Welcome to this module on a fundamental concept in coordinate geometry: the distance of a point from a line. This skill is crucial for solving various problems in competitive exams like JEE, particularly in calculus and algebra applications involving geometric constraints.

The distance of a point from a line is defined as the shortest distance between the point and any point on the line. This shortest distance is always along the perpendicular from the point to the line.

The formula for the distance of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is derived using vector methods or by finding the intersection point of the given line and the perpendicular line passing through the point.

Let's break down the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:

• $Ax_1 + By_1 + C$: This part of the numerator evaluates the line's equation at the coordinates of the given point. The sign of this expression indicates on which side of the line the point lies.
• $|Ax_1 + By_1 + C|$: The absolute value ensures that the distance is always non-negative, as distance cannot be negative.
• $\sqrt{A^2 + B^2}$: This is the magnitude of the normal vector to the line. Dividing by this normalizes the distance, making it independent of how the line's equation is written (e.g., $2x + 4y + 6 = 0$ vs. $x + 2y + 3 = 0$).

This formula is fundamental for various applications, including:

• Finding the distance from the origin (0,0) to a line: If the point is the origin, the formula simplifies to $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
• Determining if a point lies on a line: If the distance is 0, the point lies on the line.
• Calculating the distance between parallel lines: This involves finding the distance from any point on one line to the other line.
• Geometric problems involving areas and loci: Many problems require calculating distances to lines to define regions or constraints.

Find the distance of the point $(2, 3)$ from the line $3x - 4y + 5 = 0$.

Here, $(x_1, y_1) = (2, 3)$ and the line is $3x - 4y + 5 = 0$. So, $A = 3$, $B = -4$, and $C = 5$.

Using the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$:

$d = \frac{|3(2) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}}$

$d = \frac{|6 - 12 + 5|}{\sqrt{9 + 16}}$

$d = \frac{|-1|}{\sqrt{25}}$

$d = \frac{1}{5}$

To solidify your understanding, practice solving various problems involving the distance of a point from a line. Pay attention to different forms of line equations and how to convert them to the general form. Exploring problems that combine this concept with other geometric principles will enhance your problem-solving skills for competitive exams.