If #color(red)(-7y+x=3)# bisects a line segment between #color(blue)(""(1,6))# and some other point #color(green)(""(x',y'))#

with an intersection point of #color(red)(-7y+x=3)# and this line segment at #color(red)(""(barx,bary))#

Then

#color(white)("XXX")(6-color(red)(bary))=(color(red)(bary)-color(green)(y'))#

#color(white)("XXXXXX")rarr color(green)(y')=2color(red)(bary)-6#

and

#color(white)("XXX")(1-color(red)(barx))=(color(red)(barx)-color(green)(x'))#

#color(white)("XXXXXX")rarr color(green)(x')=2color(red)(barx)-1#

Specifically, we could solve #color(red)(-7y+x=3)# for a couple arbitrary solution points:

#color(white)("XXX")color(red)((barx_1,bary_1)=(-4,-1))#

and

#color(white)("XXX")color(red)((barx_2,bary_2)=(3,0))#

and obtain end-of-line-segment values:

#color(white)("XXX")color(green)((x'_1,y'_1)=(-9,-8))#

and

#color(white)("XXX")color(green)((x'_2,y'_2)=(5,-6)#

Using the slopes

#color(white)("XXX")(y-color(green)(y'_1))/(x-color(green)(x'_1))=(color(green)(y'_1-y'_2))/(color(green)(x'_1-x'_2))#

#color(white)("XXX")(y-color(green)(""(-8)))/(x-color(green)(""(-9)))=(color(green)(""(-8))-color(green)(""(-6)))/(color(green)(""(-9))-color(green)(""(-5)))#

#color(white)("XXX")(y+8)/(x+9)=1/7#

#color(white)("XXX")7y+56=x+9#

#color(white)("XXX")7y-x=-47#

or (paralleling the given form):

#color(white)("XXX")-7y+x=47#