Locus

Subject: Optional Mathematics

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Overview

To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers there corresponds a point on a plane. If we have an algebraic relation between x and y, we get a set of points in the plane for different values of x and y satisfying the relation. If we a set of points determined by some geometrical condition, we can represent it by an algebraic relation. This algebraic relation is called equation and the set of points in the plane is called Locus.

Locus
Locus
Locus

To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers, there corresponds a point on a plane. If we have an algebraic relation between x and y, we get a set of points in the plane for different values of x and y satisfying the relation. If we move a set of points determined by some geometrical condition, we can represent it by an algebraic relation. This algebraic relation is called equation and the set of points in the plane is called Locus.

The locus of a point is the path traced out by the point moving under given geometrical condition( or conditions). Alternatively, the locus is the set of all those points, which satisfy the given geometrical conditions (or conditions).

For example:

  1. Let a point P move such that its distance from a fixed line is always equal to d. The point P will trace out a straight line CD parallel to the fixed line. Thus, the locus of the moving point P is the straight line.
  2. Consider the set of all points on the x-axis. This is the locus of a point whose ordinate is zero.
  3. Let a point P move in a plane such that its distance from a fixed point, says B, is always equal to r. The point P will trace out a circle with centre B and radius r. Thus, the locus of the moving point is the circle.
  4. Consider the locus of a point whose distance from the origin is 5 units. If the coordinates of any points satisfying the condition be (x,y), then (x-0)2+(y-0)2=52 is the equation of the locus.Hence, the equation of the locus of a point whose distance from the origin is equal to 5 units is x2+y2=25.The locus obviously is a circle with centre at the origin and radius 5 units.

Remarks

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  1. Every point which satisfies the given geometrical conditions (or conditions) lies on the locus.
  2. A point which does not satisfy the given geometrical condition ( or conditions ) cannot lie on the locus.
  3. Every point which lies on the locus satisfies the given geometrical condition (or conditions).
  4. A point which does not lie on the locus cannot satisfy the given geometrical condition (or conditions).
  5. The locus of a point moving in a plane under a given geometrical conditions is always a straight or a curved line.
  6. To find the locus of moving points, plot some points satisfying the given geometrical condition, and then join these points.

To find the equation of a set of points (locus) with given geometrical conditions, the working rule is

  1. Suppose (x, y) be any points in the set,
  2. Find a relation between x and y satisfying the given conditions.
  3. The relation is the required equation of the locus.

Notes

  1. If any point belongs to the Locus, then the point satisfies the equation of the locus.
  2. Any point satisfying the equation of Locus must lie on the Locus.
  3. Any point out the locus does not satisfy the equation of the locus.

Methods of Finding the Equation of Locus

  1. Take P(x,y) as a moving point in the coordinate plane. Represent the information in a suitable figure.
  2. Find the distances according to the information provided.
  3. Simplify the expressions by using algebraic operations. Make sure that there is no common number in all the terms. Also, make sure that first term is not negative. Final result or equation so obtained is the equation of the locus.


Things to remember
  • To every point in a plane there corresponds on ordered pair of real numbers and to every ordered pair of real numbers there corresponds a point on a plane.
  • The locus of a point is the path traced out by the point moving under given geometrical condition (or conditions). 

 

  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Locus
Complex Numbers - Loci : Perpendicular bisector : ExamSolutions Maths Video Tutorials
Introduction To Locus / Coordinate Geometry / Maths Geometry
Locus
Questions and Answers

Soln: Let P(x,y) be a point which moves at a distance 5 units from the point(0,0)

figure
figure

Then, OP=5 units

or, \(\sqrt{(x-0)^2+(y-0)^2}\)=5

or, \(\sqrt{x^2+y^2}\)=5

Squaring both side,x2+y2-25=0 Ans.

Soln: Let the moving point be P(x,y) and given points

figure
figure

A(a,0) and B(0,b)

By giving ,PA=PB

or, (x-a)2+(y-0)2=(x-0)2+(y-b)2

or, x2-2ax+a2+y2=x2+y2-2by+b2

∴2ax-2by-a2+b2=0 is the required locus.Ans.

