Question

The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:

• A. \(\frac{8}{5}\)
• B. \(\frac{25}{41}\)
• C. \(\frac{2}{5}\)
• D. \(\frac{30}{41}\)

Answer 1

The Correct Option is D

To solve this problem, we must find the tangent of the angle formed by two lines, \(L_1\) and \(L_2\). These lines trisect the segment of \(4x + 5y = 20\) that lies in the first quadrant.

Step-by-step Solution

1. First, determine the intercepts of the line \(4x + 5y = 20\) within the first quadrant.
   • Setting \(y = 0\), we find \(4x = 20 \Rightarrow x = 5\) (x-intercept).
   • Setting \(x = 0\), we find \(5y = 20 \Rightarrow y = 4\) (y-intercept).
2. Next, identify the points that trisect this line segment.
   • Using the section formula, the points dividing the segment in ratios 1:2 and 2:1 are:
     • Point 1 (1:2 ratio): \[ \left(\frac{1 \times 0 + 2 \times 5}{1 + 2}, \frac{1 \times 4 + 2 \times 0}{1 + 2}\right) = \left(\frac{10}{3}, \frac{4}{3}\right) \]
     • Point 2 (2:1 ratio): \[ \left(\frac{2 \times 0 + 1 \times 5}{2 + 1}, \frac{2 \times 4 + 1 \times 0}{2 + 1}\right) = \left(\frac{5}{3}, \frac{8}{3}\right) \]
3. Now, find the equations for lines \(L_1\) and \(L_2\), which pass through the origin and these trisection points.
   • Line \(L_1\) passing through \(\left(\frac{10}{3}, \frac{4}{3}\right)\):
     The equation is \(y = mx\), where the slope \(m = \frac{4/3}{10/3} = \frac{2}{5}\).
   • Line \(L_2\) passing through \(\left(\frac{5}{3}, \frac{8}{3}\right)\):
     The equation is \(y = mx\), where the slope \(m = \frac{8/3}{5/3} = \frac{8}{5}\).
4. Finally, calculate the tangent of the angle \(\theta\) between these two lines.
   • The formula for the tangent of the angle between lines \(y = m_1x\) and \(y = m_2x\) is: \[ \tan \theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right| \]
   • Substituting \(m_1 = \frac{2}{5}\) and \(m_2 = \frac{8}{5}\) into the formula: \[ \tan \theta = \left|\frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{2}{5} \times \frac{8}{5}}\right| = \left|\frac{\frac{6}{5}}{1 + \frac{16}{25}}\right| = \left|\frac{6/5}{41/25}\right| = \left|\frac{150/5}{41/5}\right| = \left|\frac{30}{41}\right| \]

Consequently, the tangent of the angle between lines \(L_1\) and \(L_2\) is \(\frac{30}{41}\).