U6L15 Weighted Averages

To find the ​ (a)   of a line segment, we can ​ (b)   . For example, to find the midpoint of the segment from A=(0,4) to B=(6,7), average the coordinates of A and B: $$\left(\frac{0+6}{2},\ \frac{4+7}{2}\right)=\left(3,\ 5.5\right)$$ . Another way to write what we just did is $$\frac{1}{2}\left(A+B\right)\$$ or ​ (c)   .

$$A=\left(0,4\right)\ to\ B\left(6,7\right)$$  

Now, let’s find the point that is $$\frac{2}{3}$$ of the way from A to B. In other words, ​ (a)   so that segments AC and CB are in a 2:1 ​ (b)   . In the horizontal direction, segment AB stretches from x=0 to x=6. The distance from 0 to 6 is ​ (c)   , so we calculate 2/3 of 6 to get​ (d)   . Point C will be 4 horizontal units away from A, which means an ​ (e)   .

In the vertical direction, segment  stretches from $$y=4$$  to ​ (a)   . The distance from 4 to 7 is ​ (b)   , so we can calculate ​ (c)    of 3 to get 2. Point  must be 2 vertical units away from , which means a ​ (d)   .

It is possible to do this all at once by saying $$C=\frac{1}{3}A+\frac{2}{3}B$$ . This is called a ​ (a)   . Instead of finding the point in the middle, we want to find a point closer ​ (b)   to  than to $$A$$ . So we give point   $$B$$ ​ ​ ​​ ​more weight —it has a coefficient ​ of ​ (c)    rather than $$\frac{1}{3}$$  as in the midpoint calculation. To calculate , substitute and evaluate.

$$\frac{1}{3}A+\frac{2}{3}B$$

$$\frac{1}{3}$$ ​ (d)   + $$\frac{2}{3}$$ ​ (e)  

$$\left(0,4\right)+\left(4,\ \frac{14}{3}\right)$$

$$\left(4,6\right)$$

Which equation can be used to find B?

$$\frac{2}{3}A+\frac{1}{3}C$$

$$\frac{2}{3}\left(-5,2\right)+\frac{1}{3}\left(4,-10\right)$$

$$\left(-\frac{10}{2},\frac{4}{3}\right)+\left(\frac{4}{3},-\frac{10}{3}\right)$$

The coordinate of B is...

The equation to find C is...

$$\frac{1}{3}A+\frac{2}{3}B$$

$$\frac{1}{3}\left(-8,\ 4\right)+\frac{2}{3}\left(10,-2\right)$$

What comes next?

$$\frac{1}{3}A+\frac{2}{3}B$$

$$\frac{1}{3}\left(-8,\ 4\right)+\frac{2}{3}\left(10,-2\right)$$

$$\left(-\frac{8}{3},\frac{4}{3}\right)+\left(\frac{20}{3},-\frac{4}{3}\right)$$

Which is the correct equation to find Q?