Is the graph of $ r = \dfrac{3 \pi}{5} $ a diagonal line?

Question

I'm reviewing my notes and I came upon this. It says that $$ \begin{align} r \sin \theta &= 5 \tag{horizontal line} \\ r &= \dfrac{3 \pi}{5} \tag{diagonal line} \end{align}$$ How is this possible, when $\dfrac{3 \pi}{5}$ results to a constant (therefore a straight line)?

Answer 1

The equation $r=k$ is the equation of a circle, centre the origin, radius $k$. So it would be a circle of radius $\frac{3\pi}{5}$. Odd radius!

You presumably meant $\theta=\frac{3\pi}{5}$, which (depending on whether we don't allow negative $r$ or do) is a half-line or a full line.