Question

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?

Answer 1

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)`