question_answer

The equation of perpendicular bisectors of the sides AB and AC of a triangle ABC are \[x-y+5=0\] and \[x+2y=0\] respectively. If the point A is \[(1,\ -\ 2)\], then the equation of line BC is                                         [IIT 1986]

A)            \[23x+14y-40=0\]                     

B)            \[14x-23y+40=0\]

C)            \[{{\tan }^{-1}}(2)\]                  

D)            \[14x+23y-40=0\]

Correct Answer: D

Solution :

           Let the equation of perpendicular bisector FN of AB is \[x-y+5=0\]                                         ......(i)                    The middle point F of AB is \[\left( \frac{{{x}_{1}}+1}{2},\frac{{{y}_{1}}-2}{2} \right)\]lies on line (i). Therefore \[{{x}_{1}}-{{y}_{1}}=-13\]                  ?..(ii)                    Also AB is perpendicular to FN. So the product of their slopes is ?1.                    i.e. \[\frac{{{y}_{1}}+2}{{{x}_{1}}-1}\times 1=-1\]or \[{{x}_{1}}+{{y}_{1}}=-1\]                         ??(iii)                    On solving (ii) and (iii), we get \[B(-7,6)\].                    Similarly \[C\text{ }\left( \frac{11}{5},\frac{2}{5} \right)\].                    Hence the equation of BC is \[14x+23y-40=0\].