Question

A line segment is bisected by a line with the equation `-6 y - x = 3`. If one end of the line segment is at `( 8 , 3 )`, where is the other end?

Answer 1

The other end point is `(238/37,-237/37)`

Explanation:

It is assumed that the line bisecting the line segment is assumed to be a perpendicular bisector.
`-6y-x=3` Eqn (1)
`-6y=x+3`
`y=-(x/6)-(1/2)`
Slope of the equation `m1=-(1/6)`
Slope the line segment `m2=-(1/(m1))=-(1/(-1/6))=6`
Equation of the line segment is
`(y-3)=6*(x-8)`
`y-6x=-45` Eqn (2)

Solving equations (1) & (2), we get the midpoint which is also the midpoint of the line segment.
`-x-6y=3`Eqn (1)
`-36x+6y=-270` Eqn (2) * 6; Adding,
`-37x=-267`
`x=267/37`
Substituting value of x in Eqn (1),
`-6y=3+(267/37)`
`y=-(111+267)/(37*6)=-(378/(222)=-(63/3)`
Coordinate of midpoint`(267/37),-(63/37)`
Let (x2,y2) be the other end point coordinates of the line segment.
`(x2+8)/2=(267/37)`
`x2=(534/37)-8=(534-296)/37=238/37`
`(y2+3)/2=-(63/37)`
#y2+3=-(126/37) #y2=-(126+111)/37=-(237/37)#