Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).
TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)
Let P(x, y) be any point on the perpendicular bisector of AB. Then,
PA=PB
$\Rightarrow \sqrt{(x-7)^{2}+(y-1)^{2}}=\sqrt{(x-3)^{2}+(y-5)^{2}}$
$\Rightarrow(x-7)^{2}+(y-1)^{2}=(x-3)^{2}+(y-5)^{2}$
$\Rightarrow \mathrm{x}^{2}-14 x+49+y^{2}-2 y+1=\mathrm{x}^{2}-6 x+9+y^{2}-10 y+25$
$\Rightarrow-14 x+6 x+10 y-2 y+49+1-9-25=0$
$\Rightarrow-8 x+8 y+16=0$
$\Rightarrow x-y-2=0$
$\Rightarrow x-y=2$
Hence the equation of perpendicular bisector of line segment joining points $(7,1)$ and $(3,5)$ is $x-y=2$
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