1.2 Lines, Circles, and Parabolas - kau.edu.sa circles, and... · 2/17 2. Lines, Circles, and...

King Abdulaziz University

1.2 Lines, Circles, and Parabolas

Dr. Hamed Al-Sulami

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c© 2009 [email protected] http://www.kau.edu.sa/hhaalsalmi

Prepared: October 22, 2009 Presented: October 22, 2009

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http://www.kau.edu.sa

mailto:[email protected]

http://www.kau.edu.sa/hhaalsalmi

2/17

2. Lines, Circles, and Parabolas

2.1. Cartesian Coordinates

An ordered pair (x, y) of real numbers has x as its first coordinate and y as its secondcoordinate. The way for representing ordered pair is called Cartesian coordinate system.

It is consist of two real line intersecting at right angles.

The horizontal axis is often calledthe x−axis,and the vertical axis is of-ten called the y−axis.The point oftheir intersection is the origin.The twoaxes divide the plane into four quad-rants. Each point in the plane cor-responds to an ordered pair (x, y) ofreal numbers x and y.The number xis the distance from the y−axis to thepoint,and the number y is the distancefrom the x−axis to the point.For thepoint (x, y), the first coordinate is thex−coordinate, and the second coordi-nate is the y−coordinate.

�

�

x

y (x, y)

y

x

Origin

y−axis

x−axis

first Quadrantsecond Quadrant

forth Quadrantthird Quadrant

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

Lines, Circles, and Parabolas 3/17

The Figure to the right shows the plotof the points (0, 0), (2, 3),(3, 0), (0,−3),(−2, 1), (2,−2), and (−3,−1) in theCartesian coordinate system, look care-fully how is the sign of each coordinatedetermine the position of the point.

Note 1: The signs of the coordinatesof a point determine the quadrant inwhich the point lies. For example, ifx < 0 and y < 0, then the point (x, y)lies in the third quadrant.

Note 2: The notation (x, y) is used todenote a point in the plan or an openinterval on the real line, however thecontent of the problem should clarifywhether the notation is used for a pointor an interval.

1

2

3

4

-1

-2

-3

-4

1 2 3 4-1-2-3-4

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�

�

�

�

�

�

(0, 0) (3, 0)

(2, 3)

(0,−3)

(−2, 1)

(−3,−1)(2,−2)

y

x

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


2.2. Increments and Straight Lines

When a partial moves from one point in the plane to another, the net changes in its co-ordinates are called increments. They are calculated by subtracting the coordinates of thestarting point from the coordinates of the ending point. If x changes from x1 to x2, theincrement in x is Δx = x2−x1. If y changes from y1 to y2, the increment in y is Δy = y2−y1.

Example 1. Find the increments in x and y when they move from A to B.

1. A(4,−3), B(2, 5) 2. A(5, 6), B(5, 1).

Solution:

1.

Δx = x2 − x1 = 2 − 4 = −2Δy = y2 − y1 = 5 − (−3) = 8

2.

Δx = x2 − x1 = 5 − 5 = 0Δy = y2 − y1 = 1 − 6 = −5.

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❘ ➡ ➡ ➦ � � � � ➥ ➡ �


2.3. Lines

Lines rise or fall at a steady rate as we move along them from left to right, unless, they arehorizontal or vertical lines. Horizontal linesdo not fall or rise at all, and on a vertical linewe cannot move from left to right. The rate ofrise or fall, the steepness, as we move from leftto right along the line is called the slope of theline. We measure the slope in way that risinglines have a positive slopes, falling lines havenegative slopes, horizontal lines have slope 0,and vertical lines have no slope at all.

