Question

review 1.1 - 1.4
geometry
use the diagram to answer the following questions:

1. name the ray whose endpoint is point k and passes through point r.
2. how many different rays have an endpoint at point r?

a. give four different names for them.
b. name a pair of opposite rays from this list.
3 name the line in the diagram two different ways.

1. name two congruent segments.
2. fill in the missing word:

a. points j, k, and t are ______________.
b. points s, r, and m are ______________.

1. does line p contain point m?
2. line segment rt that contains s such that s is between r and t.

rs = 14 units, st = 4x + 6 units, and rt = 8x units.
draw a diagram, solve for x and find the measure of (overline{rt}).
13 draw, (overline{lm}) that bisects (angle tlc). if (angle tlc=8x + 26) and (angle clm = 3x+15), find:
x = ____
m(angle tlc)=____
m(angle tlm)=____

Explanation:

Step1: Recall ray - naming convention

A ray is named starting with its endpoint. So the ray with endpoint K passing through R is $\overrightarrow{KR}$.

Step2: Identify rays from point R

A ray has an endpoint and extends infinitely in one direction. From point R, rays are $\overrightarrow{RM}$, $\overrightarrow{RK}$, $\overrightarrow{RS}$, $\overrightarrow{RT}$. There are 4 different rays with endpoint R.

Step3: Define opposite rays

Opposite rays are two rays that have the same endpoint and form a straight - line. From the list of rays from R, $\overrightarrow{RS}$ and $\overrightarrow{RT}$ are opposite rays.

Step4: Name the line

A line can be named using any two points on it. The line in the diagram can be named as $\overleftrightarrow{ST}$ or $\overleftrightarrow{JR}$.

Step5: Find congruent segments

From the diagram, if we assume equal - length markings, $\overline{SJ}\cong\overline{JR}$.

Step6: Define collinear points

Points J, K, and T are collinear (lying on the same straight line). Points S, R, and M are non - collinear (not lying on the same straight line).

Step7: Check if M is on line p

From the diagram, line p does not contain point M.

Step8: Solve for x in line - segment problem

Since $RS + ST=RT$ (by the segment addition postulate), we substitute the given values: $14+(4x + 6)=8x$.
Simplify the left - hand side: $20 + 4x=8x$.
Subtract $4x$ from both sides: $20=4x$.
Divide both sides by 4: $x = 5$.
Then $RT=8x=8\times5 = 40$ units.

Step9: Solve for x in angle - bisector problem

Since $\overline{LM}$ bisects $\angle TLC$, then $\angle TLC = 2\angle CLM$.
Substitute the given angle expressions: $8x+26 = 2(3x + 15)$.
Expand the right - hand side: $8x+26=6x + 30$.
Subtract $6x$ from both sides: $2x+26 = 30$.
Subtract 26 from both sides: $2x=4$.
Divide both sides by 2: $x = 2$.
Then $m\angle TLC=8x + 26=8\times2+26=42^{\circ}$, and $m\angle TLM=\frac{1}{2}m\angle TLC = 21^{\circ}$.

Answer:

1. $\overrightarrow{KR}$
2. 4; $\overrightarrow{RM}$, $\overrightarrow{RK}$, $\overrightarrow{RS}$, $\overrightarrow{RT}$; $\overrightarrow{RS}$ and $\overrightarrow{RT}$
3. $\overleftrightarrow{ST}$, $\overleftrightarrow{JR}$
4. $\overline{SJ}\cong\overline{JR}$
5. a. collinear; b. non - collinear
6. No
7. $x = 5$, $RT = 40$ units
8. $x = 2$, $m\angle TLC=42^{\circ}$, $m\angle TLM = 21^{\circ}$