Parallel and Perpendicular Lines

Angle Between Two Lines

Straight Lines

Definition 1.6

The angle of inclination of a line or simply an inclination is the smallest angle, greater than or equal to 00, that the line makes with positive direction of the x-axis.

If a line is inclined upward to the right, then the inclination 1 is between 00 to 900, that is 00 < 1 < 900.

If the line is inclined upward to the left, then 900 < 2 < 1800.

The inclination of a horizontal line is 00, that is, 3 = 00. The inclination of a vertical line is 900.

1

L1

(a)

2

L2

(b)

L4

900

(c) (d)

Figure 10

L3

L4

900

Example 5:

1. Draw the line passing through (-2, 4) with = 1200.

1200

1200

2. What is the inclination if tan = -1?

1350

The trigonometric function tangent is negative on quadrant two and four. So, either = 1350 or = 3150. However inclination should be the smallest angle that the line makes with the positive x-axis. Thus, must be 1350 .

Figure 11 shows that the ratio of the differences between the y-coordinates and x-coordinates is actually the steepness of the line. So, the quotient y/x is the tangent of the inclination . Thus, the slope is equal to tangent .

L

y

x

Figure 11

Definition 1.7The slope of a line, denoted by m, is the

tangent of the inclination. That is m = tan .

In Figure 9, observe that tangent of 1 is positive and tangent of 2 is negative. So, the slopes of lines L1 and L2 are positive and negative respectively; while the slopes of a horizontal line is 0 and the slope of a vertical line is undefined since tan 900 is undefined. Theorem 2.1

The slope of a line passing through two points P1(x1, y1)

and P2(x2, y2) is m = 12

12xxyy

.

Example 6:

1. What is the slope of the line whose inclination is 600?

.

(-2, 4)

(3, -6)

Solution: The slope of this line is tangent of 600. So, m = 3

.

2. Find the slope of the line that passes through (-2, 4) and (3, -6).

Slopes of Parallel and Perpendicular Lines

Theorem 1.2Two nonvertical lines are parallel if and only

if their slopes are equal. That is, lines L1 and L2 are parallel if and only if m1 = m2.

Theorem .3Two slant lines are perpendicular if and only

if the product of their slopes is -1. That is, lines L1 and L2 are perpendicular if and only if m1 m2 = -1

Figure 12

(a) (b)

L1L2

L1

L2

1212

In figure 12 (a), 1 = 2. So tan 1 = tan 2. Thus, m1 = m2.

In figure 12 (b) , let m1 = tan 1 and m2 = tan 2 be slopes of L1 and L2 respectively. I t can be shown that 2 = 900 + 1. So, tan 2 = tan (900 + 1). Since tan 900 does not exist, take

tan 2 = )cos(

)sin(

10

10

90

90

=

10

10

10

10

9090

9090

sinsincoscos

sincoscossin

=1

1

sincos

= 1

1

tan.

A

B

D C

Example 8:Prove that the points A(-2, 7), B(5, 4), C(-1, -10) and D(-8, -7) are vertices of the rectangle ABCD.

We must show that the adjacent sides are perpendicular. That is, mABmBC = -1 and mADmDC = -1. We compute the following:

7

3

25

74

ABm7

3

18

107

DCm

3

7

28

77

ADm3

7

51

410

BCm

and

.

Thus, the result follows.Also, it can be shown that opposite sides are parallel. However, that is enough to show that the figure is already a rectangle (it may be any parallelogram). It must be shown that any of the two adjacent sides are perpendicular.

2. Draw the triangle with the vertices A(-1, 1), B(6, -2) and C(4, 3) and show that the triangle is a right triangle.

A

B

C

,

Finding the slopes of the sides yield the following:

7

3

16

12

ABm2

5

64

23

BCm5

2

14

13

ACm

and

.

Since the slopes of the sides BC and AC are negative reciprocal to each other, these sides must be perpendicular. Hence the triangle must be a right triangle.

Angle Between Two Lines:

Let be an angle between two intersecting lines L1 and L2. Suppose 1 and 2 are the inclinations of L1 and L2 respectively. Refer to Figure 13.

