Find the equation of the perpendicular bisector of the
Question:

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3,5).

Solution:

TO FIND: The equation of perpendicular bisector of line segment joining points (7, 1) and (3, 5)

Let P(xy) 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 $x-y=2$

Administrator

Leave a comment

Please enter comment.
Please enter your name.