Question

Let O be the origin and let P be a point on the line \(x + \sqrt{3}y = 10\). If OP is perpendicular to the line, then the angle between OP and the y-axis is

• A. \(15^\circ\)
• B. \(30^\circ\)
• C. \(45^\circ\)
• D. \(60^\circ\)
• E. \(75^\circ\)

Hint:

For a line in general form \(ax + by = c\), the angle its normal (perpendicular) makes with the x-axis is given by \(\tan \phi = b/a\). Here, \(\tan \phi = \sqrt{3}/1 = \sqrt{3}\), so \(\phi = 60^\circ\). The angle with the y-axis is then \(90^\circ - 60^\circ = 30^\circ\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given a line and a point P on it such that the segment from the origin to P is perpendicular to the line. The slope of the line determines the slope of the perpendicular segment OP, which in turn gives the angle it makes with the axes.

Step 2: Key Formula or Approach:
1. Slope-intercept form: \(y = mx + c\).
2. Perpendicular lines property: \(m_1 \cdot m_2 = -1\).
3. Angle with positive x-axis: \(\tan \theta = m\).

Step 3: Detailed Explanation:
The given line is \(x + \sqrt{3}y = 10\).
Rearranging into slope-intercept form:
\[ \sqrt{3}y = -x + 10 \implies y = \left(-\frac{1}{\sqrt{3}}\right)x + \frac{10}{\sqrt{3}} \]
The slope of this line is \(m_1 = -\frac{1}{\sqrt{3}}\).
Since OP is perpendicular to this line, its slope \(m_2\) is:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-1/\sqrt{3}} = \sqrt{3} \]
Let \(\theta\) be the angle OP makes with the positive x-axis. Then:
\[ \tan \theta = \sqrt{3} \implies \theta = 60^\circ \]
We need the angle OP makes with the y-axis. Since the angle with the x-axis is \(60^\circ\):
\[ \text{Angle with y-axis} = 90^\circ - 60^\circ = 30^\circ \]

Step 4: Final Answer:
The angle between OP and the y-axis is \(30^\circ\).