Question

the equation of locus of the points equidistant from the points A(-2,3) B(6,-5)
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Answer 1

Equation of the locus: x-y-3 = 0

Step-by-step explanation:

Given:  Points A(-2,3) B(6,-5)

Find: Equation of the locus of points equidistant from A and B.

Solution:  

Let C(h,k) be a point equidistant from A and B. Then AC = BC.

AC = √ (h+2)² + (k-3)²

BC = √ (h-6)² + (k+5)²

√ (h+2)² + (k-3)²  = √ (h-6)² + (k+5)²

(h+2)² + (k-3)²  = (h-6)² + (k+5)²

h² + 4h + 4 + k² -6k + 9 = h² + 36 - 12h + k² + 25 + 10k

16h - 16k = 48

  h - k = 3

Equation of the locus: x-y-3 = 0

Answer 2

Given :  points A(-2,3) B(6,-5)

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To find :  equation of locus of the points equidistant from the points

Solution:

locus of the points equidistant from the two points is perpendicular bisector of line join the point

Mid point of A ( -2 , 3)  & B ( 6 , - 5)

= ( - 2 + 6)/2  , ( 3 - 5)/2

= 2  , - 1

Slope of AB = (  -5 - 3)/(6 - (-2)) = -8/8 = - 1

=> Slope of perpendicular bisector =  1  

Equation of  locus of the points equidistant from the two points  

is  y -  (-1)  = 1(x - 2)

=> y + 1  = x - 2

=> x  - y  = 3

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

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