Question

The equation of the locus of the points equidistant from the points A(-2,3) B(6,-5) is

Answer 1

Answer:

Let an arbitrary point be P(x,y),

Then it is given that

PA=PB

or

PA  2  =PB  2

 

(x+2)  2  +(y−3)  2  =(x−6)  2  +(y+5)  2  

(x+2)  2  −(x−6)  2  =(y+5)  2  −(y−3)  2

 

(2x−4)(8)=(2y+2)(8)

2x−4=2y+2

x−2=y+1

x=y+3  or  x−y=3

The equation of the locus of the points equidistant from the points A(-2,3) B(6,-5) is x = y + 3 or x−y=3

Answer 2

Given :

Points are A ( -2 , 3 ) and B ( 6 , -5 )

To Find :

The equation of the locus of points equidistant from the points A and B.

Solution:  

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

AP = √ (x+2)² + (y-3)²

BP = √ (x-6)² + (y+5)²

Now ,

AP = BP

=> √ (x+2)² + (y-3)²  = √ (x-6)² + (y+5)²

=> (x+2)² + (y-3)²  = (x-6)² + (y+5)²

=> x² + 4x + 4 + y² -6y + 9 = x² + 36 - 12x + y² + 25 + 10y

=> 16x - 16y = 48

=> x - y = 3

The equation of the locus is x-y-3 = 0