The Portion of the line 4x + 5y = 20 in the first quadrant is trisected by the lines and passing through the origin. The tangent of an angle between the lines and is: [2024]

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Solution

Q.
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: [2024]

1 $\frac{30}{41}$

2 $\frac{25}{41}$

3 $\frac{2}{5}$

4 $\frac{8}{5}$

Ans.
(1)

Lines $L_{1}$ and $L_{2}$ trisect the line 4x + 5y = 20.

$x_{1} = \frac{5 \times 1 + 0 \times 2}{1 + 2} = \frac{5}{3}$

$y_{1} = \frac{0 \times 1 + 4 \times 2}{1 + 2} = \frac{8}{3}$

$(x_{1}, y_{1}) \equiv (\frac{5}{3}, \frac{8}{3})$

Similarly,

$x_{2} = \frac{0 \times 1 + 5 \times 2}{1 + 2} = \frac{10}{3}$

$y_{2} = \frac{4 \times 1 + 0 \times 2}{1 + 2} = \frac{4}{3}, (x_{2}, y_{2}) \equiv (\frac{10}{3}, \frac{4}{3})$

Slope of line $L_{1} : m_{1} = \frac{8}{3} \times \frac{3}{5} = \frac{8}{5}$

Slope of line $L_{2} : m_{2} = \frac{4}{3} \times \frac{3}{10} = \frac{2}{5}$

Tangent angle between the lines $L_{1}$ and $L_{2}$ :

$\tan θ =$ $| \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$ $=$ $| \frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{8}{5} \times \frac{2}{5}} |$ $= \frac{30}{41}$

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