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

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Answer 1 · 2016-09-15T17:15:53

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

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.!