section 2.2 solutions

Section 2.2 Solutions
1)
Distance between points:
𝑑 = √(6 βˆ’ 3)2 + (8 βˆ’ 4)2
𝑑 = √(3)2 + (4)2
𝑑 = √9 + 16
𝑑 = √25
1) Answer: distance = 5
3)
Distance between points:
𝑑 = √(4 βˆ’ 2)2 + (βˆ’1 βˆ’ (βˆ’5))2
𝑑 = √(2)2 + (4)2
𝑑 = √4 + 16
𝑑 = √20 = √4√5
3) Answer: distance = 2√5
5)
Distance between points:
𝑑 = √(4 βˆ’ 7)2 + (3 βˆ’ 3)2
𝑑 = √(βˆ’3)2 + (0)2
𝑑 = √9 + 0
𝑑 = √9
5) Answer: distance = 3
7)
3+6 4+8
9 12
,
)=( , )
2
2
2 2
Midpoint = (
7) Answer Midpoint = (9/2, 6) or (4.5, 6)
9)
2+4 βˆ’5+(βˆ’1)
6 βˆ’6
,
)=( , )
2
2
2 2
Midpoint = (
Answer Midpoint = (3,-3)
11)
4+7 3+3)
11 6
, 2 ) = ( 2 , 2)
2
Midpoint = (
11) Answer Midpoint = (11/2, 3) or (5.5,3)
8βˆ’4
13) π‘š = 6βˆ’3
15) π‘š =
Answer: m = 4/3
βˆ’1βˆ’(βˆ’5)
4βˆ’2
2βˆ’3
βˆ’1
0
5βˆ’5
0
17) π‘š = 7βˆ’7 =
19) π‘š = 4βˆ’3 = 1
4
=2
Answer: m = 2
Answer: m = undefined
Answer: m = 0
21) Plot the point (0,-5) then go up 2 and right 3 three times.
points labeled (0,-5) (3,-3) (6,-1) (9,1)
23) Plot the point (0,0) then go up 2 and right 3 three times.
25) Plot the point (0,-2) then go up 3 and right 2 three times.
27) First solve for y
3x + 2y = 10
-3x
-3x
2y = -3x + 10
2𝑦
2
=
𝑦=
βˆ’3π‘₯
2
βˆ’3
π‘₯
2
+
10
2
+ 5 (now plot point (0,5) and go down 3 right 2 three times)
29) First solve for y
2x + 3y = 0
3y = -2x
𝑦=
βˆ’2
π‘₯
3
(now plot point (0,0) and go down 2 right 3 three times)
31)
x-intercept let y = 0
2x + 0 = 12
2x = 12
x=6
y-intercept let x = 0
2(0) + y = 12
y = 12
Answer: x-intercept (6,0) y-intercept (0,12)
33)
x-intercept, let y = 0
4x + 2(0) = 15
4x = 15
x = 15/4 or 3.75
y-intercept, let x = 0
4(0) + 2y = 15
2y = 15
y = 15/2 or 7.5
15
Answer x-intercept ( 4 , 0) π‘œπ‘Ÿ(3.75,0) y-intercept (0,
15
) π‘œπ‘Ÿ(0,3.5)
2
35)
First clear fractions by multiplying by 6
1
2
6 βˆ— 2π‘₯ βˆ’6 βˆ—3𝑦 = 6 βˆ— 2
6
π‘₯
2
βˆ’
12
𝑦
3
= 12
3x – 4y = 12
x-intercept, let y = 0
3x – 4(0) = 12
3x = 12
x=4
y-intercept, let x = 0
3(0) – 4y = 12
-4y = 12
y = 12/-4 = - 3
Answer: x-intercept (4,0) y-intercept (0,-3)
37) First multiply by 3 to clear the fraction
2
3
3 βˆ— π‘₯ + 3 βˆ— 5𝑦 = 3 βˆ— βˆ’2
2x + 15y = -6
x-intercept, let y = 0
2x + 15(0) = -6
2x = -6
x = -3
y-intercept, let x = 0
2(0) + 15y = -6
15y = -6
y = -6/15
y = -2/5
Answer: x-intercept (-3,0) y-intercept (0,-2/5)
39) m = 4 x1 = 5 y1=-2
Use formula y – y1 = m(x – x1)
y – (-2) = 4(x – 5)
y + 2 = 4x – 20
-2
-2
Answer: y = 4x – 22
41) m =
2
3
x1 = 6 y1=5
Use formula y – y1 = m(x – x1)
y–5=
2
3
y–5=
2
π‘₯
3
(x – 6)
-
12
3
2
3
y–5= π‘₯–4
+5
+5
2
Answer: y = 3 π‘₯+1
43) m = 5 x1 = 0 y1=-2
(hint the y-intercept is the point (0,-2))
Use formula y – y1 = m(x – x1)
y – (-2) = 5(x – 0)
y + 2 = 5x
Answer y = 5x – 2
45) The m = 3 as the line y = 3x + 6 has a slope of m = 3. Our line is parallel, so it must have the same
slope.
m = 3 x1 = 4 y1=1
Use formula y – y1 = m(x – x1)
y – 1 = 3(x – 4)
y – 1 = 3x – 12
+1
+1
Answer: y = 3x – 11
4
9
47) The m = 4/9 as the line y = x +3 has a slope of m = 4/9. Our line is parallel, so it must have the
same slope.
m=
4
9
x1 = 0 y1=1
Use formula y – y1 = m(x – x1)
y–1=
4
9
y–1=
4
π‘₯
9
(x – 0)
+1
+1
4
Answer: y = 9 π‘₯+1
3
49) The line y = 3x + 6 has a slope of m = 3, or you may think of it as m = 1. The slope of our line will be
the reciprocal of this slope with the opposite sign. Our line will have a slope of m =
m=
βˆ’1
3
x1 = 4 y1=1
Use formula y – y1 = m(x – x1)
y–1=
βˆ’1
3
y–1=
βˆ’1
π‘₯
3
+1
(x – 4)
4
+3
+1
Answer: y =
βˆ’1
7
π‘₯+3
3
(hint 4/3 + 1 = 4/3 + 3/3 = 7/3)
βˆ’1
3
4
9
51) The line y = π‘₯ + 3 has a slope of m = 4/9. Our line is perpendicular so change the sign and find the
reciprocal to find our slope. The slope of our line is m = – 9/4
m=
βˆ’9
4
x1 = 0 y1=1
Use formula y – y1 = m(x – x1)
y–1=
βˆ’9
(x
4
y–1=
βˆ’9
π‘₯
4
+1
– 0)
+1
Answer: y =
βˆ’9
π‘₯+1
4
53) First I need to find the slope of the line that passes through (3,1) and (4,5)
5βˆ’1
4
π‘š = 4βˆ’3 = 1 = 4
m = 4 x1 = 3 y1=1
Use formula y – y1 = m(x – x1)
y – 1 = 4(x – 3)
y – 1 = 4x – 12
+1
+1
Answer: y = 4x – 11
55) First I need to find the slope of the line that passes through (-4,5) and (-2,1)
1βˆ’5
π‘š = βˆ’2βˆ’(βˆ’4) =
βˆ’4
2
= βˆ’2
m = -2 x1 = -4 y1=5
Use formula y – y1 = m(x – x1)
y – 5 = -2(x – (-4))
y – 5 = -2(x + 4)
y – 5 = -2x – 8
+5
+5
Answer: y = -2x – 3