(4, 5) ). The equation of the line can be rewritten as ( 0x + 1y - 3 = 0 ). Using the distance formula:
[ d = \frac{|0 \cdot 4 + 1 \cdot 5 - 3|}{\sqrt{0^2 + 1^2}} = \frac{|2|}{1} = 2 ]
The distance from the point ( P ) to the line is 2 units.
Example 2: Vertical Line
Consider the vertical line ( x = -2 ) and the point ( P(1, -1) ). The equation of the line can be rewritten as ( 1x + 0y + 2 = 0 ). Using the distance formula:
[ d = \frac{|1 \cdot 1 + 0 \cdot (-1) + 2|}{\sqrt{1^2 + 0^2}} = \frac{|3|}{1} = 3 ]
The distance from the point ( P ) to the line is 3 units.
Example 3: Oblique Line
Consider the line ( 3x + 4y - 11 = 0 ) and the point ( P(1, 2) ). Using the distance formula:
[ d = \frac{|3 \cdot 1 + 4 \cdot 2 - 11|}{\sqrt{3^2 + 4^2}} = \frac{|-1|}{5} = \frac{1}{5} ]
The distance from the point ( P ) to the line is ( \frac{1}{5} ) units or 0.2 units.
Conclusion
The distance of a point from a line is a fundamental concept in geometry that is used in various fields. By understanding the formula and how to apply it, one can solve problems related to the shortest distance between a point and a line. Remember to always use the perpendicular distance for the most accurate measurement.