Question

use the diagram. review 1.1 - 1.3

1. name the plane.
2. name three collinear points.
3. name three coplanar points.
4. name a line.
5. name a different line from #4.
6. name one ray.
7. name one line segment.
8. sketch the figure described. $overrightarrow{ab}$ and $overrightarrow{ac}$
9. sketch the figure described. plane c and plane d intersecting at $overleftrightarrow{xy}$
10. find bd.
11. find bd.
12. the endpoints of $overline{qr}$ are q(1, 6) and r(-7, 3). find the coordinates of the midpoint m.
13. the midpoint of $overline{gh}$ is m(4, -3). one endpoint is g(-2, 2). find the coordinates of endpoint h.
14. find the distance between the two points s(-5, -2) and t(-3, 4)
15. identify the segment bisector of $overline{rt}$. then find rt.

Explanation:

Step1: Solve for BD in question 10

Since BD = BC + CD and BC = 21, CD = 7, then BD=21 + 7.
BD=28

Step2: Solve for BD in question 11

Since CD = 73 and CB = 17, then BD=CD - CB.
BD=73 - 17=56

Step3: Find mid - point M of QR in question 12

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For Q(1,6) and R(-7,3), $x_1 = 1,x_2=-7,y_1 = 6,y_2 = 3$.
$x_M=\frac{1+( - 7)}{2}=\frac{-6}{2}=-3$
$y_M=\frac{6 + 3}{2}=\frac{9}{2}=4.5$
So M(-3,4.5)

Step4: Find endpoint H in question 13

Let the coordinates of H be (x,y). The mid - point formula for G(-2,2) and H(x,y) with mid - point M(4,-3) gives:
$\frac{-2 + x}{2}=4$ and $\frac{2 + y}{2}=-3$.
For $\frac{-2 + x}{2}=4$, multiply both sides by 2: - 2+x = 8, then x = 10.
For $\frac{2 + y}{2}=-3$, multiply both sides by 2: 2 + y=-6, then y=-8.
So H(10,-8)

Step5: Find distance between S and T in question 14

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 S(-5,-2) and T(-3,4), $x_1=-5,y_1=-2,x_2=-3,y_2 = 4$.
$d=\sqrt{(-3-( - 5))^2+(4-( - 2))^2}=\sqrt{(2)^2+(6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$

Step6: Find segment bisector and RT in question 15

Since point S is the segment bisector of RT, then RS = ST. So 7y-4 = 2y + 6.
Subtract 2y from both sides: 5y-4 = 6.
Add 4 to both sides: 5y=10, then y = 2.
RS=7y - 4=7(2)-4 = 10, ST = 2y + 6=2(2)+6 = 10.
RT=RS + ST=10 + 10 = 20

Answer:

1. BD = 28
2. BD = 56
3. M(-3,4.5)
4. H(10,-8)
5. $2\sqrt{10}$
6. Segment bisector: point S, RT = 20