## Thomas' Calculus 13th Edition

$y=1$
Here, $y=3, y=-1$ Formula to find the distance between two points is: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Now, $\sqrt{(x-x)^2+(y-3)^2+(z-z)^2}=\sqrt{(x-x)^2+(y-(-1))^2+(z-z)^2}$ $\implies (y+1)^2=(y-3)^2$ or, $y^2+1+2y=y^2-6y+9$ Thus, $8y=8$ Hence, $y=1$