Locus

Loci

Consider a point P (x,y) which is free to move according to certain conditions. The set of points which satisfy the conditions is called the locus of P and the equation which is satisfied by all these points is also called the locus of point P.

Distance formula

 $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)}$  

This formula is used when finding the distance from the point P(x,y) to a given point. 

Examples

1. What is the locus of a point which moves so that its distance from the point ( 3, 1) is 2 units.

P (x , y ) - is a variable point
Let A = (3, 1)
PA is the distance from the variable point P to A. (2 units) 
To find equation of locus using points (x,y) and (3,1) : 
 $\therefore PA=2$  

 $\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}=2$  
 $Squ\ both\ sides$  
 $\left(x-3\right)^2+\left(y-1\right)^2=4$  

The equation of the locus is 
 $\left(x-3\right)^2+\left(y-1\right)^2=4$  

Nte : In this question the locus is a set if points of a circle with  C(3,1) and radius 2

2.

Find the equation of the locus of a point P which moves so that it is equidistant from 2 fixed points A and B whose coordinates are (3 ,2 ) and ( 5 , -1) respectively.

Note: Equidistant means the same distance

P (x , y) is a variable point
A (3 , 2)
B ( 5 , -1)

The distance PA is equal to the distance PB. To find equation of locus: 
 $\therefore PA=PB$  

 $\sqrt{\left(x-3\right)^2+\left(y-2\right)^2}=\sqrt{\left(x-5\right)^2+\left(y+1\right)^2}$  
 $Squ\ both\ sides$  
 $\left(x-3\right)^2+\left(y-2\right)^2=\left(x-5\right)^2+\left(y+1\right)^2$  
 $x^2-6x+9+y^2-4y+4=x^2-10x+25+y^2+2y+1$  
 $-6x+10x-4y-2y+9+4-25-1=0$  
 $4x-6y-13=0$  

3.

Find the equation of a point P whose distance from the point A (-1 , 2) is twice its distance from the origin.

P (x , y )
A ( -1 ,2 )
O (0,0)
The distance PA is twice the distance of PO. To find the equation of locus : 

 $PA\ =2PO$  
 $\sqrt{\left(x+1\right)^2+\left(y-2\right)^2}\ =2\ \times\ \sqrt{\left(x-0\right)^2+\left(y-0\right)^2}$  
 $Square\ both\ sides$  
 $\left(x+1\right)^2+\left(y-2\right)^2=4\left(x^2+y^2\right)$  
 $x^2+2x+1+y^2-4y+4=4x^2+4y^2$  
 $4x^2-x^2+4y^2-y^2-2x+4y-5=0$  
 $3x^2+3y^2-2x+4y-5=0$  

4.

Find the locus of a point which is equidistant from the origin and the line x=-1.

P (x , y) | x=-1 = (-1 , y ) | O (0,0)

Distance from P to line x = -1 = Distance of P from O
 $\sqrt{\left(x+1\right)^2+\left(y-y\right)^2\ \ }=\sqrt{\left(x-0\right)^2+\left(y-0\right)^2}$  

 $Square\ both\ sides$  
 $\left(x+1\right)^2=x^2+y^2$  
 $x^2+2x+1=x^2+y^2$  
 $y^2=2x+1$  

5.

Find the locus of a point which is equidistant from the point (0 , 1) and the line y = -1.

P(x,y) | Let A = (0,1) | y= -1 = (x, -1)
 Distance from P to A =  Distance from P to y =-1

 $\sqrt{\left(x-0\right)^2+\left(y-1\right)^2}\ =\sqrt{\left(x-x\right)^2+\left(y+1\right)^2}$  
Square both side
 $x^2+\left(y-1\right)^2=\left(y+1\right)^2$  
 $x^2+y^2-2y+1=y^2+2y+1$  
 $x^2-4y=0$  
 $The\ locus\ is\ 4y=x^2$