How do you know if a perpendicular bisector passes through a point?

anonymousanonymous
in Science & Mathematics

Kay so this is the question I'm asked I need major help ):

The line segment CD had endpoints C(3,5) and D(-5,1).

a). Show that the perpendicular bisector passes through points E(6,-11)

Pleaaaaase and thank you <3

Comments

  • awms-aawms-a

    There are a couple ways to do this. My preference is actually to use geometry. The following works because the perpendicular bisector of CD is the collection of all points whose distance from C is the same as its distance from D.

    Therefore we just need to show CE = DE.

    CE = sqrt( (3-6)^2 + (5- -11)^2 )

    = sqrt( 9 + 256 )

    = sqrt( 265 )

    DE = sqrt( (-5 - 6)^2 + (1 - -11)^2 )

    = sqrt( 121 + 144 )

    = sqrt( 265 )

    Since the distances from the endpoints are the same, E lies on the perpendicular bisector.

  • tuxtux

    first you need to find the equation of a line that is the perpedicular bisector of that line

    so use midpoint formula which is (x1+x2)/2, (y1+y2)/2

    so the midpoint would be (-1, 3)

    then find the slope of the line segment CD

    formula for that is y1-y2 over x1-x2

    so 2 is slope

    then you need the opposite reciprocal of that to get the slope for the perpendicular bisector

    so that slope would be -1/2

    now you use the midpoint to find b of the line

    so 3=-1/2(-1)+b

    so 2.5=b

    now you know that the perpendicular bisector's equation is y=-1/2x+2.5

    now just plug in the point 6,-11 and if it works then it does pass through

    so -11=-1/2(6)+2.5

    so it does not work out meaning that it doesnt pass through that point

  • anonymousanonymous

    so the equation of the st line is

    (x-3)/(y-5)=(3+5)/(5-1)

    => 2y-10=(x-3)

    => x-2y+7=0

    so the equation of the st. line perpendicular to this st. line is

    2x+y=c where c is any constant

    now the mid point of the st. line x-2y+7=0 is

    ((-5+3)/2, (5+1)/2)

    = (-1,3)

    the perpendicular bisector will have the eq 2x+y=c and will pass through this mid point

    so putting the point (-1,3) in the eq 2x+y=c we get

    -2+3=c

    so, c=1

    so the equation becomes 2x+y=1

    this is the equation of the perpendicular bisector

    and now if we put the point (6,-11) in the equation we get

    12-11

    which is equal to 1

    so the perpendicular bisector passes through the point (6,-11)

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