The locus of the point which is equidistant from the point (1,1) and the line x + y +1=0 is

Step-by-step solution

Let (x₁, y₁ ) be the point which is at a distance d from the (1,1) then we have: (x₁-1 )^2+ (y₁-1 )^2 =d (x₁-1 )^2+ (y₁-1 )^2=d^2...(i) Point (x₁, y₁ ) is at the same distance from the line d= x₁+y₁+1 1+1 = x₁+y₁+1 2 ...(ii) From equations (i) & (ii) aligned & (x₁-1 )^2+ (y₁-1 )^2= 1 2 (x₁+y₁+1 )^2 & 2 [x₁^2+1-2 x₁+y₁^2+1-2 y₁ ] & =x₁^2+y₁^2+1+2 x₁ y₁+2 x₁+2 y₁ & x₁^2+y₁^2-2 x₁ y₁-6 x₁-6 y₁+3=0 & (x₁-y₁ )^2-6 (x₁+y₁ )+3=0 aligned Taking locus of point (x₁, y₁ ) , we get: (x-y)^2-6(x-y)+3=0