JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • question_answer A line through \[A(-5,-\ 4)\] meets the lines \[x+3y+2=0,\] \[2x+y+4=0\] and \[x-y-5=0\] at B, C and D respectively. If \[{{\left( \frac{15}{AB} \right)}^{2}}+{{\left( \frac{10}{AC} \right)}^{2}}={{\left( \frac{6}{AD} \right)}^{2}},\] then the equation of the line is                                                    [IIT 1993] 

    A)            \[2x+3y+22=0\]                         

    B)            \[5x-4y+7=0\]

    C)            \[3x-2y+3=0\]                           

    D)            None of these

    Correct Answer: A

    Solution :

               \[\frac{x+5}{\cos \theta }=\frac{y+4}{\sin \theta }=\frac{{{r}_{1}}}{AB}=\frac{{{r}_{2}}}{AC}=\frac{{{r}_{3}}}{AD}\]            \[({{r}_{1}}\cos \theta -5,{{r}_{1}}\sin \theta -4)\] lies on \[x+3y+2=0\].            \\[{{r}_{1}}=\frac{15}{\cos \theta +3\sin \theta }\]            Similarly \[\frac{10}{AC}=2\cos \theta +\sin \theta \] and \[\frac{6}{AD}=\cos \theta -\sin \theta \]            Putting in the given relation, we get \[{{(2\cos \theta +3\sin \theta )}^{2}}=0\]            \\[\tan \theta =-\frac{2}{3}\Rightarrow y+4=-\frac{2}{3}(x+5)\]Þ\[2x+3y+22=0.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner