Animated Solution for Mathematics - Straight Lines: 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 Journey Through the Plane

Welcome, fellow explorer of the mathematical universe. Today, we are going to dissect a classic coordinate geometry problem. It is not just about finding an answer; it is about understanding the elegant dance between lines, ratios, and angles.

Imagine you are standing on the Cartesian plane, looking at the line $4x+5y=20$. This line is our canvas.

Mapping the Canvas

Before we can trisect anything, we must know where our line lives. We find its boundaries by looking at where it kisses the axes.

Setting $y=0$ in the equation $4x+5y=20$, we find the x-intercept: $4x=20$, so $x=5$. This gives us point $P(5,0)$.

Similarly, setting $x=0$ gives us $5y=20$, so $y=4$. This is our point $Q(0,4)$. We now have a clear, finite segment $PQ$ anchored in the first quadrant.

The Art of Trisection

Now, we introduce the concept of trisection. We are not just cutting the line; we are partitioning it into three equal segments.

To achieve this, we need two points, $A$ and $B$, lying on $PQ$. Point $A$ is the first cut, dividing $PQ$ in a ratio of $1:2$. Point $B$ is the second cut, dividing $PQ$ in a ratio of $2:1$.

Using the section formula, we find the coordinates of $A$ and $B$. For point $A$, we calculate:

So, $A=(\frac{10}{3},\frac{4}{3})$. For point $B$, we shift the ratio to $2:1$:

Thus, $B=(\frac{5}{3},\frac{8}{3})$. We have successfully located our points of division.

Defining the Lines

We are told that lines $L_1$ and $L_2$ pass through the origin $(0,0)$ and these points $A$ and $B$. The slope of any line passing through the origin and a point $(x,y)$ is simply $m=\frac{y}{x}$.

For $L_1$ (passing through $A$):

For $L_2$ (passing through $B$):

The Final Convergence

We are now at the climax of our journey. We need the tangent of the angle $\theta$ between $L_1$ and $L_2$. The formula for the angle between two lines with slopes $m_1$ and $m_2$ is:

Let us substitute our slopes into this elegant expression:

The numerator simplifies beautifully: $\frac{8}{5}-\frac{2}{5}=\frac{6}{5}$. The denominator becomes $1+\frac{16}{25}=\frac{25+16}{25}=\frac{41}{25}$.

Finally, we divide:

And there it is. The tangent of the angle is $\frac{30}{41}$. It is a result born from simple, logical steps. Remember, in JEE, the complexity is often just a mask for fundamental principles waiting to be applied.