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.
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