A line segment is bisected by a line with the equation -7 y + 3 x = 2 . If one end of the line segment is at ( 2 , 4 ), where is the other end?

1 Answer
Jun 3, 2016

Infinite solutions: any of the points in the line y=(3/7)x-26/7, including the point (110/29,-424/203) when the segment is perpendicularly bisected.

Explanation:

First we should notice that the point (2,4) isn't in the line 3x-7y-2=0 since:
3*2-7*4-2=6-28-2=-24

But as the problem is stated it admits infinite solutions.

Calling AB the line segment with M the midpoint we have:
A(2,4)
M((2+x_B)/2,(4+y_B)/2)
B(x_B,y_B)

The orthogonal projection of A onto the line is also M when the segment is perpendicular bisected.
The slope of the line y=(3x-2)/7is
k=3/7
So the slope of line perpendicular to the aforementioned is
p=-1/k=-7/3
And the equation of such a perpendicular line passing through A is
y-4=-7/3(x-2) => y=(-7x+14)/3+4 => y=(-7x+26)/3

Combining the equations of the 2 lines
(3x-2)/7=(-7x+26)/3 => 9x-6=-49+162 => 58x=168 => x=84/29
-> y=(252/29-2)/7 => y=194/203
And M(84/29,194/203)

Finding B
x_M=1+x_B/2 => 84/29=1+x_B/2=>x_B=55/29*2=110/29
y_M=2+y_B/2 =>194/203=2+y_B/2 =>y_B=-212/203*2=-424/203
So when the segment AB is perpendicularly bisected by the line mentioned in the problem
B(110/29,-424/203)

But since it is not required that the bisection happens perpendicularly the set of possible points B lays in a line parallel to the bisecting line and passing through (110/29,-424/203), itself one of the possible locations of B. Then the locus in which the point B can be located is
y+424/203=(3/7)(x-110/29)
y=(3/7)x-330/203-424/203
y=(3/7)x-26/7