Locus

Locus in Coordinate Geometry:

Definition:

  • Concept: A locus represents a set of points that satisfy a particular condition or property.
  • In Coordinate Geometry: The locus is described as the path traced by a point (or points) that satisfy a given condition.

Representation:

  • Equation: The locus can often be represented by an equation involving the coordinates x and y that expresses the relationship between them.

Types of Loci:

  1. Linear Loci:

    • Examples: Lines, rays, line segments.
    • Equation: For example, the equation of a line y=mx+c represents a line's locus.
  2. Circular Loci:

    • Examples: Circles, arcs.
    • Equation: For instance, the equation x2+y2=r2 represents a circle's locus with radius r centered at the origin.
  3. Conic Sections:

    • Examples: Ellipses, parabolas, hyperbolas.
    • Equations: Equations like x2a2y2b2=1 represent hyperbolic loci.
  4. Parametric Equations:

    • Definition: Loci can be represented using parametric equations, involving additional parameters.
    • Example: Parametric equations like x=cos(t), y=sin(t) represent a locus of a point moving along a unit circle as t varies.

Application:

  • Problem Solving: Helps in solving geometric problems involving the path or trajectory of points.
  • Curve Sketching: Useful for visualizing and understanding shapes described by equations.

Example:

Find the locus of points that are equidistant from the points (-3, 4) and (5, -2).

Steps to Solve:

  1. Distance Formula:

    • Consider a point P(x, y) equidistant from (-3, 4) and (5, -2).
    • Use the distance formula to set up equations for the distances between P and each given point.
  2. Equation Setting:

    • Distance between point P(x, y) and (-3, 4): (x(3))2+(y4)2
  3. Distance between point P(x, y) and (5, -2): (x5)2+(y(2))2
    • Equate these distances to find the locus.
  1. Solve for Locus:

    • Set the two equations equal to each other and solve for the locus equation in terms of x and y.

Solution:

Let's equate the distances to find the locus equation:

(x+3)2+(y4)2=(x5)2+(y+2)2

Squaring both sides to eliminate square roots:

(x+3)2+(y4)2=(x5)2+(y+2)2

Expanding and simplifying the equation:

x2+6x+9+y28y+16=x210x+25+y2+4y+4

Simplifying further:

6x10x+925+y28y4y+164=0

4x12y4=0

Finally, rearranging the equation to isolate the locus:

x+3y+1=0

Interpretation:

The locus of points equidistant from (-3, 4) and (5, -2) is represented by the equation x+3y+1=0 in the coordinate plane. This equation describes a straight line passing through the points (-3, 4) and (5, -2), illustrating the path traced by points equidistant from these two given points.