Question

geometry quiz 1 unit 2

1. give another name for line r.
2. name the intersection of lines r and s.
3. name three collinear points.
4. give another name for plane n.
5. name a segment containing point c.
6. name a point coplanar to point k.
7. name the intersection of plane r and line jl.
8. name three non - collinear points.
9. give another name for line jk.
10. name a line containing point j.

use the figure below for questions 9 - 12.

1. if df = 61 and ef = 18, find de.
2. if de = 4x - 1, ef = 9, and df = 9x - 22, find the value of x.
3. if df = 78, de = 5x - 9, and ef = 2x + 10, find ef.
4. if de = 4x + 10, ef = 2x - 1, and df = 9x - 15, find df.

answer the following questions. draw a diagram if necessary.

1. m is the midpoint of the segment. find x. 11x - 21 8x + 15
2. m is the midpoint of $overline{pq}$. pm = 6x + 7 and mq = 9x - 8, find pq.
3. c is the midpoint of $overline{df}$. dc = 9x + 3 and cf = 11x - 17, find dc.
4. $overline{de}$ bisects $overline{ab}$ at c. if ac = 8x - 23 and cb = 3x + 22, find x.

Explanation:

Step1: Recall line - naming convention

A line can be named by any two points on it. For question 1, if line \(r\) has points \(A\) and \(B\) on it, another name could be line \(AB\).

Step2: Find intersection of lines

The intersection of two lines is a point. For question 2, find the common point of lines \(r\) and \(s\).

Step3: Identify collinear points

Collinear points lie on the same line. For question 3, look for three points on one line in the figure.

Step4: Rename a plane

A plane can be named by three non - collinear points on it. For question 4, if plane \(N\) has points \(A\), \(B\), and \(C\) on it, another name could be plane \(ABC\).

Step5: Find a segment with a given point

For question 5, look for a line segment that contains point \(C\) (for example, segment \(BC\)).

Step6: Find coplanar point

Coplanar points lie on the same plane. For question 6, if point \(K\) is in plane \(R\), point \(M\) (which is also in plane \(R\)) is a coplanar point.

Step7: Find intersection of plane and line

The intersection of a plane and a line can be a point. For question 7, find the common point of plane \(R\) and line \(JL\).

Step8: Identify non - collinear points

Non - collinear points do not lie on the same line. For question 8, pick three points not on one line in the figure.

Step9: Rename a line

For question 9, if line \(JK\) has another pair of points on it, say \(J\) and \(M\), another name could be line \(JM\).

Step10: Find a line with a given point

For question 10, a line containing point \(J\) could be line \(JK\).

Step11: Use segment addition postulate

The segment addition postulate states that if \(E\) is between \(D\) and \(F\), then \(DF=DE + EF\). So, \(DE=DF - EF\). Given \(DF = 61\) and \(EF = 18\), then \(DE=61 - 18=43\).

Step12: Apply segment addition postulate

Since \(DF=DE + EF\), we substitute the given expressions: \(9x-22=(4x - 1)+9\).

Step1: Simplify the right - hand side

\((4x - 1)+9=4x+8\). So, \(9x-22 = 4x+8\).

Step2: Solve for \(x\)

Subtract \(4x\) from both sides: \(9x-4x-22=4x-4x + 8\), which gives \(5x-22 = 8\). Then add 22 to both sides: \(5x-22 + 22=8 + 22\), so \(5x=30\). Divide both sides by 5: \(x = 6\).

Step13: Apply segment addition postulate

Since \(DF=DE + EF\), we substitute the given expressions: \(78=(5x - 9)+(2x + 10)\).

Step1: Simplify the right - hand side

\((5x - 9)+(2x + 10)=7x + 1\). So, \(7x+1 = 78\).

Step2: Solve for \(x\)

Subtract 1 from both sides: \(7x+1-1=78 - 1\), so \(7x=77\). Divide both sides by 7: \(x = 11\). Then find \(EF\): \(EF=2x + 10=2\times11+10=32\).

Step14: Apply segment addition postulate

Since \(DF=DE + EF\), we substitute the given expressions: \(9x-15=(4x + 10)+(2x - 1)\).

Step1: Simplify the right - hand side

\((4x + 10)+(2x - 1)=6x + 9\). So, \(9x-15=6x + 9\).

Step2: Solve for \(x\)

Subtract \(6x\) from both sides: \(9x-6x-15=6x-6x + 9\), which gives \(3x-15 = 9\). Then add 15 to both sides: \(3x-15 + 15=9 + 15\), so \(3x=24\). Divide both sides by 3: \(x = 8\). Then find \(DF\): \(DF=9x-15=9\times8-15=57\).

Step15: Use mid - point property

If \(M\) is the mid - point of segment \(LN\), then \(LM=MN\). So, \(11x-21=8x + 15\).

Step1: Solve for \(x\)

Subtract \(8x\) from both sides: \(11x-8x-21=8x-8x + 15\), which gives \(3x-21 = 15\). Then add 21 to both sides: \(3x-21 + 21=15 + 21\), so \(3x=36\). Divide both sides by 3: \(x = 12\).

Step16: Use mid - point property

If \(M\) is the mid - point of \(\overline{PQ}\), then \(PM = MQ\). So, \(6x+7=9x - 8\).

Step1: Solve for \(x\)

Subtract \(6x\) from both sides: \(6x-6x+7=9x-6x - 8\), which gives \(7 = 3x-8\). Then add 8 to both sides: \(7 + 8=3x-8 + 8\), so \(15=3x\). Divide both sides by 3: \(x = 5\). Then \(PM=6x + 7=6\times5+7 = 37\) and \(MQ=9x - 8=9\times5-8 = 37\). So, \(PQ=PM + MQ=37+37 = 74\).

Step17: Use mid - point property

If \(C\) is the mid - point of \(\overline{DF}\), then \(DC = CF\). So, \(9x+3=11x - 17\).

Step1: Solve for \(x\)

Subtract \(9x\) from both sides: \(9x-9x+3=11x-9x - 17\), which gives \(3 = 2x-17\). Then add 17 to both sides: \(3 + 17=2x-17 + 17\), so \(20=2x\). Divide both sides by 2: \(x = 10\). Then \(DC=9x + 3=9\times10+3 = 93\).

Step18: Use bisector property

If \(\overline{DE}\) bisects \(\overline{AB}\) at \(C\), then \(AC = CB\). So, \(8x-23=3x + 22\).

Step1: Solve for \(x\)

Subtract \(3x\) from both sides: \(8x-3x-23=3x-3x + 22\), which gives \(5x-23 = 22\). Then add 23 to both sides: \(5x-23 + 23=22 + 23\), so \(5x=45\). Divide both sides by 5: \(x = 9\).

Answer:

1. (Depends on points on line \(r\), e.g., line \(AB\))
2. (Intersection point of \(r\) and \(s\))
3. (Three points on one line in the figure)
4. (Depends on non - collinear points on plane \(N\), e.g., plane \(ABC\))
5. (e.g., segment \(BC\))
6. (e.g., point \(M\) if \(K\) is in plane \(R\))
7. (Intersection point of plane \(R\) and line \(JL\))
8. (Three non - collinear points in the figure)
9. (e.g., line \(JM\))
10. (e.g., line \(JK\))
11. 43
12. 6
13. 32
14. 57
15. 12
16. 74
17. 93
18. 9