News
Get Premium
Dashboard
Dashboard Sign out
McGraw Hill Glencoe Algebra 2, 2012
MH
McGraw Hill Glencoe Algebra 2, 2012 View details
1. Midpoint and Distance Formula
Continue to next subchapter
search

Exercise 38 Page 596

Practice makes perfect
a To determine if triangle ABC is isosceles, we need to compute the lengths of all three sides of the triangle.
To do this, we will use the Distance Formula. Let's start with side AB.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (2,1), (x_2,y_2)= (-6,5)

d=sqrt(( -6- 2)^2+( 5- 1)^2)
SubTerms

Subtract terms

d=sqrt((-8)^2+(4)^2)
CalcPow

Calculate power

d=sqrt(64+16)
AddTerms

Add terms

d=sqrt(80)
The length of the side AB is sqrt(80). Let's now find the length of side BC.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (-6,5), (x_2,y_2)= (-2,-3)

d=sqrt(( -2-( -6))^2+( -3- 5)^2)
SubNeg

a-(- b)=a+b

d=sqrt((-2+6)^2+(-3-5)^2)
AddSubTerms

Add and subtract terms

d=sqrt((4)^2+(-8)^2)
CalcPow

Calculate power

d=sqrt(16+64)
AddTerms

Add terms

d=sqrt(80)
The length of side BC is sqrt(80). Finally, let's find the length of side AC.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (2,1), (x_2,y_2)= (-2,-3)

d=sqrt(( -2- 2)^2+( -3- 1)^2)
SubTerms

Subtract terms

d=sqrt((-4)^2+(-4)^2)
CalcPow

Calculate power

d=sqrt(16+16)
AddTerms

Add terms

d=sqrt(32)
The length of the side AC is sqrt(32). Because two of the three sides have equal lengths, triangle ABC is isosceles.
b Triangle ABC is not equilateral. Only two of the three side lengths, AB and BC, are of equal length.
c To determine the type of triangle EFG is, we need to find the lengths of all three sides. We will start by using the Midpoint Formula to find the points for E, F, and G.
Let's start with point E, between AB.
Midpoint=(x_1+x_2/2,y_1+y_2/2)
SubstituteII

(x_1,y_1)= (2,1), (x_2,y_2)= (-6,5)

Midpoint=(2+( -6)/2,1+ 5/2)
AddNeg

a+(- b)=a-b

Midpoint=(2-6/2,1+5/2)
AddSubTerms

Add and subtract terms

Midpoint=(-4/2,6/2)

Simplify

Midpoint=(-2,3)
Next, let's find point F, between BC.
Midpoint=(x_1+x_2/2,y_1+y_2/2)
SubstituteII

(x_1,y_1)= (-6,5), (x_2,y_2)= (-2,-3)

Midpoint=(-6+( -2)/2,5+( -3)/2)
AddNeg

a+(- b)=a-b

Midpoint=(-6-2/2,5-3/2)
AddSubTerms

Add and subtract terms

Midpoint=(-8/2,2/2)

Simplify

Midpoint=(-4,1)
Finally, let's find point G, between AC.
Midpoint=(x_1+x_2/2,y_1+y_2/2)
SubstituteII

(x_1,y_1)= (2,1), (x_2,y_2)= (-2,-3)

Midpoint=(2+( -2)/2,1+( -3)/2)
AddNeg

a+(- b)=a-b

Midpoint=(2-2/2,1-3/2)
AddSubTerms

Add and subtract terms

Midpoint=(0/2,- 2/2)

Simplify

Midpoint=(0,-1)
We have found the location for the points E, F, and G. Now we can use the Distance Formula to find the side lengths. Let's start with EF.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (-2,3), (x_2,y_2)= (-4,1)

d=sqrt(( -4-( -2))^2+( 1- 3)^2)
SubNeg

a-(- b)=a+b

d=sqrt((-4+2)^2+(1-3)^2)
AddSubTerms

Add and subtract terms

d=sqrt((-2)^2+(-2)^2)
CalcPow

Calculate power

d=sqrt(4+4)
AddTerms

Add terms

d=sqrt(8)
The length of the side EF is sqrt(8). Next, let's find the length of FG.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (-4,1), (x_2,y_2)= (0,-1)

d=sqrt(( 0-( -4))^2+( -1- 1)^2)
SubNeg

a-(- b)=a+b

d=sqrt((0+4)^2+(-1-1)^2)
AddSubTerms

Add and subtract terms

d=sqrt(4^2+(-2)^2)
CalcPow

Calculate power

d=sqrt(16+4)
AddTerms

Add terms

d=sqrt(20)
The length of the side FG is sqrt(20). Finally, let's find the length of EG.
d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)
SubstituteII

(x_1,y_1)= (-2,3), (x_2,y_2)= (0,-1)

d=sqrt(( 0-( -2))^2+( -1- 3)^2)
SubNeg

a-(- b)=a+b

d=sqrt((0+2)^2+(-1-3)^2)
AddSubTerms

Add and subtract terms

d=sqrt(2^2+(-4)^2)
CalcPow

Calculate power

d=sqrt(4+16)
AddTerms

Add terms

d=sqrt(20)
The length of the side EG is sqrt(20). Because two of the three sides have equal length, triangle EFG is isosceles.
d The two triangles EFG and ABC are similar. This means that the ratio between corresponding sides is the same for all pairs of corresponding sides. The shortest sides EF and AC correspond to each other. Then EG corresponds to AB, and FG corresponds to BC.

EF/AC=sqrt(8)/sqrt(32)=1/2 EG/AB=sqrt(20)/sqrt(80)=1/2 FG/BC=sqrt(20)/sqrt(80)=1/2 The side lengths of triangle EFG are one-half the side lengths of triangle ABC.

Subchapter links expand_more
arrow_right Check Your Understanding p.595
1
2
3
4
5
6
7
8
9
arrow_right Practice and Problem Solving p.596-597
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Practice and Problem Solving 41 (Page 597)
arrow_right H.O.T. Problems p.597
42
43
H.O.T. Problems 44 (Page 597)
45
46
arrow_right Standardized Test Practice p.598
47
48
49
50
arrow_right Spiral Review p.598
Spiral Review 51 (Page 598)
Spiral Review 52 (Page 598)
Spiral Review 53 (Page 598)
54
55
56
57
Spiral Review 58 (Page 598)
Spiral Review 59 (Page 598)
Spiral Review 60 (Page 598)
61
62
63
arrow_right Skills Review p.598
Skills Review 64 (Page 598)
Skills Review 65 (Page 598)
Skills Review 66 (Page 598)