Straight Lines - Angle between two lines

Find the angle between the lines $y - \sqrt{3}x - 5 = 0$ and $\sqrt{3}y - x + 6 = 0$.

First, find the slopes of the lines. For the first line $y = \sqrt{3}x + 5$, the slope $m_1 = \sqrt{3}$. For the second line $\sqrt{3}y = x - 6$, which is $y = \frac{1}{\sqrt{3}}x - \frac{6}{\sqrt{3}}$, the slope $m_2 = \frac{1}{\sqrt{3}}$. Using the formula: $\tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{1 + (\sqrt{3})(\frac{1}{\sqrt{3}})} \right|$ $\tan \theta = \left| \frac{\frac{1-3}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{-2}{2\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$ Since $\tan \theta = \frac{1}{\sqrt{3}}$, the acute angle $\theta = 30^\circ$.

Slopes are extracted by converting the equations to $y = mx + c$ form. The tangent formula provides the value for the acute angle between the two slopes.

If the angle between two lines is $\frac{\pi}{4}$ and the slope of one of the lines is $\frac{1}{2}$, find the slope of the other line.

Let the slope of the first line be $m_1 = \frac{1}{2}$ and the slope of the second line be $m_2 = m$. The angle $\theta = \frac{\pi}{4}$, so $\tan \theta = \tan \frac{\pi}{4} = 1$. Using the formula: $1 = \left| \frac{m - \frac{1}{2}}{1 + \frac{m}{2}} \right|$ $1 = \left| \frac{2m - 1}{2 + m} \right|$ This gives two cases:

1. $\frac{2m - 1}{2 + m} = 1 \implies 2m - 1 = 2 + m \implies m = 3$
2. $\frac{2m - 1}{2 + m} = -1 \implies 2m - 1 = -2 - m \implies 3m = -1 \implies m = -\frac{1}{3}$. Thus, the slope of the other line is either $3$ or $-\frac{1}{3}$.

Because of the absolute value in the formula, we solve for both the positive and negative cases to find all possible slopes for the second line.

1. Convert lines to slope-intercept form $y = mx + c$: Line 1: $y = -3x + 7 \implies m_1 = -3$ Line 2: $2y = -x - 9 \implies y = -\frac{1}{2}x - \frac{9}{2} \implies m_2 = -\frac{1}{2}$

2. Use the angle formula: $\tan \theta = \left| \frac{-\frac{1}{2} - (-3)}{1 + (-3)(-\frac{1}{2})} \right|$ $\tan \theta = \left| \frac{-\frac{1}{2} + 3}{1 + \frac{3}{2}} \right| = \left| \frac{\frac{5}{2}}{\frac{5}{2}} \right| = 1$

3. Since $\tan \theta = 1$, the acute angle $\theta = 45^\circ$ (or $\frac{\pi}{4}$ radians).

Slopes are extracted from the general form $Ax+By+C=0$ using $m = -A/B$. The absolute value ensures the result is the acute angle.

1. Find slope $m_1$ of $L_1$: $m_1 = \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2}$

2. Since $L_2 \perp L_1$, $m_2 = -\frac{1}{m_1} = \frac{2}{3}$

3. Equation of $L_2$ using point $(1, 0)$: $y - 0 = \frac{2}{3}(x - 1) \implies 2x - 3y - 2 = 0$

4. Angle with x-axis (inclination $\alpha$): $\tan \alpha = m_2 = \frac{2}{3} \implies \alpha = \tan^{-1}(\frac{2}{3})$.

The perpendicularity condition $m_1 m_2 = -1$ is used to find the slope of the second line, which is then used in the point-slope form.