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(D) Let the required line passing through $(1, 2)$ be inclined at an angle $	heta$ to the $x$-axis. The equation of the line is $\frac{x-1}{\cos 	he

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) Let the required line passing through $(1, 2)$ be inclined at an angle $	heta$ to the $x$-axis. The equation of the line is $\frac{x-1}{\cos 	heta} = \frac{y-2}{\sin 	heta} = r$,where $r$ is the distance from $(1, 2)$.Any point on this line is $(1 + r \cos 	heta, 2 + r \sin 	heta)$.Given $r = \frac{\sqrt{6}}{3}$,the point is $(1 + \frac{\sqrt{6}}{3} \cos 	heta, 2 + \frac{\sqrt{6}}{3} \sin 	heta)$.Since this point lies on $x + y = 4$,we have $(1 + \frac{\sqrt{6}}{3} \cos 	heta) + (2 + \frac{\sqrt{6}}{3} \sin 	heta) = 4$.$\frac{\sqrt{6}}{3} (\cos 	heta + \sin 	heta) = 1 \implies \sin 	heta + \cos 	heta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$.Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin 	heta + \frac{1}{\sqrt{2}} \cos 	heta = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$.$\sin(	heta + 45^\circ) = \sin 60^\circ$ or $\sin 120^\circ$.$	heta + 45^\circ = 60^\circ \implies 	heta = 15^\circ$ or $	heta + 45^\circ = 120^\circ \implies 	heta = 75^\circ$.Since $75^\circ$ is one of the options,the correct answer is $75^\circ$.

In what direction should a line be drawn through the point $(1, 2)$ so that its point of intersection with the line $x + y = 4$ is at a distance of $\frac{\sqrt{6}}{3}$ from the given point (in $^\circ$)?

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