Show that the perpendicular bisectors of the sides of the triangle with vertices (7,2) ,(5,-2) & (-1,0) are concurrent . Also find the coordinates of the point of concurrence.

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Please find below the solution to the asked query:
$$\begin{gathered} \mathrm{First} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AB}, \mathrm{BC}, \mathrm{CA} \\ \mathrm{P}\left(\mathrm{x},\mathrm{y}\right)=\left[\frac{7-1}{2}, \frac{2}{2}\right]=\left(3,1\right) \\ \mathrm{Q}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=\left[\frac{7+5}{2},\frac{2-2}{2}\right]=\left(6,0\right) \\ \mathrm{R}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)=\left[\frac{5-1}{2},\frac{-2}{2}\right]=\left(2,-1\right) \\ \mathrm{Line} \mathrm{equation} \mathrm{formula} \mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \mathrm{Now} \mathrm{find} \mathrm{the} \mathrm{line} \mathrm{equation} \mathrm{BP} \\ \Rightarrow \left(\mathrm{y}-1\right)=\frac{-2-1}{5-3}\left(\mathrm{x}-3\right) \\ \Rightarrow \mathrm{y}-1=\frac{-3}{2}\left(\mathrm{x}-3\right) \\ \Rightarrow 2\mathrm{y}-2=-3\mathrm{x}+9 \\ \Rightarrow 3\mathrm{x}+2\mathrm{y}=11.....\left(1\right) \\ \mathrm{Now} \mathrm{find} \mathrm{the} \mathrm{line} \mathrm{equation} \mathrm{QC} \\ \Rightarrow \left(\mathrm{y}-0\right)=\frac{0-0}{-1-1}\left(\mathrm{x}-6\right) \\ \Rightarrow \mathrm{y}-0=0 \\ \Rightarrow \mathrm{y}=0....\left(2\right) \\ \mathrm{Now} \mathrm{find} \mathrm{the} \mathrm{line} \mathrm{equation} \mathrm{AR} \\ \Rightarrow \left(\mathrm{y}-2\right)=\frac{-1-2}{2-7}\left(\mathrm{x}-7\right) \\ \Rightarrow \mathrm{y}-2=\frac{-3}{-5}\left(\mathrm{x}-7\right) \\ \Rightarrow 5\mathrm{y}-10=3\mathrm{x}-21 \\ \Rightarrow 5\mathrm{y}-3\mathrm{x}=-11....\left(3\right) \\ \mathrm{Now} \mathrm{find} \mathrm{the} \mathrm{intersection} \mathrm{point} \mathrm{of} \mathrm{line} \mathrm{QC} \mathrm{and} \mathrm{BP} \\ \mathrm{y}=0 \\ \mathrm{x}=\frac{11-2\mathrm{y}}{3}=\frac{11}{3} \\ \mathrm{Now} \mathrm{put} \mathrm{this} \mathrm{point} \mathrm{on} \mathrm{line} \mathrm{AR} \\ \Rightarrow 5\mathrm{y}-3\mathrm{x}=-11 \\ \Rightarrow 0-3\left(\frac{11}{3}\right)=-11 \\ \Rightarrow -11=-11 \\ \mathrm{Hence} \mathrm{perpendicular} \mathrm{bisectors} \mathrm{are} \mathrm{concurrent} \mathrm{and} \mathrm{the} \mathrm{point} \mathrm{is} \left(0 , \frac{11}{3}\right) \end{gathered}$$

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