unit 1 test study guide (geometry basics)
name:____ date:____ per:____
topic 1: points, lines & planes
use the diagram to the right to answer questions 1 - 4.
1. name two points collinear to point k.____
2. give another name for line b.____
3. name the intersection of line c and plane r.____
4. name a point non - coplanar to plane r.____
use the diagram to the right to answer questions 5 - 8.
5. how many planes are shown in the figure?____
6. give another name for plane w.____
7. name the intersection of plane ade and plane w.____
8. name a point non - collinear to points a and b.____
topic 2: segment addition postulate
9. if df = 42, find de.
10. in the diagram below, if jl = 10x−2, jk = 5x−8, and kl = 7x−12, find kl.
11. if s is the midpoint of rt, rs = 5x + 17, and st = 8x−31, find rs.
12. if line y bisects ac, ab = 4−5x, and bc = 2x + 25, find ac.
13. if b is the midpoint of ac, ac = cd, ab = 3x + 4, ac = 11x−17, and ce = 49, find de.
21. given directed line segment wv, find the coordinates of r such that the ratio of wr to rv is 3:4. plot point r.
22. given ef with e( - 7,4) and f( - 4, - 5). if g lies on ef such that the ratio of eg to gf is 1:2, find the coordinates of g.
23. given jl with j( - 5, - 1) and l(19, - 9). if k lies on jl such that the ratio of jk to jl is 7:8, find the coordinates of k.

Question

Accepted Answer

1. **Answer - Question 1**:
   - # Explanation:
     ## Step1: Recall collinear points definition
     Collinear points lie on the same line. Points $H$ and $N$ lie on the same line as point $K$.
   - # Answer:
     $H$ and $N$
2. **Answer - Question 2**:
   - # Explanation:
     ## Step1: Rename a line
     A line can be named by any two points on it. Line $b$ can be named as line $HK$ (or other pairs of points on the line).
   - # Answer:
     Line $HK$
3. **Answer - Question 3**:
   - # Explanation:
     ## Step1: Find intersection
     The intersection of line $c$ and plane $R$ is point $M$ as the line pierces the plane at that point.
   - # Answer:
     Point $M$
4. **Answer - Question 4**:
   - # Explanation:
     ## Step1: Identify non - coplanar point
     A point not in plane $R$ is non - coplanar to plane $R$. Point $E$ is outside plane $R$.
   - # Answer:
     Point $E$
5. **Answer - Question 5**:
   - # Explanation:
     ## Step1: Count planes
     The planes shown are plane $R$, plane $W$, and plane $ADE$. So, there are 3 planes.
   - # Answer:
     3
6. **Answer - Question 6**:
   - # Explanation:
     ## Step1: Rename a plane
     A plane can be named by three non - collinear points on it. Plane $W$ can be named as plane $ABCD$ (or other sets of non - collinear points on the plane).
   - # Answer:
     Plane $ABCD$
7. **Answer - Question 7**:
   - # Explanation:
     ## Step1: Find intersection
     The intersection of plane $ADE$ and plane $W$ is line $AD$.
   - # Answer:
     Line $AD$
8. **Answer - Question 8**:
   - # Explanation:
     ## Step1: Identify non - collinear point
     A point not on the line passing through $A$ and $B$ is non - collinear to $A$ and $B$. Point $E$ is non - collinear to $A$ and $B$.
   - # Answer:
     Point $E$
9. **Answer - Question 9**:
   - # Explanation:
     ## Step1: Use segment addition postulate
     Since $DF=DE + EF$, and $DF = 42$, $DE=7x + 1$, $EF=4x-3$, then $42=(7x + 1)+(4x-3)$.
     ## Step2: Simplify the equation
     $42=7x + 1+4x-3$, $42 = 11x-2$.
     ## Step3: Solve for $x$
     Add 2 to both sides: $42 + 2=11x$, $44 = 11x$, so $x = 4$.
     ## Step4: Find $DE$
     Substitute $x = 4$ into $DE=7x + 1$, $DE=7\times4+1=29$.
   - # Answer:
     29
10. **Answer - Question 10**:
   - # Explanation:
     ## Step1: Use segment addition postulate
     Since $JL=JK + KL$, and $JL = 10x-2$, $JK=5x - 8$, $KL=7x-12$, then $10x-2=(5x - 8)+(7x-12)$.
     ## Step2: Simplify the equation
     $10x-2=5x - 8+7x-12$, $10x-2=12x-20$.
     ## Step3: Solve for $x$
     Subtract $10x$ from both sides: $-2=2x-20$, add 20 to both sides: $18 = 2x$, so $x = 9$.
     ## Step4: Find $KL$
     Substitute $x = 9$ into $KL=7x-12$, $KL=7\times9-12=51$.
   - # Answer:
     51
11. **Answer - Question 11**:
   - # Explanation:
     ## Step1: Use mid - point property
     Since $S$ is the mid - point of $RT$, $RS=ST$. So $5x + 17=8x-31$.
     ## Step2: Solve for $x$
     Subtract $5x$ from both sides: $17=3x-31$, add 31 to both sides: $48 = 3x$, so $x = 16$.
     ## Step3: Find $RS$
     Substitute $x = 16$ into $RS=5x + 17$, $RS=5\times16+17=97$.
   - # Answer:
     97
12. **Answer - Question 12**:
   - # Explanation:
     ## Step1: Use segment bisection property
     Since line $y$ bisects $AC$, $AB = BC$. So $4-5x=2x + 25$.
     ## Step2: Solve for $x$
     Add $5x$ to both sides: $4=7x + 25$, subtract 25 from both sides: $-21=7x$, so $x=-3$.
     ## Step3: Find $AC$
     $AC=AB + BC$, and $AB = 4-5x$, $BC=2x + 25$. Substitute $x=-3$ into either $AB$ or $BC$ and double it. $AB=4-5\times(-3)=19$, $AC = 2AB=38$.
   - # Answer:
     38
13. **Answer - Question 13**:
   - # Explanation:
     ## Step1: Use mid - point property
     Since $B$ is the mid - point of $AC$, $AB=BC$. So $3x + 4=\frac{1}{2}(11x-17)$.
     ## Step2: Solve for $x$
     Multiply both sides by 2: $6x + 8=11x-17$, subtract $6x$ from both sides: $8 = 5x-17$, add 17 to both sides: $25 = 5x$, so $x = 5$.
     ## Step3: Find $AC$
     $AC=11x-17$, substitute $x = 5$, $AC=11\times5-17=38$. Since $AC = CD$, $CD = 38$.
     ## Step4: Find $DE$
     $DE=CE - CD$, $CE = 49$, $CD = 38$, so $DE=49 - 38=11$.
   - # Answer:
     11
21. **Answer - Question 21**:
   - # Explanation:
     ## Step1: Use section formula
     Let $W(x_1,y_1)$ and $V(x_2,y_2)$. The ratio $m:n = 3:4$. The $x$ - coordinate of $R$ is $x=\frac{3x_2+4x_1}{3 + 4}$, the $y$ - coordinate of $R$ is $y=\frac{3y_2+4y_1}{3 + 4}$. (Assume coordinates of $W$ and $V$ are known from the graph).
     ## Step2: Calculate coordinates
     After substituting the values of $x_1,x_2,y_1,y_2$ and performing the arithmetic operations, we get the coordinates of $R$. (Since the actual coordinates of $W$ and $V$ are not given in text, we can't give numerical values. But the general method is as above).
   - # Answer:
     Coordinates of $R$ are $(\frac{3x_2+4x_1}{7},\frac{3y_2+4y_1}{7})$
22. **Answer - Question 22**:
   - # Explanation:
     ## Step1: Use section formula
     Given $E(-7,4)$, $F(-4,-5)$ and ratio $m:n = 1:2$. The $x$ - coordinate of $G$ is $x=\frac{1\times(-4)+2\times(-7)}{1 + 2}=\frac{-4-14}{3}=-6$.
     ## Step2: Calculate $y$ - coordinate
     The $y$ - coordinate of $G$ is $y=\frac{1\times(-5)+2\times4}{1 + 2}=\frac{-5 + 8}{3}=1$.
   - # Answer:
     $(-6,1)$
23. **Answer - Question 23**:
   - # Explanation:
     ## Step1: Use section formula
     Given $J(-5,-1)$, $L(19,-9)$ and ratio $m:n = 7:8$. The $x$ - coordinate of $K$ is $x=\frac{7\times19+8\times(-5)}{7 + 8}=\frac{133-40}{15}=\frac{93}{15}=\frac{31}{5}=6.2$.
     ## Step2: Calculate $y$ - coordinate
     The $y$ - coordinate of $K$ is $y=\frac{7\times(-9)+8\times(-1)}{7 + 8}=\frac{-63-8}{15}=\frac{-71}{15}\approx - 4.73$.
   - # Answer:
     $(6.2,-4.73)$

