A line makes intercepts 5 and 7 on the coordi | vedclass

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(B) The equation of the line with intercepts $a=5$ and $b=7$ is $\frac{x}{5} + \frac{y}{7} = 1$,which simplifies to $7x + 5y = 35$.When the axes are r

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$\frac{1}{6}$

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(B) The equation of the line with intercepts $a=5$ and $b=7$ is $\frac{x}{5} + \frac{y}{7} = 1$,which simplifies to $7x + 5y = 35$.When the axes are rotated by an angle $	heta$,the new coordinates $(x', y')$ are given by $x = x' \cos 	heta - y' \sin 	heta$ and $y = x' \sin 	heta + y' \cos 	heta$.Substituting these into the line equation: $7(x' \cos 	heta - y' \sin 	heta) + 5(x' \sin 	heta + y' \cos 	heta) = 35$.Rearranging terms: $x'(7 \cos 	heta + 5 \sin 	heta) + y'(5 \cos 	heta - 7 \sin 	heta) = 35$.The intercepts on the new axes are $a' = \frac{35}{7 \cos 	heta + 5 \sin 	heta}$ and $b' = \frac{35}{5 \cos 	heta - 7 \sin 	heta}$.Since the intercepts are equal,$a' = b'$,so $7 \cos 	heta + 5 \sin 	heta = 5 \cos 	heta - 7 \sin 	heta$.$12 \sin 	heta = -2 \cos 	heta$.$	an 	heta = -\frac{2}{12} = -\frac{1}{6}$.Therefore,$|	an 	heta| = \frac{1}{6}$.

$A$ line makes intercepts $5$ and $7$ on the coordinate axes. The axes are rotated through an angle $\theta$ in the positive direction about the origin so that the line makes equal intercepts on the new axes,then $|\tan \theta|=$

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