## University Calculus: Early Transcendentals (3rd Edition)

$y=1$
Since, we are given two equations of a plane: $y=3, y=-1$ The distance between two points can be calculated as: $\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$ Thus, $\sqrt{(x-x)^2+(y-3)^2+(z-z)^2}=\sqrt{(x-x)^2+(y-(-1))^2+(z-z)^2}$ or, $(y+1)^2=(y-3)^2$ or, $y^2+1+2y=y^2-6y+9$ or, $8y=8$ Hence, $y=1$