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

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Solution

We know that, if $y = m x + c$ is the equation of an straight line, then $m$ is called the slope of the equation.

For the straight line,
$4 x - y + 7 = 0$
$⟹ y = 4 x + 7$
Therefore, $m_{1} = 4$ is the slope of the above equation.

Similarly, for the straight line,
$k x - 5 y - 7 = 0$
$⟹ y = \frac{k}{5} x + \frac{9}{5}$
Therefore, $m_{2} = \frac{k}{5}$ is the slope of the above equation.

The angle two two straight lines is given as:
$tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$

Therefore, the angle between the given straight lines is given as
$tan 45^{0} = ∣ ∣ ∣ ∣ ∣ ∣ \frac{4 - \frac{k}{5}}{1 + 4 \times \frac{k}{5}} ∣ ∣ ∣ ∣ ∣ ∣$
$⟹ 1 = ∣ ∣ ∣ ∣ ∣ ∣ \frac{\frac{20 - k}{5}}{\frac{5 + 4 k}{5}} ∣ ∣ ∣ ∣ ∣ ∣$
$⟹ 1 = ∣ ∣ ∣ \frac{20 - k}{5 + 4 k} ∣ ∣ ∣$
$⟹ 20 - k = 5 + 4 k$
$⟹ 15 = 5 k$
$⟹ k = 3$

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