 Solution

Take a union of $d$ distinct lines in general position for projective lines, meaning that no three pass through one point. Each pair contributes one distinct singular point, and every other point lies on just one nonsingular line. Thus the union has degree of a projective curve $d$ and exactlydistinct singular points. For $d=1$, this gives a line with zero singular points, as required.

For a concrete construction, choose distinct $t_1,\dots ,t_d\in \mathbb{C}$ and setFor $i\ne j$, the intersection is $[-(t_i+t_j):-t_i{t}_j:1]$. Substitution into the equation of $L_k$ gives $-(t_k-t_i)(t_k-t_j)$, which is nonzero when $k$ is distinct from $i,j$. Thus there are no triple intersections, including at infinity.