# 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?

Jan 16, 2017

$\left(- \frac{34}{50} , - \frac{288}{50}\right)$

#### 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 $\frac{a + 1}{2} , \frac{b + 6}{2}$. Since this point would also lie on line 7y+x=1, $7 \frac{b + 6}{2} + \frac{a + 1}{2} = 1$. This on simplification would yield equation 7b+a=-41

Next, slope of line joining (1,6) and (a,b) is$\frac{b - 6}{a - 1}$ and slope of the given line is $- \frac{1}{7}$. Since both are perpendicular to each other, $\frac{b - 6}{a - 1} \cdot \left(- \frac{1}{7}\right) = - 1$. 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 $a = - \frac{34}{50} \mathmr{and} b = - \frac{288}{50}$