Locus: Definition, Theorem and Examples

Exams Prep Master

In geometry, a locus which is derived from the Latin word “location”, is a set of points that satisfies a specified condition or situation for a shape or figure. In other words, we can say that the set of the points that satisfy some property is called the locus of a point satisfying this property. The plural of the locus is loci, and the area covered by loci is referred to as the region. Geometric shapes, before the twentieth century, were considered as an entity or place where points can be located or can be moved. However, in modern Maths, the entities are considered as the set of points that satisfies the given condition. Here, we will be discussing more about the definition, theorem, examples of Locus along with some important questions.

Table of Content

What is Locus?
Six important locus theorem
Locus example in two-dimensional geometry
Things to Remember
Sample Questions

Also read: Perimeter and Area of a Circle

What is Locus?

In mathematics, a locus is a curve or shape created by all points satisfying a given equation of the connection between coordinates, or by a point, line, or moving surface. All geometries, including circles, ellipses, parabolas, and hyperbolas, are defined by the locus as a set of points. The word locus comes from the Latin word locus, which means "root." The term locus relates to a thing's location. When an object is placed somewhere or when something happens in a certain area, the locus is used to describe it.

A locus is a group of points that meet a set of criteria (typically forming a curve or surface). A circle is the locus of points in the plane that are equidistant from a given location, while a sphere is the set of points in a three-dimensional space that are equidistant from a given point.

Also read:

Circle	Area of Segment of a Circle
Semicircle	Geometric Progression

Locus of Points

In geometry, a shape is defined by the locus of points. Assume that a circle is the locus of all equidistant points from the centre. Similarly, the locus of the points defines various forms such as an ellipse, parabola, hyperbola, and so on. Only curved forms have a locus specified. These forms might be regular or asymmetrical. The forms with vertices or angles inside them are not characterised as loci.

Locus of a Circle

The locus is the collection of all points that satisfy the criteria and create geometrical forms such as a line, a line segment, a circle, a curve, and so on. So, rather than considering them as a series of points, we may think of them as locations where the point can be found or moved.

Read More: Trapezoid Formula

The circle is described in terms of the locus of the points or loci as the set of all points equidistant from a fixed point, where the fixed point is the circle's centre and the distance between the sets of points is the radius. Let's assume P is the circle's centre and r is the radius, or the distance between point P and the set of all points, or the locus of the points.

Discover about the Chapter video:

Conic Sections Detailed Video Explanation:

Also check:

Area of a Triangle	Tangent to a Circle
Minors and Cofactors	Properties of Determinants

Six Important Locus Theorems

There are six important locus theorems that are popular in geometry. Below are the six important theorems.

Read More: Trigonometry Table

Locus Theorem 1

The locus from the point “p” at the fixed distance “d” is considered as a circle with “p” as its center and “d” as its diameter.

With this theorem, we can determine the region formed by all the points which are located at the same distance from a single point

Locus Theorem 2

The locus from the line “m” at a fixed distance “d” is said to be a pair of parallel lines that are located on either side of “m” from the line “m” at a distance “d”.

We can find the region formed by all the points which are located at the same distance from the single line with the help of this theorem.

Locus Theorem 3

The locus that is equidistant from the two specified points say A and B, are considered as perpendicular bisectors of the line segment that joins the two points.

The region formed by all the points which are located at the same distance from point A and as from point B can be determined with the help of this theorem. The region formed have to be the perpendicular bisector of the line segment AB.

Read More:

Operations on Rational Numbers	Addition And Subtraction Of Integers	Multiplication And Division Of Integers

Locus Theorem 4

The locus that is equidistant from the two parallel lines say m1 and m2, is said to be a line parallel to both the lines m1 and m2 and it must be halfway between them.

This theorem helps to find out the region formed by all the points which are at the same distance from the two parallel lines.

Locus Theorem 5

The locus that is present on the interior of an angle and is equidistant from the sides of an angle is considered to be the bisector of the angle.

The region formed by all the points which are at the same distance from both sides of an angle can be determined with the help of this theorem. The region should be the angle bisector.

