Introduction

In mathematics, the concept of loci deals with the set of all points that satisfy a particular condition or set of conditions. Loci are essential in geometry as they help us visualize and understand the relationships between points, lines, and shapes. Understanding loci is crucial for solving various geometric problems and real-life applications.

Basic Concepts

1. Locus of a Point

• Definition: The locus of a point is the set of all points that satisfy a given condition.

• Example: Find the locus of points that are equidistant from points A(-2, 3) and B(4, 1).

  $$ \text{Let a point } P(x, y) \text{ be equidistant from A and B.} \ \text{Using the distance formula:} \ \sqrt{(x + 2)^2 + (y - 3)^2} = \sqrt{(x - 4)^2 + (y - 1)^2} \ (x + 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 1)^2 \ x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1 \ 12x + 4y + 12 = 0 \ 3x + y + 3 = 0 $$

2. Locus of a Line

• Definition: The locus of a line is the set of all points through which a given line passes.

• Example: Find the locus of points equidistant from the lines $2x - y + 4 = 0$ and $x + 3y - 2 = 0$.

  $$ \text{Let the point be } P(x, y). \ \text{Using the perpendicular distance formula from a point to a line:} \ \text{Distance from } P \text{ to } 2x - y + 4 = 0: \frac{|2x - y + 4|}{\sqrt{2^2 + (-1)^2}} \ \text{Distance from } P \text{ to } x + 3y - 2 = 0: \frac{|x + 3y - 2|}{\sqrt{1^2 + 3^2}} \ \frac{|2x - y + 4|}{\sqrt{5}} = \frac{|x + 3y - 2|}{\sqrt{10}} \ 2(2x - y + 4)^2 = (x + 3y - 2)^2 $$

3. Locus of Intersecting Lines

• Definition: The locus of intersecting lines is the set of all points where two given lines intersect.

• Example: Find the locus of points equidistant from the lines $3x - 4y + 5 = 0$ and $5x + 12y - 1 = 0$.

  $$ \text{Let the point be } P(x, y). \ \text{The distance between } P \text{ and the line } 3x - 4y + 5 = 0: \frac{|3x - 4y + 5|}{\sqrt{3^2 + (-4)^2}} \ \text{The distance between } P \text{ and the line } 5x + 12y - 1 = 0: \frac{|5x + 12y - 1|}{\sqrt{5^2 + 12^2}} \ \frac{|3x - 4y + 5|}{5} = \frac{|5x + 12y - 1|}{13} \ 13|3x - 4y + 5| = 5|5x + 12y - 1| $$

4. Locus of a Circle

• Definition: The locus of a circle is the set of all points that are a fixed distance (radius) from a given point (center).

• Example: Find the equation of the locus of a point moving such that its distance from the point (3, -2) is always 5 units.

  $$ \text{Let the point be } P(x, y). \ \text{The distance formula from } P \text{ to (3, -2) is } \sqrt{(x-3)^2 + (y+2)^2} = 5 \ (x-3)^2 + (y+2)^2 = 5^2 \ x^2 - 6x + 9 + y^2 + 4y + 4 = 25 \ x^2 + y^2 - 6x + 4y - 12 = 0 $$

Common Mistakes

• Forgetting to consider both positive and negative square roots when finding the locus of a point.
• Incorrectly applying the distance formula when finding the locus of a point or line.
• Failing to simplify equations correctly when finding the locus of intersecting lines or circles.

Key Points

• Loci represent sets of points that satisfy specific conditions.
• The distance formula is essential when dealing with loci of points and lines.
• Loci problems can involve finding equations of lines, circles, or other geometric shapes.

Practice Questions

1. Find the locus of points equidistant from the lines $4x + 3y - 7 = 0$ and $3x - 4y + 8 = 0$.

   Answer: $$ 7(4x + 3y - 7)^2 = 8(3x - 4y + 8)^2 $$

2. Determine the locus of points equidistant from the points A(-1, 2) and B(3, -5).

   Answer: $$ x^2 + 6x + y^2 - 3y - 15 = 0 $$

3. If the locus of points equidistant from the lines $2x + 5y - 3 = 0$ and $3x - 4y + 9 = 0$ is a circle, find its center and radius.

   Answer: $$ \text{Center: }(1, 2) \text{, Radius: } \sqrt{10} $$

4. Find the locus of points which are equidistant from the point (0, 1) and the line $2x - y + 4 = 0$.

   Answer: $$ x^2 + y^2 - 2y - 1 = 0 $$

5. Determine the locus of points equidistant from the lines $x - 2y + 3 = 0$ and $2x + y - 4 = 0$.

   Answer: $$ 5(x - 2y + 3)^2 = 4(2x + y - 4)^2 $$