Finding a centered line segment from an existing line segment Given a line segment AB, which has a center point F and a length greater than 5, I'm trying to figure out the endpoints of a new line segment CD of length 5 which also has it's center point at F and lies exactly on AB. Is there a simple way to find CD given AB?
Sorry if it sounds a little convoluted. I feel like this may have a term or pattern used to describe it that I am unaware of.
 A: Let $(x_{1},y_{1})$ and $(x_{2}, y_{2})$ be the endpoints of AB. The midpoint F has coordinates given by $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$. The equation for this line is given by:
$$(x(t),y(t)) = (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) + t(x_{2}-x_{1}, y_{2}-y_{1}), t \in \mathbb{R}.$$
One desires to find $t$ such that:
$$|(x(t),y(t)) - F| = \frac{5}{2},$$
therefore,
$$|t|\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} = \frac{5}{2}.$$
The equation above has the following two solutions:
$$t_{+} = \frac{5}{2\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}},$$
$$t_{-} = -\frac{5}{2\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}.$$
Hence, the desired endpoints of the line CD are given by:
$$(x_{3},y_{3}) = (x(t_{-}), y(t_{-})) = (\frac{x_{1}+x_{2}}{2} + t_{-}(x_{2}-x_{1}), \frac{y_{1}+y_{2}}{2} + t_{-}(y_{2}-y_{1})),$$
$$(x_{4},y_{4}) = (x(t_{+}), y(t_{+})) = (\frac{x_{1}+x_{2}}{2} + t_{+}(x_{2}-x_{1}), \frac{y_{1}+y_{2}}{2} + t_{+}(y_{2}-y_{1})).$$
A: It doesn’t really matter whether or not the original segment length $\lvert B-A\rvert$ is greater than $5$. A unit vector in the direction of the line segment is $\mathbf u = {B-A\over\lvert B-A\rvert}$. To get a different line segment with the same direction centered on $F$, you just need to move half the new segment length $d$ from $F$ in the direction of $\mathbf u$ and the same in the opposite direction. That is, the endpoints of the new segment are $F\pm\frac d2\mathbf u$.
