# 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 ${x}^{2}+{y}^{2}={r}^{2}$ represents a circle's locus with radius $r$ centered at the origin.
3. Conic Sections:

• Examples: Ellipses, parabolas, hyperbolas.
• Equations: Equations like $\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ represent hyperbolic loci.
4. Parametric Equations:

• Definition: Loci can be represented using parametric equations, involving additional parameters.
• Example: Parametric equations like $x=\mathrm{cos}\left(t\right)$, $y=\mathrm{sin}\left(t\right)$ 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): $\sqrt{\left(x-\left(-3\right){\right)}^{2}+\left(y-4{\right)}^{2}}$
3. Distance between point P(x, y) and (5, -2): $\sqrt{\left(x-5{\right)}^{2}+\left(y-\left(-2\right){\right)}^{2}}$
• 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:

$\sqrt{\left(x+3{\right)}^{2}+\left(y-4{\right)}^{2}}=\sqrt{\left(x-5{\right)}^{2}+\left(y+2{\right)}^{2}}$

Squaring both sides to eliminate square roots:

$\left(x+3{\right)}^{2}+\left(y-4{\right)}^{2}=\left(x-5{\right)}^{2}+\left(y+2{\right)}^{2}$

Expanding and simplifying the equation:

${x}^{2}+6x+9+{y}^{2}-8y+16={x}^{2}-10x+25+{y}^{2}+4y+4$

Simplifying further:

$6x-10x+9-25+{y}^{2}-8y-4y+16-4=0$

$-4x-12y-4=0$

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.