Here, Let P(x,y) be a points equidistance from the points A(-2,1) and B(4,1), then

we have

figure
figure

Now, PA=PB

PA2=PB2

or, (x2+2)+(y-1)2=(x-4)2+(y-1)2

or, x2+4x+4+y2-2y+1=x2-8x+16+y2-2y+1

or, 12x-12=0

or, 12(x-1)=0

∴ x-1=0 is the required locus.

Substituting the point (1,1) on this locus we get,

1-1=0

∴ 0=0

Hence the point (1,1) lies on the locus.

Here, Let the moving point be P(x,y)

Distance of x-axis= x

Distance of y-axis =y

Then by given 3x + 2 =y

The required locus is 3x + 2 =y

Substituing the point (1,5) in the locus, we get

3.1 - 5 + 2=0

∴ 0 =0 which is true

∴ (1,5) lies on this locus.

Soln;

figure
figure

Let the given points are A(0,-2) and B(1,0).

Let P(x,y) be a point which moves in such a way that 2AP=PB

or, 4AP2=PB2

or, 4[(x-0)2+[y-(-2)]2=(x-1)2+(y-0)2

or, 4(x2+y2+4y+4)=x2-2x+1+y2

or, 4(x2+4y2+16y+16-x2+2x-1-y2=0

or, 3x2+3y2+2x+16y+15=0

∴Required locus3x2+3y2+2x+16y+15=0 Ans.

Here,

Let the point on y-axis be B(0,y).

Let P(x,y) be a point which is equidistance from A and B.

By given,PA=PB

or, PA2=PB2

or, (x-z)2+(y-0)2=(x-0)2+(y-y)2

or, x2-2xz+x2+y2=x2+0

or, y2-2xz+z2=0

∴y2-2xz+z2=0 is required locus,Ans.

Soln:

Given point, (B,0). Let it be A(B,0). Let the points on y-axis be Q(0,y).Let the moving point be P(x,y), then by given

2PA=PQ

or, 4 PA2=PQ2

or, 4[(x-B)2+(y-0)2]

= (x-0)2+(y-y)2

or, 4(x2-2Bx+B2+y2)=x2+0

or, 4x2-8Bx+4B2+4y2=x2

∴ 3x2+4y2-8Bx+4B2=0 is the required locus. Ans.

Here, given points A(2,3) and B(-4,7) and let the moving points be,P(x,y)

PA=PB

or, PA2=PB2

or, (x-2)2+(y-3)2=(x+4)2+(y-7)2

or, x2-4x+4+y2-6y+9=x2+8x+16+y2-14y+49

or, 12x-8y+52=0

or, 4(3x-2y+13)=0

∴ 3x-2y+13=0 is required locus.Ans.

Here, PA=2PB

or, PA2=4PB2

or, (x-2)2+(y-3)2=4[(x+4)2+(y-7)2]

or, x2-4x+4+y2-6y+9=4[x2+8x+16+ y2-14y+49]

or, x2-4x+y2-6y+13=4x2+32x+64+4y2-56y+196

or, 3x2+3y2+36x-50y-247=0

∴ 3x2+3y2+36x-50y-247=0 is the required locus.Ans.

Here given, PA2=AB2

or, (x-2)2 +(y-3)2=(-4-2)2+(7-3)2

or, x2-4x+4+y2-6y+9=(-6)2+(4)2

or, x2+y2-4x-6y+13=36+16

∴ x2+y2-4x-6y-39=0 is required locus.Ans.

Here, Let the two points be A (1,-2) and B(-3,-4)

Let,P(x,y) be a point equidistance from points A and B.

By given,PA=PB

or, PA2=PB2

or, \(\sqrt {(x-1)^{2} + [y - (-2)]^{2}}\))2

=\(\sqrt {[x--(3)]^{2} + [y - (-4)]^{2}}\))2

or, (x-1)2+(y+2)2=(x+3)2+(y+4)2

or, x2-2x+1+y2+4y+4=x2+6x+9+y2+8y+16

or, x2-2x+1+y2+4y+4-x2-6x-9-y2-8y-16=0

or, -8x-4y+5-25=0

or, -8x-4y-20=0

or, -4(2x+y+5)=0

∴ Required locus is 2x+y+5=0.Ans.

Quiz

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