Find the tangent of the angle between the lines which have intercepts 3, 4 and 1, 8 on the axes respectively.

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Solution

$L i n e_{1}$ is $\frac{x}{3} + \frac{y}{4} = 1$

i.e. $4 x + 3 y = 12$

$L i n e_{2}$ is $\frac{x}{1} + \frac{y}{8} = 1$

i.e. $8 x + y = 8$

Slope of $l i n e_{1}$ and $l i n e_{2}$ is $\frac{- 4}{3}$ and $\frac{- 8}{1}$

respectively.

Thus, $t a n θ = ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣$

$= ∣ ∣ ∣ ∣ \frac{\frac{- 4}{3} - (- 8)}{1 + (\frac{- 4}{3}) (- 8)} ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ \frac{\frac{- 4}{3} + 8}{1 + \frac{32}{3}} ∣ ∣ ∣ = ∣ ∣ \frac{- 4 + 24}{3 + 32} ∣ ∣$

$= ∣ ∣ \frac{20}{35} ∣ ∣ = \frac{4}{7}$

Thus, $t a n θ = \frac{4}{7}$

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