A line segment is bisected by a line with the equation 7 y + x = 1 . If one end of the line segment is at (1 ,6 ), where is the other end?

1 Answer
bp
Jan 16, 2017

( -34/50 , -288/50)

Explanation:

There would be any number of points which when joined with (1,6) that would make a line segment, bisected by 7y+x=1.

However if the line 7y+x=1 is a perpendicular bisector, the required point would be unique.

Hence considering 7y+x=1 as a perpendicular bisector, let (a,b) be the required point. The midpoint of the line segment joining (1,6) and (a,b) would be (a+1)/2, (b+6)/2. Since this point would also lie on line 7y+x=1, 7(b+6)/2 +(a+1)/2 =1. This on simplification would yield equation 7b+a=-41

Next, slope of line joining (1,6) and (a,b) is(b-6)/(a-1) and slope of the given line is -1/7. Since both are perpendicular to each other, (b-6)/(a-1) * (-1/7)= -1. On simplification, this would yield an equation 7a-b=1.

Solving the two equations 7b+a=-41 and 7a-b=1, the result would be a= -34/50 and b= -288/50

Related questions
See all questions in Perpendicular Bisectors
Impact of this question
1959 views around the world
You can reuse this answer
Creative Commons License