If $m$ and $b$ are real numbers and $mb>0$, then the line whose equation is $y=mx+b$ cannot contain the point
$\textbf{(A)}\ (0,1997)\qquad\textbf{(B)}\ (0,-1997)\qquad\textbf{(C)}\ (19,97)\qquad\textbf{(D)}\ (19,-97)\qquad\textbf{(E)}\ (1997,0)$
Geometrically, $b$ is the y-intercept, and $m$ is the slope.
Since $mb>0$, then either we have a positive y-intercept and a positive slope, or a negative y-intercept and a negative slope.
Lines with a positive y-intercept and positive slope can go through quadrants I, II, and III.  They cannot go through quadrant IV, because if you start at $(0,b)$ for a positve $b$, you can't go down into the fourth quadrant with a positive slope.  Thus, point $C$ is a possible point.
Lines with a negative y-intercept and negative slope can go through quadrants II, III, and IV.  Thus, point $D$ is a possible point.
Looking at the axes, any point on the y-axis is possible.  Thus, $A$ and $B$ are both possible.
However, points on the positive x-axis are inpossible to reach.  If you start with a positive y-intercept, you must go up and to the right.  If you start with a negative y-intercept, you must go down and to the right.  Thus, $\boxed{E}$ cannot be reached.