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

Question

Theodore Abad · Accepted Answer

The equations are not the same.

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#

Autumn Armstrong · Answer

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

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#.

Savannah Aguilar · Answer

undefined To determine if the perpendicular bisector of YZ is the same as the median from X in triangle XYZ, we need to find the midpoint of YZ and the midpoint of XZ, then check if they coincide.

Midpoint of YZ: ((-5 - 5)/2, (-6 + 0)/2) = (-5, -3)
Midpoint of XZ: ((-3 - 5)/2, (2 + 0)/2) = (-4, 1)

The perpendicular bisector of YZ passes through the midpoint of YZ (-5, -3) and is perpendicular to YZ. The slope of YZ is (-6 - 0)/(-5 + 5) = undefined.

Therefore, the perpendicular bisector of YZ is vertical, and its equation is x = -5.

The median from X passes through X and the midpoint of YZ (-5, -3). The slope of this line is (1 - (-3))/(-4 - (-5)) = 4.

The equation of the median from X is y - 2 = 4(x + 3).

Comparing the equations of the perpendicular bisector of YZ (x = -5) and the median from X (y - 2 = 4(x + 3)), they are not the same.

Thus, the perpendicular bisector of YZ is not the same as the median from X in triangle XYZ.