A line segment is bisected by a line with the equation # - 5 y + 3 x = 1 #. If one end of the line segment is at #(6 ,4 )#, where is the other end?

1 Answer
Apr 29, 2016

#-5y+3x=1# is the bisector for a line segment between #(6,4)# and any point on the line #color(blue)(-5y+3x=4)#

Explanation:

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If #(x,y)# is the end point of a line segment with #(6,4)# as its other end; and
if #-5y+3x=1# bisects this line

Then for any point #(barx,bary)# on #-5y+3x=1#
the #Deltax# and #Deltay# between #(6,4)# and #(barx,bary)#
will be the same as #Deltax# and #Deltay# between #(barx,bary)# and the corresponding target end point.

Since #Deltax=6-barx# and #Deltay=4-bary#
for any point #(barx,bary)# on #-5y+3x=1#
the corresponding target end point will be #(2barx-6,2bary-4)#

Specifically we can see that #(barx,bary)=(2,1)# is a point on #-5y+3x=1#
with a corresponding target point of #(-2,-2)#

Note also that the target line is parallel to #-5y+3x=1#
and therefore has the same slope (namely #3/5#).

Using the slope-point form for the target line, we get
#color(white)("XXX")(y+2)=3/5(x+2)#
or
#color(white)("XXX")5y+10=3x+6#
or
#color(white)("XXX")-5y+3x=4#