1. Math 1314 HW 2(Chapter 2). Question 1. Hence the poi

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Math 1314 HW 2(Chapter 2)
Question 1
We denote the points A ( −6 ,−4 ) , B ( 0 ,−2 ) , C (−10,8)
AB= ( 0 ,−2 )− (−6 ,−4 )= (6,2 ) , Length AB= √62 + ( 2 )2= √40
BC= (−10,8 ) − ( 0 ,−2 ) = ( −10,10 ) , Length BC= √ (−10)2 + ( 10 )
2 = √ 200
AC= (−10,8 ) − ( −6 ,−4 ) = ( −4 , 12 ) , Length AB= √ 1 22+ ( −4 ) 2= √ 160
A B2+ A C2=B C2
40+ 160=200
Hence the points are vertices of a right-angled triangle.
Question 2
a) Slope= −3−4
−5−−1 =−7
−4 = 7
4 =1.75
b) Vector difference= (−5 ,−3 )− (−1,4 )=( 4 ,−7)
Distance= √ 42 + ( −7 ) 2= √ 65=8.062
c) Coordinates of midpoint = (−5 ,−3 ) + (−1,4 )
2 = (−6,1 )
2 = (−3,0.5 )
d) Let the line pass through ( −5 ,−3 ) and an arbitrary point ( x , y )
slope=1.75= y −−3
x−−5 = y+ 3
x +5
y +3= ( x+5 ) 1.75=1.75 x +8.75
y=1.75 x +8.75−3

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y=1.75 x +5.75
e) Both x and y have no extend from −∞ ¿ ∞ hence, domain=range=(−∞ , ∞)
f) The function is an “increasing function”
Question 3 part a
x2+ y2+ 6 x +8 y +9=0
We write the equation in the form, ( x−a ) 2+ ( y−b )2 =r2
x2+ y2+ 6 x +8 y +9= ( x−(−3) ) 2+ ¿ ¿
x2+ y2+ 6 x +8 y +9= ( x−(−3) )2+ ¿ ¿
( x−(−3) ) 2+ ¿ ¿
( x−(−3) )2+ ¿ ¿
Hence, it is a circle with the center ( a , b )= (−3 ,−4 )∧radius r=4
Question 3 part b
x2+ y2+2 x−6 y +14=0
We write the equation in the form, ( x−a ) 2+ ( y−b )2 =r2
x2+ y2+2 x−6 y +14= ( x−(−1) ) 2+ ¿ ¿
x2+ y2+2 x−6 y +14= ( x− ( −1 ) ) 2+ ¿ ¿
( x− (−1 ) ) 2+¿ ¿
The function does not represent a circle since r2 cannot be negative.

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Question 4 part a
The points do not define a function
Question 4 part b
The equation y= √4 x+ 1defines a function.
4 x+1 ≥ 0
4 x+1−1≥ 0−1
4 x ≥−1 , x ≥− 1
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The domain of the function is ¿
The range of the function is¿
Question 5
Both graphs define a function
The domain of function ( a ) is [−∞ ,−1 ] U [ 1 , ∞ ]
The range of the function is¿
The domain of function ( b ) is [−7 , ∞ ]
The range of the function is¿
Question 6 (1) part a
A vertical line through (−1
5 , 2
7 )has the equation x=−1
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