Two lines L₁ and L₂ passing through origin trisecting the line segment intercepted by the line 4x+ 5y = 20 between the coordinate axes. Then the tangent of angle between the lines L₁ and L₂ is

The correct option is 30/41.

Step-by-Step Explanation:

Step 1: Find the intercepts of the given line on the coordinate axes
The given equation of the line is:
$$4x+5y=20$$
To find the x-intercept, we set $y=0$:
$$4x=20\Rightarrow x=5$$
This gives the point on the x-axis, which is labeled as $P(5,0)$ in the provided image.
To find the y-intercept, we set $x=0$:
$$5y=20\Rightarrow y=4$$
This gives the point on the y-axis, which is labeled as $Q(0,4)$ in the provided image.
Thus, the line segment intercepted between the axes is $PQ$ connecting $P(5,0)$ and $Q(0,4)$.

Step 2: Determine the trisection points of the segment PQ
As observed in the image, the two lines passing through the origin $O(0,0)$ trisect the line segment $PQ$ at two points, labeled as $A$ and $B$.
Point $A$ divides the segment $QP$ in the ratio $1:2$ (closer to $Q$). Using the section formula:
$$x_A=\frac{1\cdot 5+2\cdot 0}{1+2}=\frac{5}{3}$$
$$y_A=\frac{1\cdot 0+2\cdot 4}{1+2}=\frac{8}{3}$$
So, the coordinates of point $A$ are $(\frac{5}{3},\frac{8}{3})$, which matches the label in the image.
Point $B$ divides the segment $QP$ in the ratio $2:1$ (closer to $P$). Using the section formula:
$$x_B=\frac{2\cdot 5+1\cdot 0}{2+1}=\frac{10}{3}$$
$$y_B=\frac{2\cdot 0+1\cdot 4}{2+1}=\frac{4}{3}$$
So, the coordinates of point $B$ are $(\frac{10}{3},\frac{4}{3})$, which also matches the label in the image.

Step 3: Calculate the slopes of the lines OA and OB
The line $L_1$ is the line $OA$ passing through $(0,0)$ and $A(\frac{5}{3},\frac{8}{3})$. Its slope $m_1$ is:
$$m_1=\frac{8/3}{5/3}=\frac{8}{5}$$
The line $L_2$ is the line $OB$ passing through $(0,0)$ and $B(\frac{10}{3},\frac{4}{3})$. Its slope $m_2$ is:
$$m_2=\frac{4/3}{10/3}=\frac{4}{10}=\frac{2}{5}$$

Step 4: Find the tangent of the angle between the two lines
Let $\theta$ be the angle between the lines $L_1$ and $L_2$. The formula for the tangent of the angle between two lines with slopes $m_1$ and $m_2$ is:
$$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$$
Substituting the values of $m_1$ and $m_2$:
$$\tan \theta =\left|\frac{\frac{8}{5}-\frac{2}{5}}{1+\left(\frac{8}{5}\right)\left(\frac{2}{5}\right)}\right|$$
Simplify the numerator:
$$\frac{8}{5}-\frac{2}{5}=\frac{6}{5}$$
Simplify the denominator:
$$1+\frac{16}{25}=\frac{41}{25}$$
Now divide the two terms:
$$\tan \theta =\frac{6/5}{41/25}=\frac{6}{5}\cdot \frac{25}{41}=\frac{30}{41}$$
Thus, the tangent of the angle between the lines $L_1$ and $L_2$ is $\frac{30}{41}$.