Question

if the point A (a,2) is equidistant from the point B (8,2) and C (2,-2) find the value,.​

Answer 1

➡ Since the point A isequidistant from both the points B and C, we can write AB=AC.

➡ Thus, the value of x obtained is 5 and similarly, the point A will be (5,2).

Answer 2

Answer:

let p(x,y) be any point on the locus

A(a,2),B(8,2),C(2,-2)

AP=^((x-a)^2+(y-2)^2)

AP^2=(x-a)^2+(y-2)^2

BP=^((x-8)^2+(y-2)^2)

BP^2=(x-8)^2+(y-2)^2

CP=^((x-2)^2+(y-(-2))^2)

CP^2=(x-2)^2+(y+2)^2

from the hypothesis

AP=BP+CP

Squaring on both sides

AP^2=BP^2+CP^2

=(x-a)^2+(y-2)^2=(x-8)^2+(y-2)^2+(x-2)^2+(y+2)^2

=x^2+a^2+2xa+y^2+4-4y=x^2+64-16x+y^2+4-4y+x^2+4-4x+y^2+4+4y

=x^2+y^2-4y+4+a^2+2xa=2x^2+2y^2-20×+76

that implies a^2+2xa=x^2+y^2+4y+72