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:
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:
\(\frac{8}{5}\)
\(\frac{25}{41}\)
\(\frac{2}{5}\)
\(\frac{30}{41}\)
The Correct Option is D
Solution and Explanation
Step 1. Identify the Points Where the Line Intersects the Axes: The line \( 4x + 5y = 20 \) intersects the x-axis when \( y = 0 \):
\(4x = 20 \Rightarrow x = 5\)
So, the x-intercept is \( (5, 0) \). The line intersects the y-axis when \( x = 0 \):
\(5y = 20 \Rightarrow y = 4\)
So, the y-intercept is \( (0, 4) \).
Step 2. Determine the Coordinates of the Trisection Points: The line segment from \( (5, 0) \) to \( (0, 4) \) in the first quadrant is trisected at two points, dividing it into three equal parts. Using the section formula, the trisection points \( P \) and \( Q \) are:
\(P = \left( \frac{2 \cdot 0 + 1 \cdot 5}{3}, \frac{2 \cdot 4 + 1 \cdot 0}{3} \right) = \left( \frac{5}{3}, \frac{8}{3} \right)\)
\(Q = \left( \frac{1 \cdot 0 + 2 \cdot 5}{3}, \frac{1 \cdot 4 + 2 \cdot 0}{3} \right) = \left( \frac{10}{3}, \frac{4}{3} \right)\)
Step 3. Find the Slopes of Lines \( L_1 \) and \( L_2 \):
- Line \( L_1 \) passes through the origin and point \( P \left( \frac{5}{3}, \frac{8}{3} \right) \), so its slope \( m_1 \) is:
\(m_1 = \frac{8/3}{5/3} = \frac{8}{5}\)
- Line \( L_2 \) passes through the origin and point \( Q \left( \frac{10}{3}, \frac{4}{3} \right) \), so its slope \( m_2 \) is:
\(m_2 = \frac{4/3}{10/3} = \frac{2}{5}\)
Step 4. Calculate the Tangent of the Angle Between \( L_1 \) and \( L_2 \): The tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\(\tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}\)
- Substituting \( m_1 = \frac{8}{5} \) and \( m_2 = \frac{2}{5} \):
\(\tan \theta = \frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{8}{5} \cdot \frac{2}{5}} = \frac{\frac{6}{5}}{1 + \frac{16}{25}} = \frac{6/5}{41/25} = \frac{30}{41}\)
Thus, the tangent of the angle between the lines \( L_1 \) and \( L_2 \) is \( \frac{30}{41} \).
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Tangents and Normals
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Diagram Explaining Tangents and Normal:
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