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Partitioning a Segment in a Given Ratio

Suppose you have a line segment $\overline{PQ}$ on the coordinate plane, and you need to find the point on the segment $\frac{1}{3}$  of the way from $P$ to $Q$.

Let’s first take the easy case where $P$ is at the origin and line segment is a horizontal one.

The length of the line is $6$ units and the point on the segment $\frac{1}{3}$  of the way from $P$ to $Q$ would be $2$ units away from $P$, $4$ units away from $Q$ and would be at $\left(2,0\right)$.

Consider the case where the segment is not a horizontal or vertical line.

The components of the directed segment $\overline{PQ}$ are $\langle 6,3\rangle$  and we need to find the point, say $X$ on the segment $\frac{1}{3}$  of the way from $P$ to $Q$.

Then, the components of the segment $\overline{PX}$ are $\langle \left(\frac{1}{3}\right)\left(6\right),\left(\frac{1}{3}\right)\left(3\right)\rangle =\langle 2,1\rangle$.

Since the initial point of the segment is at origin, the coordinates of the point $X$ are given by $\left(0+2,0+1\right)=\left(2,1\right)$.

Now let’s do a trickier problem, where neither $P$ nor $Q$ is at the origin.

Use the end points of the segment $\overline{PQ}$  to write the components of the directed segment.

$\begin{array}{l}\langle \left(x_2-x_1\right),\left(y_2-y_1\right)\rangle =\langle \left(7-1\right),\left(2-6\right)\rangle \\ =\langle 6,-4\rangle \end{array}$

Now in a similar way, the components of the segment $\overline{PX}$  where $X$ is a point on the segment $\frac{1}{3}$  of the way from $P$ to $Q$are $\langle \left(\frac{1}{3}\right)\left(6\right),\left(\frac{1}{3}\right)\left(-4\right)\rangle =\langle 2,-1.25\rangle$.

To find the coordinates of the point $X$ add the components of the segment $\overline{PX}$  to the coordinates of the initial point $P$.

So, the coordinates of the point $X$ are $\left(1+2,6-1.25\right)=\left(3,4.75\right)$.

Note that the resulting segments, $\overline{PX}$ and $\overline{XQ}$, have lengths in a ratio of $1:2$.

In general: what if you need to find a point on a line segment that divides it into two segments with lengths in a ratio $a:b$?

Consider the directed line segment $\overline{XY}$  with coordinates of the endpoints as $X\left(x_1,y_1\right)$  and $Y\left(x_2,y_2\right)$.

Suppose the point $Z$ divided the segment in the ratio $a:b$, then the point is $\frac{a}{a+b}$ of the way from $X$ to $Y$.

So, generalizing the method we have, the components of the segment $\overline{XZ}$ are $\langle \left(\frac{a}{a+b}\left(x_2-x_1\right)\right),\left(\frac{a}{a+b}\left(y_2-y_1\right)\right)\rangle$.

Then, the $X$-coordinate of the point $Z$ is

$\begin{array}{l}{x}_1+\frac{a}{a+b}\left(x_2-x_1\right)=\frac{x_1\left(a+b\right)+a\left(x_2-x_1\right)}{a+b}\\ =\frac{b{x}_1+a{x}_2}{a+b}\end{array}$.

Similarly, the $Y$-coordinate is

$\begin{array}{l}{y}_1+\frac{a}{a+b}\left(y_2-y_1\right)=\frac{y_1\left(a+b\right)+a\left(y_2-y_1\right)}{a+b}\\ =\frac{b{y}_1+a{y}_2}{a+b}\end{array}$.

Therefore, the coordinates of the point $Z$ are $\left(\frac{b{x}_1+a{x}_2}{a+b},\frac{b{y}_1+a{y}_2}{a+b}\right)$.

You can note that the Midpoint Formula is a special case of this formula when $a=b=1$.