JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • 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)            \[{{75}^{o}}\]

    Correct Answer: D

    Solution :

               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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