# Find the equation of the perpendicular bisector of `bar(AB)` .Coordinates of AB are A (-√147/3, 10/√50)  B (√1/√3, √18)

embizze | High School Teacher | (Level 1) Educator Emeritus

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Find the equation of the perpendicular bisector of `bar(AB)` with the coordinates `A((-sqrt(147))/3,10/sqrt(50))` and `B(sqrt(1)/(sqrt(3)),sqrt(18))`

`(-sqrt(147))/3=(-sqrt(49*3))/3=(-7sqrt(3))/3`

`10/sqrt(50)=10/sqrt(25*2)=10/(5sqrt(2))=2/sqrt(2)=sqrt(2)`

`sqrt(1)/sqrt(3)=1/sqrt(3)=sqrt(3)/3`

`sqrt(18)=sqrt(9*2)=3sqrt(2)`

So the points are `A((-7sqrt(3))/3,sqrt(2)),B(sqrt(3)/3,3sqrt(2))`

(a) The perpendicular bisector goes throught the midpoint of `bar(AB)` ; the midpoint is `M(((-7sqrt(3))/3+sqrt(3)/3)/2,(sqrt(2)+3sqrt(2))/2)`

`((-7sqrt(3))/3+sqrt(3)/3)/2=((-6sqrt(3))/3)/2=(-2sqrt(3))/2=-sqrt(3)`

`(sqrt(2)+3sqrt(2))/2=(4sqrt(2))/2=2sqrt(2)`

So the midpoint is `M(-sqrt(3),2sqrt(2))`

(b) The slope of the perpendicular line is the opposite reciprocal of the slope through A and B (or the product of their slopes is -1)

The slope from A to B is `m=(3sqrt(2)-sqrt(2))/(sqrt(3)/3-(-7sqrt(3))/3)=(2sqrt(2))/((8sqrt(3))/3)=sqrt(6)/4`

Then the slope of the perpendicular line is `(-4)/sqrt(6)=(-2sqrt(6))/3`

(c) We have a point `M(-sqrt(3),2sqrt(2))` and a slope `m=(-2sqrt(6))/3` so the equation of the line is:

`y-2sqrt(2)=(-2sqrt(6))/3 (x+sqrt(3))` or

`y=(-2sqrt(6))/3 x`

The graph:

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