The portion of the line 4x + 5y = 20 in the first quadrant is tri… — Solution | JEE Mathematics

MCQMediumJEE 2024Straight Line Equations

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

A
$\frac{8}{5}$
B
$\frac{25}{41}$
C
$\frac{2}{5}$
D
$\frac{30}{41}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The line segment of $4x + 5y = 20$ lying in the first quadrant is trisected by lines $L_1$ and $L_2$ drawn from the origin.

Find: The tangent of the angle between $L_1$ and $L_2$ .

First find the intercepts of the line on the coordinate axes:

\begin{aligned} y &= 0 \Rightarrow 4x = 20 \Rightarrow x = 5 \\ x &= 0 \Rightarrow 5y = 20 \Rightarrow y = 4 \end{aligned}

So the relevant segment joins $$ (5,0) \text{ and } (0,4)

The trisection points divide the segment in the ratios $1:2$ and $2:1$ . Using section formula:

\left(\frac{1\cdot 0 + 2\cdot 5}{1+2}, \frac{1\cdot 4 + 2\cdot 0}{1+2}\right) = \left(\frac{10}{3}, \frac{4}{3}\right)

and

\left(\frac{2\cdot 0 + 1\cdot 5}{2+1}, \frac{2\cdot 4 + 1\cdot 0}{2+1}\right) = \left(\frac{5}{3}, \frac{8}{3}\right)

Using Slopes of the Two Lines

Since both lines pass through the origin, their slopes are obtained from the trisection points:

m_1 = \frac{\frac{4}{3}}{\frac{10}{3}} = \frac{2}{5}, \qquad m_2 = \frac{\frac{8}{3}}{\frac{5}{3}} = \frac{8}{5}

Now use the formula for angle between two lines:

\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|

Substituting the values,

\tan\theta = \left|\frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{2}{5}\cdot\frac{8}{5}}\right| = \left|\frac{\frac{6}{5}}{1 + \frac{16}{25}}\right| = \left|\frac{\frac{6}{5}}{\frac{41}{25}}\right| = \frac{30}{41}

Therefore, the tangent of the angle between $L_1$ and $L_2$ is $\frac{30}{41}$ . Hence, the correct option is D.

Common mistakes

Using the intercepts $(5,0)$ and $(0,4)$ directly as the required points is incorrect because $L_1$ and $L_2$ pass through the trisection points of the segment, not the endpoints. First find the two internal trisection points, then form the lines from the origin.
Applying the section formula with reversed ratios can interchange the two trisection points. Although the final angle remains the same, the slopes get swapped. Write both ratios carefully as $1:2$ and $2:1$ before substitution.
Using the wrong formula for angle between lines, such as $\frac{m_2-m_1}{1-m_1m_2}$ , is incorrect here. For the tangent of the angle between two lines, use $\tan\theta = \left|\frac{m_2-m_1}{1+m_1m_2}\right|$ .

Practice more Straight Line Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Straight Line Equations

Answer

Step-by-step solution

Common mistakes

Practice more Straight Line Equations questions

Related questions