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?

1 Answer
sankarankalyanam
Oct 4, 2017

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)#

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