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

1 Answer
Jul 15, 2016

The other end could be any point on the line #y=2x-1#

Explanation:

For convenience, I will rearrange the given equation
#color(white)("XXX")-3y+6x=6#
as
#color(white)("XXX")y=2x-2#

Consider the vertical line through #(3,3)#.
Since #x# is constant for all points on a vertical line,
this vertical line will intersect #y=2x-2# at #(3,4)#

The distance between #(3,3)# and #(3,4)# is #1# unit.

The point #(3,5)# is also on this vertical line at a distance of #2# units from #(3,3)#.

Therefore #y=2x-2# (originally given as #-3y+6x=6#) bisects the line segment joining #(3,3)# and #(3,5)#.

Therefore one possible endpoint would be at #(3,5)#

enter image source here

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Perhaps less obviously, any point on a line through #(3,5)# and parallel to the given line #y=2x-2# (or in its original but less convenient form #-3y+6x=6#)
will also be a line segment endpoint bisected by the given equation.
enter image source here
From the image above we can see that for an arbitrary point #E# on a line parallel to the given bisector line
#triangle ABD ~ triangle ACE#
and since #abs(AB)=1/2abs(AC)#
#rarr abs(AD)=1/2abs(AE)#
(i.e. the given line is a bisector for #AE#)

Since #y=2x-2# has a slope of #2#
the line parallel to it and through #(3,5)# will also have a slope of #2#
and using the slope-point form:
#color(white)("XXX")y-5=2(x-3)#
Which can be simplified as
#color(white)("XXX")y=2x-1#