Question

Find the angle between the lines root 3x+y=1 and x+root 3y=1

Answer 1

To find the angle between the lines given by the equations \( \sqrt{3}x + y = 1 \) and \( x + \sqrt{3}y = 1 \), we first need to rewrite both equations in slope-intercept form \( y = mx + b \), where \( m \) is the slope.

1. For the first line \( \sqrt{3}x + y = 1 \):
\[
y = -\sqrt{3}x + 1
\]
The slope \( m_1 \) of the first line is \( -\sqrt{3} \).

2. For the second line \( x + \sqrt{3}y = 1 \):
\[
\sqrt{3}y = 1 - x \quad \Rightarrow \quad y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}
\]
The slope \( m_2 \) of the second line is \( -\frac{1}{\sqrt{3}} \).

Next, we can find the angle \( \theta \) between the two lines using the formula:
\[
\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\]

Substituting \( m_1 = -\sqrt{3} \) and \( m_2 = -\frac{1}{\sqrt{3}} \):
\[
\tan(\theta) = \left| \frac{-\sqrt{3} - \left(-\frac{1}{\sqrt{3}}\right)}{1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)} \right|
\]
Simplifying the numerator:
\[
-\sqrt{3} + \frac{1}{\sqrt{3}} = -\frac{3}{\sqrt{3}} + \frac{1}{\sqrt{3}} = -\frac{2}{\sqrt{3}}
\]

Simplifying the denominator:
\[
1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 + 1 = 2
\]

Now substituting back:
\[
\tan(\theta) = \left| \frac{-\frac{2}{\sqrt{3}}}{2} \right| = \left| -\frac{2}{2\sqrt{3}} \right| = \frac{1}{\sqrt{3}}
\]

The angle \( \theta \) can be found using the arctangent:
\[
\theta = \tan^{-1}\left( \frac{1}{\sqrt{3}} \right)
\]
This corresponds to:
\[
\theta = 30^\circ
\]

Thus, the angle between the lines \( \sqrt{3}x + y = 1 \) and \( x + \sqrt{3}y = 1 \) is \( \theta = 30^\circ \).

**Final answer: \( 30^\circ \)**