If the equations of the perpendicular bisectors of the sides AB and AC of a ΔABC are x−y+5=0 and x+2y=0respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

# If the equations of the perpendicular bisectors of the sides AB and AC of a $\Delta ABC\text{\hspace{0.17em}}are\text{\hspace{0.17em}}x-y+5=0\text{\hspace{0.17em}}and\text{\hspace{0.17em}}x+2y=0$respectively and if A is (1, -2) then the equation of the perpendicular bisector of the side BC is

1. A

$3x+3y+5=0$

2. B

$9x-23y+40=0$

3. C

$6x+15y=5$

4. D

$23x-14y+100=0$

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### Solution:

$\begin{array}{l}If\text{\hspace{0.17em}}x-y+5=0\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\perp r\text{\hspace{0.17em}}bi\mathrm{sec}tor\text{\hspace{0.17em}}of\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}then\text{\hspace{0.17em}}\\ B\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\mathrm{Im}age\text{\hspace{0.17em}}of\text{\hspace{0.17em}}A\left(1,\text{\hspace{0.17em}}-2\right)\text{\hspace{0.17em}}=\left({x}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{y}_{1}\right)\\ \frac{h-1}{1}=\frac{k+2}{-1}=\frac{-2\left(1+2+5\right)}{1+1}\\ \frac{h-1}{1}=\frac{k+2}{-1}=-8\\ B\left(h\text{\hspace{0.17em}\hspace{0.17em}}k\right)=\text{\hspace{0.17em}}\left(-7,\text{\hspace{0.17em}\hspace{0.17em}}6\right)\\ If\text{\hspace{0.17em}}x+2y=0\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\perp r\text{\hspace{0.17em}}bi\mathrm{sec}tor\text{\hspace{0.17em}}of\text{\hspace{0.17em}}AC\text{\hspace{0.17em}}then\\ c\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\mathrm{Im}age\text{\hspace{0.17em}}of\text{\hspace{0.17em}}A\left(1,\text{\hspace{0.17em}}-2\right)\\ \frac{h-1}{1}=\frac{k+2}{2}=\frac{-2\left(1-4\right)}{1+4}\\ \frac{h-1}{1}=\frac{k+2}{2}=\frac{6}{5}\end{array}$

$\begin{array}{l}h=\frac{6}{5}+1⇒\text{\hspace{0.17em}}\frac{11}{5},k=\frac{12}{5}-2,\text{\hspace{0.17em}}⇒\frac{2}{5}\\ \therefore \text{\hspace{0.17em}}C\left(h\text{\hspace{0.17em}\hspace{0.17em}}k\right)=\left(\frac{11}{5},\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{5}\right)\\ slope\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\overline{BC}\text{\hspace{0.17em}}=\frac{\frac{2}{5}-6}{\frac{11}{5}+7}=\frac{-28}{46}=\frac{-14}{23}\\ Mid\text{\hspace{0.17em}}po\mathrm{int}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}BC\text{\hspace{0.17em}}=\left(\frac{\frac{11}{5}-7}{2},\text{\hspace{0.17em}\hspace{0.17em}}\frac{\frac{2}{5}+6}{2}\right)⇒\text{\hspace{0.17em}}\left(\frac{-12}{5},\text{\hspace{0.17em}\hspace{0.17em}}\frac{16}{5}\right)\\ Equation\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\perp r\text{\hspace{0.17em}}bi\mathrm{sec}tor\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\overline{BC}\text{\hspace{0.17em}}is\text{\hspace{0.17em}}\\ y-\frac{16}{5}=\frac{23}{14}\left(x+\frac{12}{5}\right)\\ ⇒\text{\hspace{0.17em}}5y-16=\frac{23}{14}\left(5x+12\right)\\ ⇒\text{\hspace{0.17em}}70y-224=115x+276\\ ⇒\text{\hspace{0.17em}}115x-70y+500=0\end{array}$

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