Question

Let $O$ be the origin and $P$ be a point on a line such that $OP$ is perpendicular to that line. If $OP$ makes an obtuse angle $\alpha$ with the $x$-axis, $OP = 5$ and $\sin \alpha = \frac{3}{5}$, then the equation of the line is

• A. $3x - 4y - 25 = 0$
• B. $4x + 3y + 25 = 0$
• C. $3x - 4y + 25 = 0$
• D. $4x - 3y - 25 = 0$
• E. $4x - 3y + 25 = 0$

Hint:

Always pay attention to the term "obtuse". It defines the sign of the trigonometric functions. In the second quadrant, $\sin$ is positive but $\cos$ is negative.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The equation of a straight line in normal form is given by \( x \cos \alpha + y \sin \alpha = p \), where \( p \) is the length of the perpendicular from the origin to the line, and \( \alpha \) is the angle that this perpendicular makes with the positive $x$-axis.
Key Formula or Approach:
1. Use the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \) to find \( \cos \alpha \).
2. Consider the quadrant of the angle \( \alpha \) (obtuse means second quadrant).
3. Substitute \( p \), \( \cos \alpha \), and \( \sin \alpha \) into the normal form equation.

Step 2: Detailed Explanation:
Given \( p = OP = 5 \) and \( \sin \alpha = \frac{3}{5} \).
Since \( \alpha \) is an obtuse angle, it lies in the second quadrant ($90^\circ < \alpha < 180^\circ$).
In the second quadrant, cosine is negative.
\[ \cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \]
Now, substitute these into the normal form:
\[ x \cos \alpha + y \sin \alpha = p \]
\[ x \left(-\frac{4}{5}\right) + y \left(\frac{3}{5}\right) = 5 \]
Multiply the entire equation by 5:
\[ -4x + 3y = 25 \]
Rearranging terms to match the options:
\[ 4x - 3y + 25 = 0 \]

Step 3: Final Answer:
The equation of the line is \( 4x - 3y + 25 = 0 \).