1

2

3

-1

-2

1-1-2x

y

The slope is navigate

slope=: -2.0


slope=: -1.94


slope=: -1.89


slope=: -1.83


slope=: -1.78


slope=: -1.72


slope=: -1.67


slope=: -1.61


slope=: -1.56


slope=: -1.5


slope=: -1.44


slope=: -1.39


slope=: -1.33


slope=: -1.28


slope=: -1.22


slope=: -1.17


slope=: -1.11


slope=: -1.06


slope=: -1.0


slope=: -0.94


slope=: -0.89


slope=: -0.83


slope=: -0.78


slope=: -0.72


slope=: -0.67


slope=: -0.61


slope=: -0.56


slope=: -0.5


slope=: -0.44


slope=: -0.39


slope=: -0.33


slope=: -0.28


slope=: -0.22


slope=: -0.17


slope=: -0.11

The slope is zero

slope=: -0.06

The slope is positive

slope=: 0.0


slope=: 0.06


slope=: 0.11


slope=: 0.17


slope=: 0.22


slope=: 0.28


slope=: 0.33


slope=: 0.39


slope=: 0.44


slope=: 0.5


slope=: 0.56


slope=: 0.61


slope=: 0.67


slope=: 0.72


slope=: 0.78


slope=: 0.83


slope=: 0.89


slope=: 0.94


slope=: 1.0


slope=: 1.06


slope=: 1.11


slope=: 1.17


slope=: 1.22


slope=: 1.28


slope=: 1.33


slope=: 1.39


slope=: 1.44


slope=: 1.5


slope=: 1.56


slope=: 1.61


slope=: 1.67


slope=: 1.72


slope=: 1.78


slope=: 1.83


slope=: 1.89


slope=: 1.94

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


2.3.1. slopes

Let P1(x1, y1) and P2(x2, y2) be any twopoints on a nonvertical straight line l. Wecall the increment in y, Δy = y2 − y1, therise from P1 to P2 and the increment in x,Δx = x2 − x1, the run from P1 to P2. Sincethe line is nonvertical, Δx �= 0 and we de-fined the slope of the line to be Δy/Δx, theamount of rise per unit of run. We denoted theslope by m. Hence the slope of the line passingthrough the points P1(x1, y1) and P2(x2, y2) is

m =Δy

Δx=

y2 − y1

x2 − x1. For vertical line we can

NOT use the formula m = Δy/Δx becauseΔx = 0. We express this by saying verticallines have no slope.

x

y

� � � � ��

�

�

(x1, y1)

(x2, y2)

Δx = x2 − x1

= rise

= run

Δy = y2 − y1

l

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


An equation of the line with slope m passing throughthe point (x1, y1) is y − y1 = m(x − x1).(Point-Slope equation) The equation y = mx + b is theSlope-Intercept equation of the line of slope m andy-intercept b(0, b).

Example 2. Find an equation of the line that passesthrough the points (−1, 5) and (3,−3).Solution:

m =y2 − y1

x2 − x1slope-formula

=−3 − 5

3 − (−1)

m =−84

= −2

y − y1 = m(x − x1) Point-Slope equation

y − (−1) = −2(x − 2) Substitute −1 for y1,

2 for x1, and −2 for m.

y + 1 = −2x + 4 Simplify

y = −2x + 3 Solve for y.

�

1

2

3

4

5

6

-1

-2

-3

-4

1 2 3 4-1-2

x

y

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�

�

Δx = 3 − (−1) = 4Δy

=−

3−

5=

−8

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 3. Find an equation of the line witha slope of 2 and y-intercept −3.

Solution:

y = mx + b Slope-Intercept equation

y = 2x + (−3) Substitute 2 for m, and −3 for b

y = 2x − 3

�

Example 4. Find an equation of the line thathas a slope of 2 and passes through the point(2,−1).Solution:

y − y1 = m(x − x1) Point-Slope equation

y − (−1) = 2(x − 2) Substitute −1 for y1,

2 for x1, and 2 for m.

y + 1 = 2x − 4 Simplifyy = 2x − 5 Solve for y.

