Locus and its Equation

In coordinate geometry, a locus refers to the set of all points that satisfy a given geometric condition or property. Think of it as tracing a path or a shape formed by points that behave in a specific way. The equation that represents this set of points is called the equation of the locus. Understanding loci is fundamental to describing curves, lines, and other geometric figures using algebraic expressions.

Understanding the Concept of Locus

Imagine a point that moves according to certain rules. The path traced by this moving point is its locus. For example, consider a point that always maintains a fixed distance from a fixed point. The locus of this moving point is a circle. The fixed point is the center, and the fixed distance is the radius. Similarly, a point equidistant from two fixed points traces a straight line, which is the perpendicular bisector of the segment joining the two fixed points.

The geometric condition can be simple or complex. It might involve distances, angles, or relationships between different points and lines. The key is that all points on the locus must satisfy the given condition, and any point not satisfying the condition will not be on the locus.

Steps to Find the Equation of a Locus

To find the equation of the locus of a point P, we follow a systematic procedure:

1. Assume a General Point: Let P(x, y) be any arbitrary point on the locus. This point P must satisfy the given geometric condition.
2. Express the Geometric Condition Algebraically: Translate the given geometric condition into an algebraic equation involving the coordinates (x, y) of point P and the coordinates of any fixed points or lengths involved. This usually involves using distance formulas, section formulas, slope formulas, or other relevant geometric relationships.
3. Simplify the Equation: Simplify the algebraic equation obtained in the previous step. Eliminate any parameters or variables that are not x or y. The resulting equation, which relates x and y, is the equation of the locus.

Examples of Locus Problems

Example 1: Locus of a point equidistant from two fixed points.

Find the locus of a point P which is equidistant from two fixed points A(2, 3) and B(4, 1).

Step 1: Let P(x, y) be any point on the locus.

Step 2: The condition is that P is equidistant from A and B. This means the distance PA is equal to the distance PB. We use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

$PA = \sqrt{(x-2)^2 + (y-3)^2}$

$PB = \sqrt{(x-4)^2 + (y-1)^2}$

Since PA = PB, we have $PA^2 = PB^2$ (squaring both sides to avoid square roots).

$(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-1)^2$

Step 3: Simplify the equation.

Expand both sides:

$(x^2 - 4x + 4) + (y^2 - 6y + 9) = (x^2 - 8x + 16) + (y^2 - 2y + 1)$

$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1$

Cancel out $x^2$ and $y^2$ terms from both sides:

$-4x + 4 - 6y + 9 = -8x + 16 - 2y + 1$

Combine constant terms:

$-4x - 6y + 13 = -8x - 2y + 17$

Move all terms to one side to get the equation in the standard form:

$(-4x + 8x) + (-6y + 2y) + (13 - 17) = 0$

$4x - 4y - 4 = 0$

Divide by 4:

$x - y - 1 = 0$

This is the equation of the locus. Geometrically, this is the perpendicular bisector of the line segment AB.

Example 2: Locus of a point moving such that its distance from a fixed point is constant.

Find the locus of a point P that moves such that its distance from a fixed point C(h, k) is always equal to a constant 'r'.

Step 1: Let P(x, y) be any point on the locus.

Step 2: The distance PC is equal to r. Using the distance formula:

$PC = \sqrt{(x-h)^2 + (y-k)^2}$

So, $\sqrt{(x-h)^2 + (y-k)^2} = r$

Step 3: Simplify the equation. Square both sides:

$(x-h)^2 + (y-k)^2 = r^2$

This is the equation of a circle with center (h, k) and radius r. This confirms our initial geometric intuition.

Example 3: Locus of a point equidistant from a fixed point and a fixed line.

Find the locus of a point P which moves such that its distance from the fixed point A(a, 0) is always equal to its distance from the fixed line x = -a.

Step 1: Let P(x, y) be any point on the locus.

Step 2: The distance from P to the fixed point A(a, 0) is $PA = \sqrt{(x-a)^2 + (y-0)^2}$.

The distance from P(x, y) to the fixed line $x = -a$ (or $x+a=0$) is given by the formula for the distance from a point to a line: $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Here, $A=1, B=0, C=a$, and $(x_1, y_1) = (x, y)$.

Distance from P to line $x+a=0$ is $\frac{|1 \cdot x + 0 \cdot y + a|}{\sqrt{1^2 + 0^2}} = \frac{|x+a|}{1} = |x+a|$.

The condition is $PA = |x+a|$. So, $\sqrt{(x-a)^2 + y^2} = |x+a|$.

Step 3: Simplify the equation. Square both sides:

$(x-a)^2 + y^2 = (x+a)^2$

$x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$

Cancel out $x^2$ and $a^2$ terms:

$-2ax + y^2 = 2ax$

Rearrange the terms:

$y^2 = 2ax + 2ax$

$y^2 = 4ax$

This is the equation of a parabola with focus (a, 0) and directrix x = -a.

Types of Loci and Their Equations

The locus of a point can form various geometric shapes, each with a characteristic equation. Here are some common ones:

Geometric Shape	Geometric Condition	Equation of Locus	Notes
Straight Line	Point equidistant from two fixed points.	$ax + by + c = 0$	Perpendicular bisector.
Straight Line	Point equidistant from two parallel lines.	Equation of the line exactly in the middle.	Parallel to the given lines.
Circle	Point at a fixed distance from a fixed point.	$(x-h)^2 + (y-k)^2 = r^2$	Center (h, k), radius r.
Parabola	Point equidistant from a fixed point (focus) and a fixed line (directrix).	$y^2 = 4ax$ (standard form)	Opens right if 'a' is positive.
Ellipse	Sum of distances from the point to two fixed points (foci) is constant.	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (standard form)	Sum of distances = 2a.
Hyperbola	Difference of distances from the point to two fixed points (foci) is constant.	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (standard form)	Difference of distances = 2a.

