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(C) The slopes of the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ are $m_1 = 4$ and $m_2 = \frac{k}{5}$,respectively.Given the angle $	heta = 45^{\c

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$3$

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(C) The slopes of the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ are $m_1 = 4$ and $m_2 = \frac{k}{5}$,respectively.Given the angle $	heta = 45^{\circ}$ between the lines,we use the formula:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$$	an 45^{\circ} = \left| \frac{4 - \frac{k}{5}}{1 + 4 \cdot \frac{k}{5}} ight|$$1 = \left| \frac{20 - k}{5 + 4k} ight|$This implies $\frac{20 - k}{5 + 4k} = 1$ or $\frac{20 - k}{5 + 4k} = -1$.Case $1$: $20 - k = 5 + 4k$ $\Rightarrow 5k = 15$ $\Rightarrow k = 3$.Case $2$: $20 - k = -5 - 4k$ $\Rightarrow 3k = -25$ $\Rightarrow k = -\frac{25}{3}$.Since $k > 0$,the only valid solution is $k = 3$.

The value of $k (> 0)$,such that the angle between the lines $4x - y + 7 = 0$ and $kx - 5y - 9 = 0$ is $45^{\circ}$,is

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