Find the equation of line passing through the point \((\sqrt{3},-1)\) and at a distance \(\sqrt{2}\) units from the origin. (a) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}-(\sqrt{3}+1) \mathrm{y}=4\) (b) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (c) \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\) (d) \((\sqrt{3}-1) \mathrm{x}+(\sqrt{3}-1) \mathrm{y}=4\) or \((\sqrt{3}+1) \mathrm{x}+(\sqrt{3}+1) \mathrm{y}=4\)

Understanding the point-slope form of a line is essential when attempting to write an equation of a line in coordinate geometry. Simply put, if you know a single point through which the line passes and the slope of the line, you can write its equation. The general form of the point-slope equation is given by \( y - y_1 = m(x - x_1) \) where \( (x_1, y_1) \) are the coordinates of the known point on the line, and \( m \) is the slope of the line. This formulation provides a straightforward way to build the equation of a line, as it directly incorporates a point and the slope—two fundamental characteristics defining a unique line.

In the context of JEE Maths problems, the point-slope form is invaluable when the question involves finding an equation of a line with limited information. Once you identify the slope and a point, you're just a substitution away from getting your line's equation. Plus, the beauty of the point-slope form is its versatility; it can quickly adapt to different forms, such as slope-intercept or standard form, with some algebraic manipulation.

The distance formula is your go-to tool when it comes to finding the length of a segment connecting two points in the coordinate plane. It's derived from the Pythagorean Theorem and it is expressed as \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points. This formula is especially useful in numerous geometry problems where you need to establish relationships between different parts of geometric figures.

In JEE Maths, mastering the distance formula is crucial for solving problems related to coordinate geometry. It enables you to determine the lengths of sides within polygons, the radius of circles, and the distances between points—critical for proofs and for solving complex geometry problems with a coordinate approach.

The concept of perpendicular distance in coordinate geometry deals with finding the shortest distance from a point to a line. This is always given by the length of the perpendicular segment from the point to the line. The formula for this distance is given by \( d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \) where \( ax + by + c = 0 \) is the equation of the line and \( (x_1, y_1) \) is the point in question.

For students preparing for JEE Maths, understanding how to apply this formula is vital because it allows them to solve a range of problems, including those concerning loci, optimization, and even some aspects of calculus where the shortest distance can be involved. Knowing the perpendicular distance formula empowers students to analyze geometric situations involving points and lines more deeply, and to solve for unknown parameters that define the position and nature of geometric figures.