Important Considerations and Techniques

When solving locus problems, keep these points in mind:

• Square Roots: Always try to eliminate square roots by squaring both sides of the equation. This simplifies the algebra significantly.
• Absolute Values: When dealing with distances to lines or other situations involving absolute values, squaring both sides is also a common technique to remove the absolute value. For example, $|x| = a$ becomes $x^2 = a^2$.
• Parameters: If the geometric condition involves parameters (like distances, slopes, etc.), ensure your final equation of the locus does not contain these parameters. They should be eliminated during simplification.
• Degenerate Cases: Sometimes, the locus might turn out to be a point, a pair of lines, or even empty. Be aware of these possibilities. For example, if the condition leads to $(x-1)^2 + (y-2)^2 = 0$, the only point satisfying this is (1, 2).
• Geometric Interpretation: Always try to visualize the locus. Knowing the geometric shapes associated with common conditions can help you anticipate the form of the equation.

Locus Problems Involving Ratios and Slopes

Example 4: Locus involving a ratio of distances.

Find the locus of a point P such that its distance from point A(1, 2) is twice its distance from point B(3, 4).

Step 1: Let P(x, y) be the point on the locus.

Step 2: The condition is PA = 2 * PB.

$PA^2 = (x-1)^2 + (y-2)^2$

$PB^2 = (x-3)^2 + (y-4)^2$

So, $PA^2 = (2 \cdot PB)^2 = 4 \cdot PB^2$.

$(x-1)^2 + (y-2)^2 = 4[(x-3)^2 + (y-4)^2]$

Step 3: Simplify.

$x^2 - 2x + 1 + y^2 - 4y + 4 = 4[x^2 - 6x + 9 + y^2 - 8y + 16]$

$x^2 - 2x + y^2 - 4y + 5 = 4[x^2 - 6x + y^2 - 8y + 25]$

$x^2 - 2x + y^2 - 4y + 5 = 4x^2 - 24x + 4y^2 - 32y + 100$

Move all terms to the right side:

$0 = (4x^2 - x^2) + (-24x + 2x) + (4y^2 - y^2) + (-32y + 4y) + (100 - 5)$

$0 = 3x^2 - 22x + 3y^2 - 28y + 95$

$3x^2 + 3y^2 - 22x - 28y + 95 = 0$

This is the equation of a circle.

Example 5: Locus involving slopes.

Find the locus of a point P(x, y) such that the slope of the line segment PA is twice the slope of the line segment PB, where A = (1, 3) and B = (2, 4).

Step 1: P(x, y), A(1, 3), B(2, 4).

Step 2: Slope of PA ($m_{PA}$) = $\frac{y-3}{x-1}$.

Slope of PB ($m_{PB}$) = $\frac{y-4}{x-2}$.

The condition is $m_{PA} = 2 \cdot m_{PB}$.

$\frac{y-3}{x-1} = 2 \left( \frac{y-4}{x-2} \right)$

Step 3: Simplify.

$(y-3)(x-2) = 2(y-4)(x-1)$

Expand both sides:

$xy - 2y - 3x + 6 = 2(xy - y - 4x + 4)$

$xy - 2y - 3x + 6 = 2xy - 2y - 8x + 8$

Move all terms to the right side:

$0 = (2xy - xy) + (-2y + 2y) + (-8x + 3x) + (8 - 6)$

$0 = xy - 5x + 2$

So, the equation of the locus is $xy - 5x + 2 = 0$.

Locus in Different Coordinate Systems

While we've focused on Cartesian coordinates (x, y), loci can also be described in polar coordinates (r, θ) or other systems. The method of finding the equation remains similar:

1. Polar Coordinates: Let P be a point with polar coordinates (r, θ). Express the geometric condition using r, θ, and any fixed polar coordinates or distances/angles.

For example, the locus of points at a fixed distance 'a' from the origin is $r = a$ in polar coordinates, which is a circle centered at the origin with radius 'a'.

Common Pitfalls and How to Avoid Them

1. Incorrectly applying distance formula: Double-check the coordinates and the subtraction order. $(x_1-x_2)^2 = (x_2-x_1)^2$, so the order within the square doesn't matter, but ensure you subtract corresponding coordinates. 2. Algebraic errors during expansion or simplification: Be meticulous with expanding brackets and combining like terms. It's helpful to write down each step clearly. 3. Forgetting to square both sides: When dealing with distances, you'll often have square roots. Squaring is essential to get a polynomial equation. 4. Not removing parameters: Ensure the final equation only contains x and y, not any arbitrary constants introduced by the problem statement (unless they are fixed points/lines). 5. Misinterpreting the geometric condition: Carefully read the condition. "Equidistant" means equal distances. "Twice the distance" means multiply one distance by 2.

Locus and Conic Sections

Many important loci in mathematics are conic sections: circles, ellipses, parabolas, and hyperbolas. Understanding the definitions of these curves in terms of loci is crucial.

• Circle: Locus of points equidistant from a fixed point.
• Parabola: Locus of points equidistant from a fixed point and a fixed line.
• Ellipse: Locus of points where the sum of distances to two fixed points is constant.
• Hyperbola: Locus of points where the difference of distances to two fixed points is constant.

The ability to derive the standard equations of these conics from their locus definitions is a key skill tested in exams.

Summary of the Locus Method

To solve any locus problem:

Mastering this process will allow you to confidently tackle a wide range of locus problems in your examinations.