Question

1. complete the steps below for constructing a bisector of $overline{jk}$, shown below, using paper - folding.

step 1: use a compass and straightedge to copy $overline{jk}$ on a piece of paper.
step 2: fold the paper so that point __________ is on top of point __________.
step 3: open the paper. label the point where the crease intersects the segment as point n.
point n is the __________ of $overline{jk}$ and the crease is a __________ of $overline{jk}$.

1. the weights of the points a and b, shown on the number line, are $\frac{5}{6}$ and $\frac{1}{6}$, respectively.

which statement is true?
a. the weighted average of a and b lies beyond point b.
b. the weighted average of a and b is closer to point a than to point b.
c. the weighted average of a and b is closer to point b than to point a.
d. the weighted average of a and b lies exactly midway between the points a and b.

1. find the coordinate of p that represents the weighted average of points f and h if point f weights 30% and point h weights 70%.
2. find the coordinate of point p along the directed line segment ab so that a is located at (4,0), b is at (9,10), and the ratio of ap to pb is 1 to 1.
3. draw the segment pq with endpoints p(-2,-1) and q(2,4) on the coordinate plane. then find the length and midpoint of $overline{pq}$.

Explanation:

Step1: Solve question 4

Fold the paper so that point $J$ is on top of point $K$. Point $N$ is the mid - point of $\overline{JK}$ and the crease is a perpendicular bisector of $\overline{JK}$.

Step2: Solve question 5

The weighted average formula is $w_1x_1 + w_2x_2$, where $w_1=\frac{5}{6}$, $x_1 = 2$, $w_2=\frac{1}{6}$, $x_2=7$.
\[

$$\begin{align*} \text{Weighted average}&=\frac{5}{6}\times2+\frac{1}{6}\times7\\ &=\frac{10 + 7}{6}\\ &=\frac{17}{6}\approx2.83 \end{align*}$$

\]
Since $\frac{17}{6}\approx2.83$ is closer to point $A(2)$ than to point $B(7)$, the answer is B.

Step3: Solve question 6

Let the coordinate of $F=-2$ and the coordinate of $H = 5$. Using the weighted - average formula $w_1x_1+w_2x_2$, where $w_1 = 0.3$, $x_1=-2$, $w_2 = 0.7$, $x_2 = 5$.
\[

$$\begin{align*} \text{Weighted average}&=0.3\times(-2)+0.7\times5\\ &=- 0.6+3.5\\ &=2.9 \end{align*}$$

\]

Step4: Solve question 7

If the ratio of $AP$ to $PB$ is $1$ to $1$, then $P$ is the mid - point of $\overline{AB}$.
The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1 = 4,y_1 = 0,x_2 = 9,y_2 = 10$.
\[

$$\begin{align*} x_P&=\frac{4 + 9}{2}=\frac{13}{2}=6.5\\ y_P&=\frac{0 + 10}{2}=5 \end{align*}$$

\]
The coordinate of $P$ is $(6.5,5)$.

Step5: Solve question 8

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For $P(-2,-1)$ and $Q(2,4)$,
\[

$$\begin{align*} d&=\sqrt{(2+2)^2+(4 + 1)^2}\\ &=\sqrt{4^2+5^2}\\ &=\sqrt{16 + 25}\\ &=\sqrt{41}\approx6.4 \end{align*}$$

\]
The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1+x_2}{2},\frac{y_1 + y_2}{2})$.
\[

$$\begin{align*} x_{mid}&=\frac{-2+2}{2}=0\\ y_{mid}&=\frac{-1 + 4}{2}=\frac{3}{2}=1.5 \end{align*}$$

\]
The length of $\overline{PQ}$ is $\sqrt{41}\approx6.4$ and the mid - point is $(0,1.5)$.

Answer:

1. $J$, $K$, mid - point, perpendicular bisector
2. B
3. $2.9$
4. $(6.5,5)$
5. Length: $\sqrt{41}\approx6.4$, Mid - point: $(0,1.5)$