# A line segment is bisected by a line with the equation  -6 y - x = 3 . If one end of the line segment is at ( 5 , 1 ), where is the other end?

Nov 23, 2017

Coordinates of the other end are $\left(\frac{167}{37}\right) , - \left(\frac{84}{37}\right)$

#### Explanation:

Assumption : The line bisecting the segment is its perpendicular bisector.

Eqn of line $x + 6 y = - 3$ Eqn (1)
$6 y = - x - 3$
$y = - \left(\frac{x}{6}\right) - \left(\frac{1}{2}\right)$
Slope of the line is -(1/6)#

Slope of altitude the line segment is 6
Eqn of line segment is
$y - 1 = 6 \left(x - 5\right)$
$y - 6 x = 1 - 30 = - 29$ Eqn (2)
Solving Eqns (1) & (2) we get the intersection of the lines or the midpoint of the line segment.
Coordinates of mid point are $\left(\frac{171}{37}\right) , - \left(\frac{47}{37}\right)$

$\frac{171}{37} = \frac{5 + {x}_{2}}{2} , - \left(\frac{47}{37}\right) = \frac{1 + {y}_{2}}{2}$ where ${x}_{2} , {y}_{2}$are the coordinates of the other end point.

${x}_{2} = \left(\frac{342}{37}\right) - 5 = \frac{167}{37}$
${y}_{2} = - \left(\frac{47}{37}\right) - 1 = - \left(\frac{84}{37}\right)$

Nov 23, 2017

$\left(\frac{157}{37} , - \frac{131}{37}\right)$

#### Explanation:

First we need to find the equation of the line that is perpendicular to the line $- 6 y - x = 3$ and passes through the point $\left(5 , 1\right)$. Since these lines are perpendicular we can find the gradient of the required line by using the fact that, if ${m}_{1}$is the gradient of the known equation, then ${m}_{1} \cdot {m}_{2} = - 1$

$\therefore$

$- \frac{1}{6} \cdot {m}_{2} = - 1 \implies {m}_{2} = 6$

So second equation is:

$y - 1 = 6 \left(x - 5\right) \implies y = 6 x - 29$

The point of intersection of these lines is the coordinates of the midpoint of the line segment.

Solving simultaneously:

$- \frac{1}{6} x - \frac{1}{2} = 6 x - 29 \implies x = \frac{171}{37}$

$y = 6 \left(\frac{171}{37}\right) - 29 = - \frac{47}{37}$

We know for a line segment with coordinates $\left({x}_{1} , {y}_{2}\right)$ and $\left(5 , 1\right)$, that the coordinates of the midpoint are $\left(\frac{5 + {x}_{1}}{2} , \frac{1 + {y}_{1}}{2}\right)$.

Midpoint coordinates:

$\left(\frac{171}{37} , - \frac{47}{37}\right)$

$\therefore$

$\frac{5 + {x}_{1}}{2} = \frac{171}{37} \implies {x}_{1} = \frac{157}{37}$

$\frac{1 + {y}_{1}}{2} = - \frac{47}{37} \implies {y}_{1} = - \frac{131}{37}$

Coordinates of end point of line segment:

$\left(\frac{157}{37} , - \frac{131}{37}\right)$

Plot: