Is the angle between $px+qy+12=0$ and $px+qy+20=0$ $0$ or $\pi$?

Question

This question came in the Rajshahi University admission exam 2010-11

Q) Angle between $px+qy+12=0$ and $px+qy+20=0$ is-

(a) $\frac{\pi}{3}$

(b) $\frac{\pi}{2}$

(c) $\pi$

(d) None of the above

My attempt:

$$\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}$$

$$\tan\theta=\pm\frac{-\frac{p}{q}+\frac{p}{q}}{1+\frac{p^2}{q^2}}$$

$$\tan\theta=0$$

$$\tan\theta=n\pi,n\in \mathbb{Z}$$

Now, $\theta=0,\pi,...$

There is $\pi$ in (c), so I'll go with (c). The third-party question bank agrees with me that (c) is the correct answer. However, I saw the following on the internet:

An angle is formed when one line intersect another line in some specific manner . And the degree of the angle is determined by how the lines intersect each other .

Two parellel line never intersect each other . So no angle is formed between them .

So its safe to say the angle between two lines is undefined or zero degrees.

As "undefined" or "0" has not been given in the options, should I go with (d)?

Answer 1

From my Answer to your previous Question:

“when two straight lines intersect, the angle between them is uniquely specified by some angle in $\mathbf{[0,90^\circ].}$ This is in fact the standard way to specify the angle between two straight lines.”

Adding to that:

• the angle between two coincident lines is $\mathbf{0^\circ},$ while
• the angle between two parallel lines is undefined.

On the other hand, the angle between two nonzero Euclidean vectors sharing a common $n$-dimensional space is always defined (and lying in $[0,180^\circ]$), since these vectors are “portable” in space.

Answer 2

Yes. The lines are parallel, so the answer is "none of the above".