Question

1. use the figure below for questions a - d.

a. give another name for lines r
b. name the intersection of line r and line s.
c. name three collinear points.
d. give another name for plane n.
you must show work fully and neatly to receive credit.

1. if rs = 23 - 2x, st = 9x - 5, and rt = 39, find rs.
2. if ln = 6x - 5, lm = x + 7, and mn = 3x + 20, find mn.
3. if k is the midpoint of jl, jk = 8x + 11 and kl = 14x - 1, find jl. (hint: draw a picture)
4. if line n bisects ce, find cd.

Explanation:

Step1: Answer for 6A

Lines can be named by any two points on the line. Another name for line $r$ could be line $AB$.

Step2: Answer for 6B

The intersection of two lines is a point. The intersection of line $r$ and line $s$ is point $B$.

Step3: Answer for 6C

Collinear points lie on the same line. Three collinear points are $A$, $B$, $C$ (since they lie on line $r$).

Step4: Answer for 6D

A plane can be named by three non - collinear points on the plane. Another name for plane $N$ could be plane $ABC$.

Step5: Answer for 7

Since $RS + ST=RT$, we substitute the given expressions: $(23 - 2x)+(9x - 5)=39$.
Simplify the left - hand side: $23-2x + 9x-5=39$, which gives $7x + 18=39$.
Subtract 18 from both sides: $7x=39 - 18=21$.
Divide both sides by 7: $x = 3$.
Then $RS=23-2x=23-2\times3=23 - 6 = 17$.

Step6: Answer for 8

Since $LN=LM + MN$, we substitute the given expressions: $6x-5=(x + 7)+(3x + 20)$.
Simplify the right - hand side: $6x-5=x + 7+3x + 20$, so $6x-5=4x + 27$.
Subtract $4x$ from both sides: $6x-4x-5=4x-4x + 27$, which gives $2x-5=27$.
Add 5 to both sides: $2x=27 + 5=32$.
Divide both sides by 2: $x = 16$.
Then $MN=3x + 20=3\times16+20=48 + 20=68$.

Step7: Answer for 9

If $K$ is the mid - point of $JL$, then $JK=KL$. So $8x + 11=14x-1$.
Subtract $8x$ from both sides: $8x-8x + 11=14x-8x-1$, which gives $11 = 6x-1$.
Add 1 to both sides: $11 + 1=6x-1+1$, so $12=6x$.
Divide both sides by 6: $x = 2$.
$JK=8x + 11=8\times2+11=16 + 11=27$, $KL=14x-1=14\times2-1=28 - 1=27$.
$JL=JK + KL=27+27 = 54$.

Step8: Answer for 10

If line $n$ bisects $\overline{CE}$, then $CD=DE$. So $x + 6=4x-21$.
Subtract $x$ from both sides: $x-x + 6=4x-x-21$, which gives $6=3x-21$.
Add 21 to both sides: $6 + 21=3x-21+21$, so $27=3x$.
Divide both sides by 3: $x = 9$.
$CD=x + 6=9+6=15$.

Answer:

A. Line $AB$
B. Point $B$
C. $A$, $B$, $C$
D. Plane $ABC$

1. $RS = 17$
2. $MN=68$
3. $JL = 54$
4. $CD = 15$