• # 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)            $14x+23y-40=0$

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 $14x+23y-40=0$.