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1

2

3

4

5

-1

-2

-3

-4

-5

-6

-7

1 2 3 4 5-1-2

x

y

�

�

x y

0.0 -5.0

0.0 -3.0

�

�

x y

0.06 -4.89

0.06 -2.89

�

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x y

0.11 -4.77

0.11 -2.77

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x y

0.17 -4.66

0.17 -2.66

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x y

0.23 -4.54

0.23 -2.54

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x y

0.29 -4.43

0.29 -2.43

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x y

0.34 -4.31

0.34 -2.31

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x y

0.4 -4.2

0.4 -2.2

�

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x y

0.46 -4.09

0.46 -2.09

�

�

x y

0.51 -3.97

0.51 -1.97

�

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x y

0.57 -3.86

0.57 -1.86

�

�

x y

0.63 -3.74

0.63 -1.74

�

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x y

0.69 -3.63

0.69 -1.63

�

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x y

0.74 -3.51

0.74 -1.51

�

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x y

0.8 -3.4

0.8 -1.4

�

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x y

0.86 -3.29

0.86 -1.29

�

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x y

0.91 -3.17

0.91 -1.17

�

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x y

0.97 -3.06

0.97 -1.06

�

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x y

1.03 -2.94

1.03 -0.94

�

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x y

1.09 -2.83

1.09 -0.83

�

�

x y

1.14 -2.71

1.14 -0.71

�

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x y

1.2 -2.6

1.2 -0.6

�

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x y

1.26 -2.49

1.26 -0.49

�

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x y

1.31 -2.37

1.31 -0.37

�

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x y

1.37 -2.26

1.37 -0.26

�

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x y

1.43 -2.14

1.43 -0.14

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x y

1.49 -2.03

1.49 -0.03

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x y

1.54 -1.91

1.54 0.09

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x y

1.6 -1.8

1.6 0.2

�

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x y

1.66 -1.69

1.66 0.31

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x y

1.71 -1.57

1.71 0.43

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x y

1.77 -1.46

1.77 0.54

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x y

1.83 -1.34

1.83 0.66

�

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x y

1.89 -1.23

1.89 0.77

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x y

1.94 -1.11

1.94 0.89

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x y

2.0 -1.0

2.0 1.0

�

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x y

2.06 -0.89

2.06 1.11

�

�

x y

2.11 -0.77

2.11 1.23

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x y

2.17 -0.66

2.17 1.34

�

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x y

2.23 -0.54

2.23 1.46

�

�

x y

2.29 -0.43

2.29 1.57

�

�

x y

2.34 -0.31

2.34 1.69

�

�

x y

2.4 -0.2

2.4 1.8

�

�

x y

2.46 -0.09

2.46 1.91

�

�

x y

2.51 0.03

2.51 2.03

�

�

x y

2.57 0.14

2.57 2.14

�

�

x y

2.63 0.26

2.63 2.26

�

�

x y

2.69 0.37

2.69 2.37

�

�

x y

2.74 0.49

2.74 2.49

�

�

x y

2.8 0.6

2.8 2.6

�

�

x y

2.86 0.71

2.86 2.71

�

�

x y

2.91 0.83

2.91 2.83

�

�

x y

2.97 0.94

2.97 2.94

�

�

x y

3.03 1.06

3.03 3.06

�

�

x y

3.09 1.17

3.09 3.17

�

�

x y

3.14 1.29

3.14 3.29

�

�

x y

3.2 1.4

3.2 3.4

�

�

x y

3.26 1.51

3.26 3.51

�

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x y

3.31 1.63

3.31 3.63

�

�

x y

3.37 1.74

3.37 3.74

�

�

x y

3.43 1.86

3.43 3.86

�

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x y

3.49 1.97

3.49 3.97

�

�

x y

3.54 2.09

3.54 4.09�

�

x y

3.6 2.2

3.6 4.2�

�

x y

3.66 2.31

3.66 4.31�

�

x y

3.71 2.43

3.71 4.43�

�

x y

3.77 2.54

3.77 4.54�

�

x y

3.83 2.66

3.83 4.66�

�

x y

3.89 2.77

3.89 4.77�

�

x y

3.94 2.89

3.94 4.89�

�

x y

4.0 3.0

4.0 5.0

�

�

�

Δx = 1

Δy = 2

y-intercept

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


2.4. Summary of Equations of Lines

General form: Ax + By + C = 0Vertical line: x = a

Horizontal line: y = b

Point-Slope equation: y − y1 = m(x − x1)Slope-Intercept equation: y = mx + b

Example 5. Find the slope and y−intercept of theline 2y + 3x − 6 = 0.

