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In a \mathrm{\triangle A B C}, the equation of the perpendicular bisector of \mathrm{A C} is \mathrm{3 x-2 y+8=0}.If the coordinates of the points \mathrm{A} and \mathrm{B} are (1,-1) and (3,1) respectively, then the circumcentre of the \mathrm{\triangle A B C} is

Option: 1

\left(\frac{4}{5}, \frac{14}{5}\right)


Option: 2

\left(-\frac{4}{5}, \frac{14}{5}\right)


Option: 3

\left(\frac{4}{5},-\frac{14}{5}\right)


Option: 4

none of these


Answers (1)

best_answer

\mathrm{E}  is mid of \mathrm{A B \Rightarrow E(2,0)}
\mathrm{Slope\: of \: A B=\frac{1+1}{3-1}=1, E N \perp A B}
\mathrm{\therefore \quad Equation\: of \: E N\: is \: y-0=-1(x-2)}

\mathrm{\Rightarrow x+y-2=0\quad \cdots(i)}

\mathrm{Equation \: of \: D N\: is \: 3 x-2 y+8=0}\quad \cdots(ii)


Solving (i) and (ii), we get \mathrm{N\left(-\frac{4}{5}, \frac{14}{5}\right)}, which is the circumcentre.

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Pankaj

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