Locus Theorem 6

The locus that is equidistant from the two intersecting lines say m1 and m2, is considered to be a pair of lines that bisects the angle produced by the two lines m1 and m2.

This theorem helps to find the region formed by all the points which are located at the same distance from the two intersecting lines. The region formed should be a pair of lines that bisect the angle formed.

Also read:

Inverse Trigonometric Function	Area of Parallelogram	Introduction to Trigonometry

Locus Examples in Two-Dimensional Geometry

Here are some of the locus examples in two-dimensional geometry:

Perpendicular Bisector

The collection of points which bisects the line, moulded by joining two points and equally distant from two points is called perpendicular bisector.

Angle Bisector

A locus of collection of points that bisect an angle and are equally distant from two intersecting lines, which forms an angle is known as an angle bisector.

Ellipse

The ellipse is defined as a collection of points that fulfil the condition where the sum of the distances of two focal points is fixed.

Parabola

It is the collection of points that are equally distant from a fixed point and a line is known as a parabola. The fixed point is represented as the locus and the line is represented as the directrix of the parabola.

Hyperbola

A hyperbola has two distinct focal points which are equally distant from the centre of the semi-major axis. The collection of points fulfil the condition where the absolute value of the difference between the distances to two given foci is constant.

Things to Remember

Locus is a curve or shape created by all points satisfying a given equation of the connection between coordinates, or by a point, line, or moving surface.
All geometries, including circles, ellipses, parabolas, and hyperbolas, are defined by the locus as a set of points.
The locus of the points defines various forms such as an ellipse, parabola, hyperbola, and so on. Only curved forms have a locus specified. These forms might be regular or asymmetrical.
The circle is described in terms of the locus of the points or loci as the set of all points equidistant from a fixed point, where the fixed point is the circle's center and the distance between the sets of points is the radius.
A locus of collection of points that bisect an angle and are equally distant from two intersecting lines, which forms an angle is known as an angle bisector.

Also Read:

Limit Formula	Surface Area of a Cone Formula
Midpoint Formula	Geometry Formula

Sample Question

Ques: The sum of the intercept cut off from the axes of coordinates by a variable straight line is 10 units. Find the locus of the point which divides internally the part of the straight line intercepted between the axes of coordinates in the ratio 2 : 3. (4 marks)

Ans: Let us assume that the variable straight line at any position intersects the x-axis at A (a, 0) and the y-axis at B (0, b).

clearly, AB is the part of the line intercepted between the coordinates axes. Further assume that the point (h, k) divides the line-segment AB internally in the ratio 2 : 3. Then we have,

H = (2 · 0 + 3 · a)/(2 + 3)

or, 3a = 5h

or, a = 5h/3

And k = (2 · b + 3 · a)/(2 + 3)

or, 2b = 5k

or, b = 5k/2

Now, by problem,

A + b = 10

or, 5h/3 + 5k/2 = 10

or, 2h + 3k = 12

Therefore, the required equation to the locus of (h, k) is 2x + 3y = 12.

Read More:

Quadrilateral Formula	Trapezoid Formula	Tan2x Formula

Ques: For all values of the coordinates of a moving point P are (a cos θ, b sin θ); find the equation to the locus of P. (2 marks)

Ans: Let (x, y) be the coordinates of any point on the locus traced out by the moving point P. then we shall have,

x = a cos θ

or, x/a = cos θ

and y = b sin θ

or, y/b = sin θ

x²/a² + y²/b² = cos² θ + sin² θ

or, x²/a² + y²/b² = 1

Which is the required equation to the locus of P.

Ques: The co-ordinates of any position of a moving point P are given by {(7t – 2)/(3t + 2)}, {(4t + 5)/(t – 1)}, where t is a variable parameter. Find the equation to the locus of P. (4 marks)

Ans: Let (x, y) be the coordinates of any point on the locus traced out by the moving point P. then, we shall have,

x = (7t – 2)/(3t + 2)

or, 7t – 2 = 3tx + 2x

or, t(7 – 3x) = 2x + 2

or, t = 2(x + 1)/(7 – 3x) …………………………. (1)

And

y = (4t + 5)/(t – 1)

or, yt – y = 4t + 5

Or, t (y – 4) = y +5

or , t = (y + 5)/(y – 4)………………………….. (2)

From (1) and (2) we get,

(2x + 2)/(7 – 3x) = (y + 5)/( y – 4)

or, 2xy - 8x + 2y – 8 = 7y – 3xy + 35 – 15x

or, 5xy + 7x -5y = 43, which is the required education to the locus of the moving point P.

