Question

The angle between the lines $3x-4y=1$ and $2x+3y=5$ is

• A. $tan^{-1}(\frac{-17}{6})$
• B. $tan^{-1}(\frac{-1}{6})$
• C. $tan^{-1}(\frac{1}{6})$
• D. $tan^{-1}(\frac{17}{6})$

Answer 1

Answer: (D)

$tan^{-1}(\frac{17}{6})$

Answer 2

To find the angle between two lines, we first determine their slopes. The general equation of a straight line is $Ax + By + C = 0$. Its slope $m$ can be found by rearranging it into the slope-intercept form $y = mx + c$, where $m = -\frac{A}{B}$.

Step 1: Find the slope of the first line. The equation of the first line is $3x - 4y = 1$. Rearrange it to solve for $y$: $4y = 3x - 1$ $y = \frac{3}{4}x - \frac{1}{4}$ So, the slope of the first line, $m_1 = \frac{3}{4}$.

Step 2: Find the slope of the second line. The equation of the second line is $2x + 3y = 5$. Rearrange it to solve for $y$: $3y = -2x + 5$ $y = -\frac{2}{3}x + \frac{5}{3}$ So, the slope of the second line, $m_2 = -\frac{2}{3}$.

Step 3: Use the formula for the angle between two lines. The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula: $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$ This formula gives the acute angle between the lines.

Substitute the values of $m_1$ and $m_2$ into the formula: $m_1 = \frac{3}{4}$ $m_2 = -\frac{2}{3}$

Calculate the numerator $m_2 - m_1$: $m_2 - m_1 = -\frac{2}{3} - \frac{3}{4} = \frac{-2 \times 4 - 3 \times 3}{12} = \frac{-8 - 9}{12} = -\frac{17}{12}$

Calculate the denominator $1 + m_1 m_2$: $1 + m_1 m_2 = 1 + \left(\frac{3}{4}\right) \left(-\frac{2}{3}\right) = 1 - \frac{6}{12} = 1 - \frac{1}{2} = \frac{1}{2}$

Now, substitute these values into the $\tan \theta$ formula: $\tan \theta = \left| \frac{-\frac{17}{12}}{\frac{1}{2}} \right|$ $\tan \theta = \left| -\frac{17}{12} \times \frac{2}{1} \right|$ $\tan \theta = \left| -\frac{17}{6} \right|$ $\tan \theta = \frac{17}{6}$

Therefore, the angle $\theta$ is $\tan^{-1}\left(\frac{17}{6}\right)$.