JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • question_answer The vertices of a triangle are (2, 1), (5, 2) and    (4, 4). The lengths of the perpendicular from these vertices on the opposite sides are [IIT 1962]

    A)            \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}}\]        

    B)            \[\frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}}\]

    C)            \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}}\]        

    D)            \[\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}\]

    Correct Answer: D

    Solution :

               \[{{L}_{12}}\equiv x-3y+1=0\]                    \[{{L}_{23}}\equiv 2x+y-12=0\]                    \[{{L}_{13}}\equiv 3x-2y-4=0\]                    Therefore, the required distances are                    \[{{D}_{3}}=\left| \frac{4-3\times 4+1}{\sqrt{10}} \right|=\frac{7}{\sqrt{10}}\]                    \[{{D}_{1}}=\left| \frac{4+1-12}{\sqrt{5}} \right|=\frac{7}{\sqrt{5}}\]                    \[{{D}_{2}}=\left| \frac{3\times 5-2\times 2-4}{\sqrt{9+4}} \right|=\frac{7}{\sqrt{13}}\].


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