1. **State the problem:** We are given points $A=(-5,5)$ and $B=(3,7)$ and must find point $P$ on the y-axis such that $P$ is equidistant from $A$ and $B$. 2. **Analyze the problem:** Since $P$ lies on the y-axis, its $x$-coordinate is 0. Let $P = (0,y)$. 3. **Set up the distance equations:** The distance from $P$ to $A$ is $$\sqrt{(0-(-5))^2+(y-5)^2} = \sqrt{25+(y-5)^2}.$$ The distance from $P$ to $B$ is $$\sqrt{(0-3)^2+(y-7)^2} = \sqrt{9+(y-7)^2}.$$ 4. **Set distances equal:** Since $P$ is equidistant from $A$ and $B$, set the distances equal: $$\sqrt{25+(y-5)^2} = \sqrt{9+(y-7)^2}.$$ 5. **Square both sides:** $$25+(y-5)^2 = 9+(y-7)^2.$$ Expanding both squares: $$25 + (y^2 -10y + 25) = 9 + (y^2 -14y + 49).$$ 6. **Simplify equation:** $$25 + y^2 -10y + 25 = 9 + y^2 -14y + 49$$ Combine like terms: $$y^2 -10y + 50 = y^2 -14y + 58.$$ Subtract $y^2$ from both sides: $$-10y + 50 = -14y + 58.$$ 7. **Solve for $y$:** Bring variables to one side, constants to the other: $$-10y + 14y = 58 - 50$$ $$4y = 8$$ $$y = 2.$$ 8. **Answer:** The coordinates of point $P$ on the y-axis equidistant from $A$ and $B$ are $$P = (0,2).$$