imc

Let $h$ be the length of the altitude of $△ABC.$ Since segment $BC$ is the base of the triangle and cannot change, the area of the triangle is $\frac{1}{2}(BC)(h)=1$ and $h=\frac{2}{BC}.$ Thus $S$ consists of two lines parallel to $BC$ and is $\frac{2}{BC}$ units away from it. $⟶\boxed{\text{two parallel lines}}$