Read More:

Calculus Formula	Definite Integral Formula	Determinant Formula

Ques: Find the locus of a point that moves at a constant distance of two units above the x-axis. (2 marks)

Ans: Two units from the x-axis. In geometry, a locus is a set of all points whose location satisfies or is determined by one or more specified conditions.

The equation of a line that is 2 units above the x−axis for all points is y=2.

Ques: What is the equation of the locus of a point P, the square of the whose distance from the origin is 4 times its y coordinate. (3 marks)

Ans: Let the given origin be A ( 2,0)

Let us assume the point on the locus as P ( x,y)

The distance of P from X-axis = y

Given that OP² = 4 PM

√(X - 0)² + (Y - 0)²)² = 4y²

x² + y² - 4y = 0

Therefore, the equation of the locus of P (x,y) is

x²+ y² - 4y = 0

Ques: Prove that locus of centres of circles passing through points A and B is a perpendicular bisector of line segment AB. (4 marks)

Ans:

Let us assume that P and Q are the centers of two circle C and C, each passing through two given points A and B.

Then, PA = PB (radii of the circle C) ⇒ P that lies on the perpendicular bisector of AB ….(i)

Again QA = QB ⇒ Q also lies on the perpendicular bisector of AB … (ii)

From (i) and (ii) it can be summed up that P and Q both. lie on the perpendicular bisector of AB.

Therefore, the locus of the centres of all the circles passing through A and B is the perpendicular bisector of AB.

Read More:

Bayes Theorem Formula	Conditional Probability Formula	Differentiation and Integration Formula

Ques: In an Isosceles triangle show that the bisector of the angle formed between two similar sides is also an altitude, a median, and a perpendicular bisector of the side of that triangle. (3 marks)

Ans: In ΔABD and ACD,

AB = AC

∠BAD = ∠CAD [\(\because\) AD, is bisector of ∠A]

Again, AD = AD

Therefore, ΔABD ≅ ΔACD

Thus, BD = CD

AD is also a median

and ∠ADB = ∠ADC = 90°

Hence, AD is the perpendicular bisector of BC.

Read More: Differential Equation

Ques: In the figure given below, two isosceles triangles ΔPBC and ΔQBC lie on both sides of BC at common base BC. Prove that line joining P and Q bisects line BC at 90°. (5 marks)

Ans: Given = ΔPBC and ΔQBC are the two isosceles triangles which lie on both sides of base BC.

Here, PB = PC and BQ = CQ

To prove: ∠POB = ∠POC = 90°

or, ∠QOB = ∠QOC = 90°

In ΔPBC, PB = PC (Given)

Therefore, ∠PBO = ∠PCO (Equal sides)

PO = PO Common by S.A.S. congruence of

ΔPOB ≅ ΔPOC ⇒ ∠PBO = ∠POC ….(i)

As we know that ∠PBO + ∠POC = 180°

∠PBO + ∠POB = 180° [From equation (i)]

2∠POB = 180°

∠POB = 180°/2 = 90°

∠PBO = ∠POC = 90°

Similarly, ∠QOB = ∠QOC = 90°

Hence, PQ, bisects BC at 90°

Also Read:

Sin Squared x Formula with Solved Examples	Factorial Formula
Centroid of a Triangle	Algebra formulas

CBSE CLASS XII Related Questions

1.
Find:
Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).
- \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
- \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
- \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
- \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)
View Solution
2.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).
View Solution
3.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).
View Solution
4.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:
- \(-\frac{\pi}{2}\)
- \(-\frac{\pi}{4}\)
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
View Solution
5.
Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).
View Solution
6.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]
View Solution