How do you find the equation of the perpendicular bisector of the segment joining the points A (6, -3) and B (-2, 5)?

1 Answer
Ratnaker Mehta
Sep 15, 2016

"The eqn. of p.b. is "x-y-1=0.

Explanation:

From Geometry, we know that the perpendicular bisector (p.b.) of a

line segment (sgmt.) is the line passing through its mid-point (m.p.) &

perpendicular to the sgmt.

Let us denote, by bar(AB), the line sgmt. joining the pts.

A(6,-3), and, B(-2,5).

M is the m.p. of bar(AB)

rArr M=M((6-2)/2,(-3+5)/2)=M(2,1).

Slope of bar(AB)=(5-(-3))/(-2-6)=8/-8=-1.

:." The Slope of p.b. of "bar(AB)" must be 1."

Using the Slope-Point Form of line for the p.b., we get its eqn.

y-1=1(x-2) rArr x-y-1=0.

Method II:=

We use the following Geometrical Property of the p.b. of a line sgmt.

"Any pt. on the p.b. of a line sgmt. is equidistant from the "

"end-points of the sgmt."

Let P(x,y) be any pt. on the p.b. of bar(AB).

"Then, dist. PA= dist. PB" rArr PA^2=PB^2.

rArr (x-6)^2+(y+3)^2=(x+2)^2+(y-5)^2

rArr x^2-12x+36+y^2+6y+9

=x^2+4x+4+y^2-10y+25

rArr -16x+16y+16=0

rArr x-y-1=0., as before!

Enjoy Maths.!

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