A line segment is bisected by a line with the equation # - 6 y + 2 x = 3 #. If one end of the line segment is at #( 4 , 8 )#, where is the other end?

1 Answer
Jun 21, 2016

Any point on the line #3y-x=-27#

Explanation:

Consider a horizontal line segment from #(4,8)# bisected by #-6y+2x=3#
The equation of the horizontal line segment from #(4,8)# is
#color(white)("XXX")y=8#

Noting that #-6y+2x=3 rArr x=(6y+3)/2#

The intersection of this horizontal line segment with the given bisector line will occur at
#color(white)("XXX")((6(8)+3)/2,8)=(51/2,8)#

Continuing to travel horizontally a point twice as far away from #(4,8)# as #(51/2,8)#
will be at #(51,8)#

That is #(51,8)# is one possible endpoint for a line segment form #(4,8)# bisected by #-6y+2x=3# would be #(51,8)#

For any point which could be such an endpoint, a line through this point parallel to the bisecting line will provide all possible bisected line segment endpoints.

#-6y+2x=3# (and all lines parallel to it) has a slope of #1/3#

Using the previously determined possible endpoint #(51,8)# and this slope of #1/3#
we can write the slope-point version:
#color(white)("XXX")y-8=1/3(x-51)#

#color(white)("XXX")3y-24=x-51#

#color(white)("XXX")3y-x=-27#

Any solution to the equation #3y-x=-27# will provide a valid endpoint for a line segment from #(4,8)# which will be bisected by #6y-2x=3#

Here is a graph of the point and the two lines in question:
graph{(-6y+2x-3)(3y-x+27)((x-4)^2+(y-8)^2-0.02)=0 [-25.65, 25.64, -12.83, 12.81]}