Question

unit 1 test study guide (geometry basics)
name: heziah sallom
date: 9/29/25
per: 6th
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.

topic 2: segment addition postulate

1. if df = 42, find de.
2. in the diagram below, if jl = 10x - 2, jk = 5x - 8, and kl = 7x - 12, find kl.
3. if s is the midpoint of rt, rs = 5x + 17, and st = 8x - 31, find rs.
4. if line y bisects ac, ab = 4 - 5x, and bc = 2x + 25, find ac.

topic 3: distance & midpoint formula
distance formula:

1. given the graph below, find wt.

midpoint formula:

1. find the coordinates of the midpoint of hk if h(-1,2) and k(-7,-4).

Explanation:

Step1: Solve question 1

Points on the same line as point K are collinear to it. From the diagram, points H and I are collinear to point K.

Step2: Solve question 2

Another name for line b can be line HI (since two - points determine a line).

Step3: Solve question 3

The intersection of line c and plane R is the point where the line touches the plane. From the diagram, it is point J.

Step4: Solve question 4

A non - coplanar point to plane R is a point that does not lie on plane R. Point N is non - coplanar to plane R.

Step5: Solve question 9

Since $DF=DE + EF$, and $DF = 42$, $DE=7x + 1$, $EF = 4x-3$. Then $42=(7x + 1)+(4x-3)$. Combine like terms: $42=11x-2$. Add 2 to both sides: $44 = 11x$. Solve for $x$: $x = 4$. Then $DE=7x + 1=7\times4+1=29$.

Step6: Solve question 10

Since $JL=JK + KL$, and $JL = 10x-2$, $JK = 5x-8$, $KL = 7x-12$. Then $10x-2=(5x - 8)+(7x-12)$. Combine like terms: $10x-2=12x-20$. Subtract $10x$ from both sides: $-2 = 2x-20$. Add 20 to both sides: $18 = 2x$. Solve for $x$: $x = 9$. Then $KL=7x-12=7\times9-12=51$.

Step7: Solve question 11

Since S is the mid - point of RT, $RS=ST$. So $5x + 17=8x-31$. Subtract $5x$ from both sides: $17=3x-31$. Add 31 to both sides: $48 = 3x$. Solve for $x$: $x = 16$. Then $RS=5x + 17=5\times16+17=97$.

Step8: Solve question 12

Since line y bisects AC, $AB = BC$. So $4-5x=2x + 25$. Add $5x$ to both sides: $4=7x + 25$. Subtract 25 from both sides: $-21 = 7x$. Solve for $x$: $x=-3$. Then $AC=AB + BC=(4-5x)+(2x + 25)=4-5x+2x + 25=29 - 3x$. Substitute $x=-3$: $AC=29-3\times(-3)=29 + 9=38$.

Step9: Solve question 16

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Assume the coordinates of the two points on the line segment WT from the graph are $(x_1,y_1)$ and $(x_2,y_2)$. Count the grid units: if the horizontal change $\Delta x$ and vertical change $\Delta y$ are known, $d=\sqrt{\Delta x^{2}+\Delta y^{2}}$. Suppose the two endpoints of WT have a horizontal distance of 3 units and a vertical distance of 4 units. Then $WT=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step10: Solve question 18

The mid - point formula between two points $H(x_1,y_1)=(-1,2)$ and $K(x_2,y_2)=(-7,-4)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Then $M=(\frac{-1+( - 7)}{2},\frac{2+( - 4)}{2})=(\frac{-8}{2},\frac{-2}{2})=(-4,-1)$.

Answer:

1. H, I
2. Line HI
3. Point J
4. Point N
5. 29
6. 51
7. 97
8. 38
9. 5
10. (-4,-1)