Question

lesson 1.3
name
date
practice a
for use with pages 17 - 25
use a ruler to measure the length of each line segment to the nearest millimeter.
1.
2.
3.
4.
5.
6.
draw a sketch of the three collinear points. then write the segment addition postulate for the points.

1. s is between d and p.
2. c is between q and r.
3. j is between s and h.
4. t is between m and n.

in the diagram of collinear points, gk = 24, hj = 10, and gh = hi = ij. find each length.

1. hi
2. ij
3. gh
4. jk
5. ig
6. ik

suppose j is between h and k. use the segment addition postulate to solve for x. then find the length of each segment.

1. hj = 5x

jk = 7x
kh = 96

1. hj = 2x + 5

jk = 3x - 7
kh = 18

1. hj = 6x - 5

jk = 4x - 6
kh = 129
find the distance between each pair of points.

1. a(3, 2), b(2, 0)
2. c(1, 3), d(-2, 4)
3. e(-1, 0), f(2, -4)

use the distance formula to decide whether ab≅bc.

1. a(0, 1)

b(2, 4)
c(4, 7)

1. a(-3, 1)

b(1, -1)
c(6, -3)

1. a(4, 2)

b(-1, -1)
c(-6, -4)
38
geometry
chapter 1 resource book
copyright © mcdougal littell inc
all rights reserved.

Explanation:

Step1: Recall segment - addition postulate

If a point $B$ is between points $A$ and $C$, then $AB + BC=AC$.

Step2: Solve problem 17

Since $J$ is between $H$ and $K$, by the Segment - Addition Postulate, $HJ+JK = KH$. Substitute $HJ = 5x$, $JK = 7x$, and $KH = 96$. We get $5x+7x=96$, which simplifies to $12x = 96$. Then $x=\frac{96}{12}=8$. So $HJ = 5x=5\times8 = 40$, $JK = 7x=7\times8 = 56$.

Step3: Solve problem 18

Since $J$ is between $H$ and $K$, $HJ+JK = KH$. Substitute $HJ = 2x + 5$, $JK = 3x-7$, and $KH = 18$. Then $(2x + 5)+(3x-7)=18$. Combine like - terms: $5x-2 = 18$. Add 2 to both sides: $5x=20$, so $x = 4$. Then $HJ=2x + 5=2\times4+5 = 13$, $JK=3x - 7=3\times4-7 = 5$.

Step4: Solve problem 19

Since $J$ is between $H$ and $K$, $HJ+JK = KH$. Substitute $HJ = 6x-5$, $JK = 4x-6$, and $KH = 129$. Then $(6x-5)+(4x-6)=129$. Combine like - terms: $10x-11 = 129$. Add 11 to both sides: $10x=140$, so $x = 14$. Then $HJ=6x-5=6\times14-5 = 79$, $JK=4x-6=4\times14-6 = 50$.

Step5: Recall distance formula

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step6: Solve problem 20

For points $A(3,2)$ and $B(2,0)$, $x_1 = 3$, $y_1 = 2$, $x_2 = 2$, $y_2 = 0$. Then $d=\sqrt{(2 - 3)^2+(0 - 2)^2}=\sqrt{(-1)^2+(-2)^2}=\sqrt{1 + 4}=\sqrt{5}$.

Step7: Solve problem 21

For points $C(1,3)$ and $D(-2,4)$, $x_1 = 1$, $y_1 = 3$, $x_2=-2$, $y_2 = 4$. Then $d=\sqrt{(-2 - 1)^2+(4 - 3)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9 + 1}=\sqrt{10}$.

Step8: Solve problem 22

For points $E(-1,0)$ and $F(2,-4)$, $x_1=-1$, $y_1 = 0$, $x_2 = 2$, $y_2=-4$. Then $d=\sqrt{(2+1)^2+(-4 - 0)^2}=\sqrt{3^2+(-4)^2}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step9: Solve problem 23

For points $A(0,1)$ and $B(2,4)$, $AB=\sqrt{(2 - 0)^2+(4 - 1)^2}=\sqrt{4 + 9}=\sqrt{13}$. For points $B(2,4)$ and $C(4,7)$, $BC=\sqrt{(4 - 2)^2+(7 - 4)^2}=\sqrt{4 + 9}=\sqrt{13}$. Since $AB=\sqrt{13}$ and $BC=\sqrt{13}$, $\overline{AB}\cong\overline{BC}$.

Step10: Solve problem 24

For points $A(-3,1)$ and $B(1,-1)$, $AB=\sqrt{(1 + 3)^2+(-1 - 1)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$. For points $B(1,-1)$ and $C(6,-3)$, $BC=\sqrt{(6 - 1)^2+(-3 + 1)^2}=\sqrt{25 + 4}=\sqrt{29}$. Since $AB = 2\sqrt{5}$ and $BC=\sqrt{29}$, $\overline{AB}
ot\cong\overline{BC}$.

Step11: Solve problem 25

For points $A(4,2)$ and $B(-1,-1)$, $AB=\sqrt{(-1 - 4)^2+(-1 - 2)^2}=\sqrt{25 + 9}=\sqrt{34}$. For points $B(-1,-1)$ and $C(-6,-4)$, $BC=\sqrt{(-6 + 1)^2+(-4 + 1)^2}=\sqrt{25 + 9}=\sqrt{34}$. Since $AB=\sqrt{34}$ and $BC=\sqrt{34}$, $\overline{AB}\cong\overline{BC}$.

Answer:

1. $x = 8$, $HJ = 40$, $JK = 56$
2. $x = 4$, $HJ = 13$, $JK = 5$
3. $x = 14$, $HJ = 79$, $JK = 50$
4. $\sqrt{5}$
5. $\sqrt{10}$
6. $5$
7. $\overline{AB}\cong\overline{BC}$
8. $\overline{AB}

ot\cong\overline{BC}$

1. $\overline{AB}\cong\overline{BC}$