A line segment is bisected by a line with the equation  - y + 7 x = 1 . If one end of the line segment is at (1 ,3 ), where is the other end?

Jun 1, 2016

Any point on the line $- y + 7 x = - 2$

Explanation:

Note that $\left(x , y\right) = \left(1 , 6\right)$ is a point on the line -y+7x=1

The distance from $\left(1 , 3\right)$ to $\left(1 , 6\right)$ is $3$

The distance from $\left(1 , 3\right)$ to $\left(1 , 6 + 3\right) = \left(1 , 9\right)$ is $6$

$\left(1 , 3\right) , \left(1 , 6\right) , \mathmr{and} \left(1 , 9\right)$ are co-linear.

Therefore the line segment from $\left(1 , 3\right)$ to $\left(1 , 9\right)$ is bisected by the line $- y + 7 x = 1$

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Furthermore a line segment between $\left(1 , 3\right)$ and any point on a line through $\left(1 , 9\right)$ parallel to $- y + 7 x = 1$
will also be bisected by $- y + 7 x = 1$

The equation of this line is
$\textcolor{w h i t e}{\text{XXX}} y - 9 = 7 \left(x - 1\right)$
or
$\textcolor{w h i t e}{\text{XXX}} y - 7 x = 2$
or, in a form similar to the given equation
$\textcolor{w h i t e}{\text{XXX}} - y + 7 x = - 2$

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See similar problem solution at:
https://socratic.org/questions/a-line-segment-is-bisected-by-a-line-with-the-equation-2-y-x-1-if-one-end-of-the272239
for a more detailed solution with diagrams