A line segment is bisected by a line with the equation # - 2 y - x = 1 #. If one end of the line segment is at #( 8 , 3 )#, where is the other end?

1 Answer
May 31, 2016

#-2y-x=1# bisects line segments between #(8,3)# and any point on the line #-2y-x=16#

Explanation:

Consider a horizontal line segment from #(8,3)# extending to the left
with a total length twice that of the horizontal distance from #(8,3)# to the line #-2y-x=1#.
From the diagram below we can see that this horizontal line will intersect #-2y-x=1# at #(-7,3)#
and will have its left-most end at #(-22,3)#
enter image source here
Obviously this horizontal line is bisected by #-2y-x=1#

Consider a line through this left-most point with the same slope as #-2y-x=1#.
Using the slope-point form we can see that the equation of this line can be written as #-2y-x=16#

Assertion
for any point #(x',y')# on this new line, the line segment between #(x',y')# and #(8,3)# will be bisected by #-2y-x=1#

This Assertion follows immediately from considering a modified version of the above diagram.
enter image source here

Since #QR||ST#
#color(white)("XXX")trianglePQR ~ trianglePST#

#color(white)("XXX")rArr PR:RT = PQ:QS = 1:1#

#rArr QR# bisects #PT# for any point #T=(x',y')# on #-2y-x=16#