If Q 0 , 1 is equidistant from the points P 5 , - 3 and R x , 6 , find the value of x .

Given Q 0 , 1 is equidistant from P 5 , - 3 and R x , 6 So, P Q = Q R 5 - 0 2 + - 3 - 1 2 = 0 - x 2 + 1 - 6 2 5 2 + - 4 2 = - x 2 + - 5 2 25 + 16 = x 2 + 25 41 = x 2 + 25 16 = x 2 x = ± 4 Hence, the value of x = ± 4

Find the relation between x and y such that point x , y is equidistant from the points 7 , 1 and 3 , 5 .

The distance between two points x 1 , y 1 and x 2 , y 2 is given by the formula, d = x 2 - x 1 2 + y 2 - y 1 2 Let the point be P x , y , A 7 , 1 and B 3 , 5 Distance between the points A 7 , 1 and P x , y is P A = x - 7 2 + y - 1 2 Distance between the points B 3 , 5 and P x , y is P B = x - 3 2 + y - 5 2 According to question, P A = P B So, x - 7 2 + y - 1 2 = x - 3 2 + y - 5 2 ⇒ x - 7 2 + y - 1 2 = x - 3 2 + y - 5 2 ⇒ x 2 + 49 - 14 x + y 2 + 1 - 2 y = x 2 + 9 - 6 x + y 2 + 25 - 10 y ⇒ 50 - 14 x - 2 y = 34 - 6 x - 10 y ⇒ 50 - 34 = 14 x + 2 y - 6 x - 10 y ⇒ 8 x - 8 y = 50 - 34 ⇒ 8 x - y = 16 ⇒ x - y = 16 8 ⇒ x - y = 2 Hence, the required relation is x - y = 2 .