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What is the angle made by the line [tex]\( x \cos \alpha + y \sin \alpha = p \)[/tex] with the positive [tex]\( x \)[/tex]-axis?

A. [tex]\(\alpha\)[/tex]
B. [tex]\(180^{\circ} - \alpha\)[/tex]
C. [tex]\(90^{\circ} - \alpha\)[/tex]
D. [tex]\(90^{\circ} + \alpha\)[/tex]


Sagot :

To determine the angle that the line [tex]\( x \cos \alpha + y \sin \alpha = p \)[/tex] makes with the positive [tex]\( x \)[/tex]-axis, we need to carefully consider the standard form of a linear equation and the associated trigonometric relationships.

First, recall that the general form of a line equation in Cartesian coordinates is:
[tex]\[ Ax + By + C = 0 \][/tex]

In our given line equation, we can rewrite it to the standard form:
[tex]\[ x \cos \alpha + y \sin \alpha = p \\ or, x \cos \alpha + y \sin \alpha - p = 0 \][/tex]

Here, [tex]\( A = \cos \alpha \)[/tex] and [tex]\( B = \sin \alpha \)[/tex].

The angle [tex]\( \theta \)[/tex] that a line makes with the positive [tex]\( x \)[/tex]-axis can be found using the relationship:
[tex]\[ \tan \theta = -\frac{A}{B} \][/tex]

For the given line equation:
[tex]\[ \tan \theta = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha \][/tex]

The line's slope [tex]\( m \)[/tex] is therefore:
[tex]\[ m = -\cot \alpha \][/tex]

Thus, the angle [tex]\( \theta \)[/tex] can be determined as:
[tex]\[ \theta = \alpha \][/tex]

Therefore, the correct option which indicates the angle this line makes with the positive [tex]\( x \)[/tex]-axis is:

[tex]\[ \boxed{\alpha} \][/tex]

Hence, the correct answer is (6).