Question

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?

Answer 1

`( -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`