Solution:

2y + 3x − 6 = 0 Isolate y term.2y = − 3x + 6 Solve for y.

y = −3/2x + 3 Point-Slope equationm = −3/2, b = 3

�

1

2

3

4

-1

-2

-3

-4

1 2 3 4-1-2

x

y

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

�

�

y-intercept

y = −3/2x + 3

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


2.5. Parallel Lines

Two distinct nonvertical line are parallel❍? if they havethe same slope. That is two lines with slopes m1 andm2 are parallel if m1 = m2. Any two vertical lines areparallel.

Example 6. Find an equation for the line that is par-allel to 2x − 3y = 5 and passes through (2,−1).Solution: we find the slope of the line 2x − 3y = 5

−3y = −2x + 5 Isolate y term.y = −2/ − 3x + 5/ − 3 Solve for y.

y = 2/3x− 5/3 Point-Slope form

Now, use the m = 2/3 and the point (2,−1).

y − y1 = m(x − x1) Point-Slope formy − (−1) = 2/3(x − 2) Substitute

y + 1 = 2/3x − 4/3 Simplify3y + 3 = 2x − 4 Simplify

y = 2/3x − 7/3

�

1

2

3

4

-1

-2

-3

-4

1 2 3 4-1-2

x

y

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2x − 3y = 5

2x − 3y = 7

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! �"��


2.6. Perpendicular Lines

Two nonvertical line are perpendicular❍? if the product oftheir slopes equal to −1. That is two lines with slopes m1

and m2 are perpendicular if m1 = −1/m2. Any vertical andany horizontal lines are perpendicular.

Example 7. Find an equation for the line that is perpendic-ular to 2x − 3y = 5 and passes through (2,−1).

Solution: we find the slope of the line 2x − 3y = 5

−3y = −2x + 5 Isolate y term.

y = −2/ − 3x + 5/ − 3 Solve for y.

y = 2/3x − 5/3 Point-Slope form

Now, use the m = −3/2 and the point (2,−1).

y − y1 = m(x − x1) Point-Slope form

y − (−1) = −3/2(x − 2) Substitute

y + 1 = −3/2x + 3 Simplify

2y + 2 = −3x + 6 Simplify

y = −3/2x + 2

�

1

2

3

4

-1

-2

-3

-4

1 2 3 4-1-2

x

y

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y = 2/3x − 5/3

y = −3/2x + 2

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

! �#��


2.7. The Distance

On the real line distance between tonumbers a, b is given by d = |a − b| =|b − a|. It follows from this that thedistance between two points P1(x1, y)and P2(x2, y) on any horizontal line isd = |x1 − x2| and the distance betweentwo points P1(x, y1) and P2(x, y2) onany vertical line is d = |y1 − y2|.To find a formula for the distance be-tween any two pint on the Cartesiancoordinate system, we use this informa-tion and the Pythagorean Theorem.❍?

Now, if P1(x1, y1) and P2(x2, y2) beany two points in the plane. The linesy = y1 and x = x2 will intersect atthe point (x2, y1) to form a right trian-gle with vertices (x1, y1), (x2, y1), and(x2, y2). Using Pythagorean Theoremfor the triangle in Figure we have

d2 = (y2 − y1)2 + (x2 − x1)2.

x

y

�

�

�

(x1, y1)

(x2, y2)

(x2, y1)

d

x2 − x1

y2 − y1

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

The Pythagorean Theorem says a tri-angle is right triangle, if and only if thehypotenuse c and the sides a and b arerelated by c2 = a2 + b2.

a

bc


Definition 2.1: [Distance Formula]Then the distance between P (x1, y1) and Q(x2, y2) isd =

√(Δx)2 + (Δy)2 =

√(x2 − x1)2 + (y2 − y1)2.

