Q 1 Find relationship between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). so

∴ AP = BP

         AP 2 = BP 2 

(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2

x2 – 4x + 4 + y2 – 10y + 25

                 =

x2 + 6x + 9 + y2 – 14y + 49

⇒ -4x – 10y +29 =  6x - 14y +58 

⇒ -10x + 4y = 29

 is the required relation.