53. Find a point on the \( y \)-axis that is equidistant from \( (5,-5) \) and \( (1,1) \).

To find a point on the \( y \)-axis that is equidistant from \( (5,-5) \) and \( (1,1) \), let's denote the point on the \( y \)-axis as \( (0, y) \). The distance from \( (0, y) \) to \( (5, -5) \) is given by: \[ d_1 = \sqrt{(0 - 5)^2 + (y - (-5))^2} = \sqrt{25 + (y + 5)^2} \] The distance from \( (0, y) \) to \( (1, 1) \) is given by: \[ d_2 = \sqrt{(0 - 1)^2 + (y - 1)^2} = \sqrt{1 + (y - 1)^2} \] Setting these distances equal, we have: \[ \sqrt{25 + (y + 5)^2} = \sqrt{1 + (y - 1)^2} \] Squaring both sides, we get: \[ 25 + (y + 5)^2 = 1 + (y - 1)^2 \] Expanding both sides: \[ 25 + (y^2 + 10y + 25) = 1 + (y^2 - 2y + 1) \] \[ 25 + y^2 + 10y + 25 = 1 + y^2 - 2y + 1 \] \[ 50 + 10y = 2 - 2y \] Now, let's combine like terms: \[ 50 + 12y = 2 \] Subtracting 50 from both sides gives: \[ 12y = 2 - 50 \rightarrow 12y = -48 \] Dividing by 12: \[ y = -4 \] Thus, the point on the \( y \)-axis that is equidistant from \( (5,-5) \) and \( (1,1) \) is \( (0, -4) \).