Example 8. Find the distance between P (−1, 2) and Q(3, 4)

Solution:

d =√

(x2 − x1)2 + (y2 − y1)2

=√

(3 − (−1))2 + (4 − 2)2

=√

(4)2 + (2)2

=√

16 + 4 =√

20

=√

4 · 5 = 2√

5

1

2

3

4

-11 2 3-1-2

y

x

�

��

(−1, 2)

(3, 4)

�

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Circles

Let C(h, k) be the center of circle of radius r andP (x, y) be any point on the circle then the distancebetween C(h, k) and P (x, y) is

r = d(C, P )

=√

(x − h)2 + (y − k)2

Then, r2 = (x − h)2 + (y − k)2

Definition 2.2: [Equation of the circle ]An equation of the circle with center (h, k) and radiusr is (x − h)2 + (y − k)2 = r2.

�

�

(h, k)

(x, y)

r

x

y

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 9. Find an equation of the of the circle that has center (3, 4) and radius 2

Solution: Using the equation of the circle with radius r = 2 and center (3, 4). We get

(x − h)2 + (y − k)2 = r2

(x − 3)2 + (y − 4)2 = 22

(x − 3)2 + (y − 4)2 = 4.❍? �

Example 10. Find the center and radius of the circle x2 + y2 + 4x − 6y − 3 = 0.

Solution:

x2 + y2 + 4x − 6y − 3 = 0 Gather terms and move the constant

to the other side

x2 + 4x + y2 − 6y = 3 complete the square for x and y

x2 + 4x +(

42

)2

+ y2−6y +(−62

)2

= 3 +(

42

)2

+(−62

)2

x2 + 4x + 4 + y2 − 6y + 9 = 3 + 4 + 9 we have completed the square for x and y

(x + 2)2 + (y − 3)2 = 16

The center is (−2, 3) and r =√

16 = 4. �

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

1

2

3

4

5

6

-1

-2

1 2 3 4 5-1-2

�

�

x

y


Parabolas

Definition 2.3: [Parabola ]

The graph of the equation y = ax2 + bx + c, a �= 0, is a parabola. It can be written as

y = a(x − h)2 + k where h =−b

2aand k = c − b2

4a. The point (h, k) is called the vertex of

the parabola and the line x = h =−b

2ais called the axis of symmetry. The graph of the

parabola opens up if a > 0 and opens down if a < 0.

� (h, k)

y = a(x − h)2 + k, a > 0

y = a(x − h)2 + k, a < 0

h

k x

y

❘ ➡ ➡ ➦ � � � � ➥ ➡ �


Example 11. Identify the type of the curve and sketch the graph y − x2 − 2x = 0

Solution: First we write the equation as the following y = x2 + 2x and complete the square forx. Hence

y = x2 + 2x complete the square for x

= x2 + 2 x+1− 1 by adding and subtract

(2

2

)2

= (x2 + 2x + 1) − 1 this is a perfect square

y = (x + 1)2 − 1.

This is an equation of a parabola with vertex (−1,−1) and it will be open up. �

Solution: We have y = x2 + 2x, hence a = 1, b = 2, and c = 0. Now, h =−b

2a=

−22

= −1

and k = (−1)2 + 2(−1) = −1 Hence y = x2 + 2x = (x − (−1))2 − 1 = (x + 1)2 − 1.❍? Thus

the vertex (−1,−1) and it opens up. �

❘ ➡ ➡ ➦ � � � � ➥ ➡ �

�

x

y = (x + 1)2 − 1

y

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