The points $(-1, q)$ and $(-3, r)$ are on a line parallel to $y=\frac{3}{2} x+1$. What is the value of $r-q$ ?
(A) 3
(B) $\frac{4}{3}$
(C) $-\frac{3}{4}$
(D) $-\frac{4}{3}$
(E) -3
The line with equation $y=\frac{3}{2} x+1$ has slope $\frac{3}{2}$.

Since the line segment joining $(-1, q)$ and $(-3, r)$ is parallel to the line with equation $y=\frac{3}{2} x+1$, then the slope of this line segment is $\frac{3}{2}$.

Therefore, $\frac{r-q}{(-3)-(-1)}=\frac{3}{2}$ or $\frac{r-q}{-2}=\frac{3}{2}$.

Thus, $r-q=(-2) \cdot \frac{3}{2}=-3$.

ANSWER: (E)

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