Question

use the diagram to the right for questions 1 - 4:

1. name two points collinear to point d. ________
2. give another name for line p. ________
3. name the intersection of line r and plane k. ________
4. name a point non - coplanar to plane k. ________

use the diagram to the right for questions 5 - 8:

1. how many planes appear in the figure? ________
2. name four coplanar points. ________
3. name a point non - coplanar to plane z. ________
4. name the intersection of plane jps and plane z. ________
5. if rs = 23 - 2x, st = 9x - 5, and rt = 39, find rs.
6. if ln = 6x - 5, lm = x + 7, and mn = 3x + 20, find mn.
7. if k is the midpoint of $overline{jl}$, jk = 8x + 11 and kl = 14x - 1, find jl.
8. if line n bisects $overline{ce}$, find cd.

Explanation:

Step1: Recall collinear - points on same line

Points A and E are on the same line as point D.

Step2: Rename line by two - points

Line p can be named as line AE.

Step3: Find intersection of line and plane

The intersection of line r and plane K is point X.

Step4: Identify non - coplanar point

Point W is non - coplanar to plane K.

Step5: Count planes in figure

There are 5 planes in the second figure (planes Z, JPS, JKL, PQMS, RQLN).

Step6: Name coplanar points

Points J, K, L, M are coplanar (lying on plane Z).

Step7: Identify non - coplanar point

Point P is non - coplanar to plane Z.

Step8: Find intersection of planes

The intersection of plane JPS and plane Z is line JK.

Step9: Use segment addition postulate

Since \(RS + ST=RT\), we have \((23 - 2x)+(9x - 5)=39\).
Simplify the left - hand side: \(23-2x + 9x-5=18 + 7x\).
Set up the equation \(18 + 7x=39\).
Subtract 18 from both sides: \(7x=39 - 18=21\).
Divide both sides by 7: \(x = 3\).
Then \(RS=23-2x=23-2\times3=23 - 6 = 17\).

Step10: Use segment addition postulate

Since \(LM+MN=LN\), we have \((x + 7)+(3x + 20)=6x-5\).
Simplify the left - hand side: \(x + 7+3x + 20=4x+27\).
Set up the equation \(4x + 27=6x-5\).
Subtract \(4x\) from both sides: \(27=6x-5-4x=2x-5\).
Add 5 to both sides: \(2x=27 + 5=32\).
Divide both sides by 2: \(x = 16\).
Then \(MN=3x + 20=3\times16+20=48 + 20=68\).

Step15: Use mid - point property

Since K is the mid - point of \(\overline{JL}\), \(JK = KL\).
Set up the equation \(8x + 11=14x-1\).
Subtract \(8x\) from both sides: \(11=14x-1-8x=6x-1\).
Add 1 to both sides: \(6x=11 + 1=12\).
Divide both sides by 6: \(x = 2\).
\(JK=8x + 11=8\times2+11=16 + 11=27\), \(KL = 14x-1=14\times2-1=28 - 1=27\).
\(JL=JK + KL=27+27 = 54\).

Step16: Use bisector property

Since line n bisects \(\overline{CE}\), \(CD=DE\).
Set up the equation \(x + 6=4x-21\).
Subtract \(x\) from both sides: \(6=4x-21-x=3x-21\).
Add 21 to both sides: \(3x=6 + 21=27\).
Divide both sides by 3: \(x = 9\).
\(CD=x + 6=9+6=15\).

Answer:

1. A, E
2. Line AE
3. X
4. W
5. 5
6. J, K, L, M
7. P
8. Line JK
9. 17
10. 68
11. 54
12. 15