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 \(x\cos\alpha + y\sin\alpha = p\), where \(p\) is the perpendicular distance from the origin to the line, and \(\alpha\) is the angle the perpendicular makes with the positive x-axis.
Step 2: Key Formula or Approach:
We are given \(p = 5\) and \(\sin\alpha = \frac{3}{5}\).
Since \(\alpha\) is an obtuse angle, it lies in the second quadrant. In the second quadrant, cosine is negative.
Use \(\cos^2\alpha + \sin^2\alpha = 1\) to find \(\cos\alpha\).
Step 3: Detailed Explanation:
Find \(\cos\alpha\):
\[ \cos^2\alpha = 1 - \sin^2\alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \cos\alpha = \pm\frac{4}{5} \] Since \(\alpha\) is obtuse (2nd quadrant), \(\cos\alpha = -\frac{4}{5}\).
Now, substitute \(p, \sin\alpha\), and \(\cos\alpha\) into the normal form equation:
\[ x\left(-\frac{4}{5}\right) + y\left(\frac{3}{5}\right) = 5 \] Multiply the entire equation by 5 to eliminate the denominator:
\[ -4x + 3y = 25 \] Rearrange to standard form \(Ax + By + C = 0\):
\[ 4x - 3y + 25 = 0 \] Step 4: Final Answer:
The equation of the line is \(4x - 3y + 25 = 0\).