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How do we find the equation of perpendicular bisector of a segment whose end points are #(-2,-1)# and #(-18,-5)#?

Answer 1

Equation of perpendicula bisector is #4x+y+43=0#

A perpendicular bisector of a line segment whose endpoints are given, is locus of a point, which moves so that its distance from the two points is always equal.

Let the point be #(x,y)#. Its distance from #(-2,-1)# is #sqrt((x+2)^2+(y+1)^2)# and its distance from #(-18,-5)# is #sqrt((x+18)^2+(y+5)^2)#.

Hence equation of line is

#(x+2)^2+(y+1)^2=(x+18)^2+(y+5)^2#

or #x^2+4x+4+y^2+2y+1=x^2+36x+324+y^2+10y+25#

or #32x+8y+344=0#

or#4x+y+43=0#

Alternate way is to find slope of line joining #(-2,-1)# and #(-18,-5)#, which is #(-1-(-5))/(-2-(-18))=4/16=1/4#. As such slope of perpendicular bisector is #-1/(-1/4)=-4#.

As perpendicular bisector passes through mid point i.e. #(-2-18)/2,(-1-5)/2)# i.e. #(-10,-3)# and hence using point slope form, its equation is

#y+3=-4(x+10)# or #4x+y+43=0#

graph{(4x+y+43)((x+2)^2+(y+1)^2-0.1)((x+18)^2+(y+5)^2-0.1)(x-4y-2)=0 [-29.66, 10.34, -11.84, 8.16]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.