• # question_answer In what direction a line be drawn through the point (1, 2) so that its points of intersection with the line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point                             [IIT 1966; MNR 1987] A)            ${{30}^{o}}$                             B)            ${{45}^{o}}$ C)            ${{60}^{o}}$                             D)            ${{75}^{o}}$

Let the required line through the point (1,2) be inclined at an angle $\theta$ to the axis of x. Then its equation is     $\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=r$                         .....(i)                    where r is distance of any point (x, y) on the line from the point (1, 2).                    The coordinates of any point on the line (i) are $(1+r\cos \theta ,\text{ }2+r\sin \theta )$. If this point is at a distance $\frac{\sqrt{6}}{3}$ form (1, 2), then $r=\frac{\sqrt{6}}{3}.$                    Therefore, the point is $\left( 1+\frac{\sqrt{6}}{3}\cos \theta ,\text{ }2+\frac{\sqrt{6}}{3}\sin \theta \right)$.                    But this point lies on the line $x+y=4$.                    Þ $\frac{\sqrt{6}}{3}(\cos \theta +\sin \theta )=1$ or $\sin \theta +\cos \theta =\frac{3}{\sqrt{6}}$                    Þ $\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{\sqrt{3}}{2}$, {Dividing both sides by$\sqrt{2}$}                    Þ $\sin (\theta +{{45}^{o}})=\sin {{60}^{o}}$or sin ${{120}^{o}}$                    Þ $\theta ={{15}^{o}}$or ${{75}^{o}}$.
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