Lesson 15

Weighted Averages

Let’s split segments using averages and ratios.

For the questions in this activity, use the coordinate grid if it is helpful to you.

What is the midpoint of the segment connecting $(1,2)$ and $(5,2)$ ?
What is the midpoint of the segment connecting $(5,2)$ and $(5,10)$ ?
What is the midpoint of the segment connecting $(1,2)$ and $(5,10)$ ?

Point $A$ has coordinates $(2,4)$ . Point $B$ has coordinates $(8,1)$ .

Line segment AB, A= 2 comma 4, B=8 comma 1.

Find the point that partitions segment $AB$ in a $2:1$ ratio.
Calculate $C=\frac 13 A + \frac 23 B$ .
What do you notice about your answers to the first 2 questions?
For 2 new points $K$ and $L$ , write an expression for the point that partitions segment $KL$ in a $3:1$ ratio.

Are you ready for more?

Consider the general quadrilateral $QRST$ with $Q=(0,0),R=(a,b),S=(c,d),$ and $T=(e,f)$ .

Find the midpoints of each side of this quadrilateral.
Show that if these midpoints are connected consecutively, the new quadrilateral formed is a parallelogram.

Here is quadrilateral $ABCD$ .

Polygon drawn in a coordinate plane, first quadrant. Vertices at A, 2 comma 3, B, 6 comma 3, C, 8 comma 9, D, 4 comma 9.

Find the point that partitions segment $AB$ in a $1:4$ ratio. Label it $B’$ .
Find the point that partitions segment $AD$ in a $1:4$ ratio. Label it $D’$ .
Find the point that partitions segment $AC$ in a $1:4$ ratio. Label it $C’$ .
Is $AB’C’D’$ a dilation of $ABCD$ ? Justify your answer.

To find the midpoint of a line segment, we can average the coordinates of the endpoints. For example, to find the midpoint of the segment from $A=(0,4)$ to $B=(6,7)$ , average the coordinates of $A$ and $B$ : $\left(\frac{0 + 6}{2}, \frac{4+7}{2}\right) = (3,5.5)$ . Another way to write what we just did is $\frac12 (A+B)$ or $\frac12 A + \frac12 B$ .

Now, let’s find the point that is $\frac23$ of the way from $A$ to $B$ . In other words, we’ll find point $C$ so that segments $AC$ and $CB$ are in a $2:1$ ratio.

In the horizontal direction, segment $AB$ stretches from $x=0$ to $x=6$ . The distance from 0 to 6 is 6 units, so we calculate $\frac23$ of 6 to get 4. Point $C$ will be 4 horizontal units away from $A$ , which means an $x$ -coordinate of 4.

Coordinate plane, first quadrant. Line from A, 0 comma 4 through C, 4 comma 6, to B, 6 comma 7. Lines drawn from A to 6 comma 4, from C to 6 comma 4, from C to 4 comma 4, and from B to 6 comma 4.

In the vertical direction, segment $AB$ stretches from $y=4$ to $y=7$ . The distance from 4 to 7 is 3 units, so we can calculate $\frac23$ of 3 to get 2. Point $C$ must be 2 vertical units away from $A$ , which means a $y$ -coordinate of 6.

It is possible to do this all at once by saying $C = \frac13 A + \frac23 B$ . This is called a weighted average. Instead of finding the point in the middle, we want to find a point closer to $B$ than to $A$ . So we give point $B$ more weight—it has a coefficient of $\frac23$ rather than $\frac12$ as in the midpoint calculation. To calculate $C = \frac13 A + \frac23 B$ , substitute and evaluate.

$\frac13 A + \frac23 B$

$\frac13 (0,4) + \frac23 (6,7)$

$\left(0,\frac43 \right) + \left(4, \frac{14}{3} \right)$

$(4,6)$

Either way, we found that the coordinates of $C$ are $(4,6)$ .

opposite

Two numbers are opposites of each other if they are the same distance from 0 on the number line, but on opposite sides.

The opposite of 3 is -3 and the opposite of -5 is 5.
point-slope form

The form of an equation for a line with slope $m$ through the point $(h,k)$ . Point-slope form is usually written as $y-k = m(x-h)$ . It can also be written as $y = k + m(x-h)$ .
reciprocal

If $p$ is a rational number that is not zero, then the reciprocal of $p$ is the number $\frac{1}{p}$ .

Lesson 15

15.1: Part Way: Points

15.2: Part Way: Segment

15.3: Part Way: Quadrilateral

Summary

Glossary Entries