In a #DeltaXYZ# having #X(-3,2)#, #Y(-5,-6)# and #Z(-5,0)#, is perpendicular bisector of #YZ# same as median from #X#?

2 Answers
Mar 27, 2017

The equations are not the same.

Explanation:

Find the midpoint of YZ using the midpoint formula #((x_1+x_2)/2, (y_1+y_2)/2) = ((-5+ -5)/2, (0+ -6)/2) = (-5, -3)#

Since YZ is a vertical line: #x = -5#,
The perpendicular bisector is a horizontal line: #y = -3#

The median from X goes from #(-3, 2)# to the midpoint of YZ #(-5, -3)#:
#m_"median" = (y_2 - y_1)/(x_2 - x_1) = (-3 - 2)/(-5 - -3) = (-5)/(-2) = 5/2#

Equation of the median #y = mx + b#:
#y = 5/2 x +b#

Use either the midpoint or point X to find b:
#2 = 5/2 * -3/1 +b#
#2 = -15/2 + b#
#4/2 + 15/2 = 19/2 = b#

Equation of the median: #y = 5/2 x + 19/2#

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Mar 27, 2017

Perpendicular bisector of #YZ# is not the same line as the median from #X#.

Explanation:

As the abscissa (i.e. #x#-coordinate) of #Y(-5,-6)# and #Z(-5,0)# both are #-5#, #YZ# is parallel to #y#-axis

and as such its perpendicular bisector will be parallel to #x#-axis.

As midpoint of #YZ# is #P((-5-5)/2,(0+6)/2)# i.e. #(-5,-3)#,

equation of perpendicular bisector is #y+3=0#.

and median from #X(-3,2)# is #XP# and its equation is

#(y-2)/(3-2)=(x-(-3))/(-5-(-3))#

or #(y-2)/1=(x+3)/(-2)#

or #x+3=-2y+4#

or #x+2y-1=0#

Note it is different from the equation of median given in question.

Obviously perpendicular bisector of #YZ# is not the same line as the median from #X#.