question_answer

The equation of perpendicular bisectors of sides\[AB\]and\[AC\]of a\[\Delta ABC\]are\[x-y+5=0\] and\[x+2y=0\]respectively.   If  the coordinates of vertex\[A\]are\[(1,\,\,-2)\], the equation of\[BC\]is

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

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

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

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

Correct Answer: A

Solution :

Let\[B({{x}_{1}},\,\,{{y}_{1}})\]and\[C({{x}_{2}},\,\,{{y}_{2}})\]be two vertices and \[P\left( \frac{{{x}_{1}}+1}{2},\,\,\frac{{{y}_{1}}-2}{2} \right)\]lies on perpendicular bisector\[x-y+5=0\] \[\therefore \]  \[\frac{{{x}_{1}}+1}{2}-\frac{{{y}_{2}}-2}{2}=-5\] \[\Rightarrow \]               \[{{x}_{1}}-{{y}_{1}}=-13\]                                           ... (i) Also,\[PN\]is perpendicular to\[AB\] \[\therefore \]  \[\frac{{{y}_{1}}+2}{{{x}_{1}}-1}\times 1=-1\] \[\Rightarrow \]               \[{{y}_{1}}+2={{x}_{1}}+1\]                                         \[\Rightarrow \]               \[{{x}_{1}}+{{y}_{1}}=-1\]                                             ... (ii) On solving Eqs. (i) and (ii), we get                 \[{{x}_{1}}=-7,\,\,{{y}_{1}}=6\] \[\therefore \]The coordinates of\[B\] are\[(-7,\,\,6)\]. Similarly, the coordinates of\[C\]are\[\left( \frac{11}{5},\,\,\frac{2}{5} \right)\]. Hence, the equation of\[BC\]is                 \[y-6=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}(x+7)\] \[\Rightarrow \]               \[y-6=\frac{-14}{23}(x+7)\] \[\Rightarrow \]               \[14x+23y-40=0\]