A line segment is bisected by line with the equation # 6 y - 7 x = 3 #. If one end of the line segment is at #(7 ,2 )#, where is the other end?

1 Answer
Oct 9, 2017

Other end point is #(103/51,130/17)#

Explanation:

Assumption : It’s a perpendicular bisector.
#6y-7x=3# Eqn (1)
#y=(7/6)x+cancel(3/6)(1/2)#
It’s in the form #y=mx+c# where m is the slope.
#:.m=7/6. Slope of perpendicular line is -(6/7)#
Equation of line segment is
#y-2=-(6/7)(x-7)#
#7y-14=-6x+42#
#7y+6x=56# Eqn (2)

Solving Eqns (1) & (2)
#36y-42x=18#
#49y+42x=392# Adding,
#85y=410#
#y=410/85=82/17#
#x=(56-7y)/6=(56-(6*82)/17))/6=(952-492)/102=460/102=230/51#

Mid point #=(230/51,82/17) One end point #=(7,2)# Other end point be #(x1,y1)
#(x1+7)/2=230/51#
#x1=(460/51)-7=(460-357)/51=103/51#
#(y1+2)/2=82/17#
#y1=(164/17)-2=(164-34)/17=130/17#
Other end point is #(103/51,130/17)#