If the equation of a line having a slope m (m | vedclass

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(B) The slope of the line is $m$. The slope of the line $x + y - 3 = 0$ is $m_2 = -1$. Given $	an 	heta = \left|\frac{m - m_2}{1 + m m_2}ight| = \

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$-42$

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(B) The slope of the line is $m$. The slope of the line $x + y - 3 = 0$ is $m_2 = -1$. Given $	an 	heta = \left|\frac{m - m_2}{1 + m m_2}ight| = \frac{5}{7}$. Substituting $m_2 = -1$,we get $\left|\frac{m + 1}{1 - m}ight| = \frac{5}{7}$. This implies $7|m + 1| = 5|1 - m|$. Case $1$: $7(m + 1) = 5(1 - m)$ $\Rightarrow 7m + 7 = 5 - 5m$ $\Rightarrow 12m = -2$ $\Rightarrow m = -1/6$. Case $2$: $7(m + 1) = -5(1 - m)$ $\Rightarrow 7m + 7 = -5 + 5m$ $\Rightarrow 2m = -12$ $\Rightarrow m = -6$. Since $m \in \mathbb{Z}$,we must have $m = -6$. The equation of the line passing through $(1, 1)$ with slope $m = -6$ is $y - 1 = -6(x - 1)$. $y - 1 = -6x + 6 \Rightarrow 6x + y - 7 = 0$. Comparing with $ax + y + c = 0$,we get $a = 6$ and $c = -7$. Therefore,$ac = 6 	imes (-7) = -42$.

If the equation of a line having a slope $m$ $(m \in \mathbb{Z})$,passing through $(1, 1)$ and making an angle of $\tan^{-1}\left(\frac{5}{7}\right)$ with the line $x + y - 3 = 0$ is $ax + y + c = 0$,then $ac =$

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