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:
Remarks
To find the equation of a set of points (locus) with given geometrical conditions, the working rule is
Notes
Methods of Finding the Equation of Locus
.
Soln: Let P(x,y) be a point which moves at a distance 5 units from the point(0,0)
Then, OP=5 units
or, \(\sqrt{(x-0)^2+(y-0)^2}\)=5
or, \(\sqrt{x^2+y^2}\)=5
Squaring both side,x^{2}+y^{2}-25=0 Ans.
Examine which of the points (5,7) and (6,-2) lie on the locus with equation x^{2}+y^{2}-4x-6y=12.
(7,8)
(9,2)
(2,2)
(5,7)
Find the value of k so that the point (4,5) lies in the locus of x^{2}+y^{2}+kx-8y-25=0
Find the value of k so that the point (3,2) lies in the locus of x^{2}+y^{2}+ky=21
Find the value of m so that the point (-2,0) lies in the locus of x^{2_}mx +5y=18.
4
8
7
2
Find the equation of a locus of a point, which is equidistant from the point (1,2) and (2,-3).
x-5y=3
x-5y=4
x-9y=8
x-4y=8
A point P moves so that its distance from the two points (3,4) and (5,-2) are equal to one another.Find the equation of the locus of P.
x+3y-1=0
x-3y-2=0
x-3y+1=0
x-3y-1=0
A point moves so that its distance from the point (3,2) is always twice its distance from the point (2,1).Find the equation of a locus.
3x^{2}+3y^{2}-9x-4y+7=0
3x^{2}+3y^{2}-10x-4y+7=0
3x^{2}+3y^{2 }+10x-4y+7=0
4x^{2}+3y^{2}-10x-4y+7=0
Find the equation of the locus of a point whose distance from (-1,1) is equal to the twice its distance from the X-axis.
x^{2}-3y^{2}+2x-2y-2=0
x^{2}-3y^{2}+2x-2y+2=5
x^{2}-2y^{2}+2x-2y+2=0
x^{2}-3y^{2}+2x-2y+2=0
Find the equation of the locus of a point which moves so that its distance from the point (0,2) is one-third of its distance from the point (3,0).
7x^{2}+8y^{2}+6x-36y+7=0
8x^{2}+8y^{2}+6x-36y+27=0
8x^{2}+5y^{2}+6x+36y+27=0
8x^{2}+8y^{2}+6x-36y-22=0
A point moves so that the ratio of its distance from the point( -a,0) and (a,0) is 2:3.Find the equation of its locus.
5x^{2}+5y^{2}+26ax+5a^{2}=0
5x^{2}+5y^{2}-26ax+5a^{2}=0
5x^{2}+5y^{2}+26ax+4a^{2}=0
2x^{2}+2y^{2}+26ax+5a^{2}=0
You must login to reply
Sep 03, 2017
0 Replies
Successfully Posted ...
Please Wait...
Sep 03, 2017
0 Replies
Successfully Posted ...
Please Wait...
Ajk
There's a question which I find rather confusing Prove that the locus of a point which moves in equidistance from points (3,0) and(-3,0) lies on y axis
Jan 18, 2017
1 Replies
Successfully Posted ...
Please Wait...