Find the value of $k$ if the angle between St. lines $4 x - y + 7 = 0, k x - 5 y - 9 = 0$ is $45^{\circ}$ .

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Solution

Line $1$ is $4 x - y + 7 = 0$ or $y = 4 x + 7$ is of the form $y = m_{1} x + c$
Line $2$ is $k x - 5 y - 9 = 0$ or $5 y = k x - 9$ or $y = \frac{k x}{5} - \frac{9}{5}$ is of the form $y = m_{2} x + c$
Slope of line $1$ is $m_{1} = 4$ and that of line $2$ is $\frac{k}{5}$
We have $tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$
$\Rightarrow tan 45^{\circ} = \frac{4 - \frac{k}{5}}{1 + \frac{4 k}{5}}$
$\Rightarrow 1 = \frac{20 - k}{5 + 4 k}$
$\Rightarrow 5 + 4 k = 20 - k$
$\Rightarrow 5 k = 20 - 5 = 15$
$∴ k = \frac{15}{5} = 3$
Hence $k = 3$

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