QUESTION IMAGE

Question

Explanation:

Step1: Recall collinear points definition

Step1: Rename a line

Step1: Find intersection

Step1: Identify non - coplanar point

Step1: Count planes

Step1: Rename a plane

Step1: Find intersection

Step1: Identify non - collinear point

Step1: Use segment addition postulate

Step2: Simplify the equation

Step3: Solve for \(x\)

Step4: Find \(DE\)

Step1: Use segment addition postulate

Step2: Simplify the equation

Step3: Solve for \(x\)

Step4: Find \(KL\)

Step1: Use mid - point property

Step2: Solve for \(x\)

Step3: Find \(RS\)

Step1: Use segment bisection property

Step2: Solve for \(x\)

Answer:

Step1: Recall collinear points definition

Step1: Rename a line

Step1: Find intersection

Step1: Identify non - coplanar point

Step1: Count planes

Step1: Rename a plane

Step1: Find intersection

Step1: Identify non - collinear point

Step1: Use segment addition postulate

Step2: Simplify the equation

Step3: Solve for \(x\)

Step4: Find \(DE\)

Step1: Use segment addition postulate

Step2: Simplify the equation

Step3: Solve for \(x\)

Step4: Find \(KL\)

Step1: Use mid - point property

Step2: Solve for \(x\)

Step3: Find \(RS\)

Step1: Use segment bisection property

Step2: Solve for \(x\)

Step3: Find \(AC\)

Step1: Use mid - point property

Step2: Solve for \(x\)

Step3: Find \(AC\)

Step4: Find \(DE\)

Step1: Use section formula

Step2: Calculate coordinates

Step1: Use section formula

Step2: Calculate \(y\) - coordinate

Step1: Use section formula

Step2: Calculate \(y\) - coordinate