Figure 13

L1

1 2

L2

It can be checked that + 1 = 2. So, = 2 - 1 and tan = tan (2 - 1). Thus,

tan = .

But tan 1 is the slope of L1 and tan 2 is the slope of L2. Hence,

tan = ,

where is obtained by taking the inverse tangent of both sides.

It must be noted that m1 is the slope of the initial line and m2 is the slope of the terminal side.

12

121

tantantantan

12

121 mm

mm

Example 9:1.The slope of one line is 3/2. Find the slope of the

other line if the angle between the two lines is 1350.

1

1

2

31

2

3

m

m

1

1

2

31

2

3

m

m

Solution:From Definition 1.8, either tan 1350 =

or tan 1350 = .

Definition 1.8If m1 and m2 are the slopes of lines L1 and

L2 respectively, then the angle from L1 to L2 is obtained from the equation

tan = .12

121 mm

mm

L2

L1

1350

1350

L2

L1

2. Find all the angles of a triangle whose vertices are at (-4, 2), (3, 3), and (2, -3).

(-4, 2)

L3

L1

L2

(2, -3)

(3, 3)

The slopes of L1, L2, and L3 are m1 = 6, m2 = 1/7 and m3 = -5/6 respectively.

Let 1 be the angle from L2 to L1, 2 be the angle from L1 to L3, and 3 be the angle from L3 to L2.

By Definition 1.8, tan 1 =

1341

671

1

71

6

and

tan 2 =

24

41

66

51

66

5

. Also, 3 = 1800 – (1 + 2).

Thus, 1 = tan-1

1341

= 72.40 ,

2 = tan -1

24

41= 59.70, and 3 = 1800 – (72.40 + 59.70) = 49.90.

Example: Page 19

A television camera is located along the 40-yd line at a football game. If the camera is 20 yds back from the sideline, through what angle should it be able to pan in order to cover the entire field of play, including the end zones, which are 10 yds deep?

Area of a Triangle Given the Vertices

Let A(x1, y1), B(x2, y2), and C(x3, y3) be the vertices of a triangle. Figure below shows that the area of the triangle ABC is equal to the sum of the areas of the trapezoids ADEC and CEFB minus the area of the trapezoid ADFB. The area of a trapezoid is one-half the sum of the parallel sides times the altitude. So,

A

B

C

D E F

Thus,

AABC = )xx(yy21

1331 + )xx(yy21

3223

- )xx(yy21

2121

AADEC = DECEAD21

= )xx(yy21

1331 ,

ACEFB = EFBFCE21

= )xx(yy21

3223 ,

AADFB = DFBFAD21

= )xx(yy21

1221 ,

Equivalently, we can write

AABC = 1321

1321

yyyy

xxxx

21

where the points

are chosen in a counterclockwise order.

which is also equal to the determinant

AABC =

1yx

1yx

1yx

21

33

22

11

.

Example:Find the area of the triangle with vertices (2, 2), (-4, -1), and (6, -5).

(2, 2)

(-4, -1)

(6, -5)

Solution:

A =

156

114

122

21

= 108620122

21

A = 27 square units.

Assignment: (p22 fuller)

54. A television camera is 30 ft from the sideline of a basketball court 94 ft long. The camera is located 7 ft from the midcourt. Through what angle must it sweep in order to cover all action on the court?55. A bridge is trussed as shown below. Find the slopes and inclinations of the sections AB and BC.

The Straight Lines

The Straight LinesTheorem 2.1 The equation of every straight line is

expressible in terms of the first degree. Conversely, the graph of a first-degree equation is a straight line.

Equations of Lines

Vertical Lines and Horizontal Lines

x = a

a

y = b b

Standard Forms of the Equation of a Line

1. Slope-Intercept Form

Let m be the slope and b be the y-intercept of the line.

(0, b)

(x, y)

Using slope given two points (0, b) and (x, y), we get

m = .

Thus, mx = y – b.So, y = mx + b.

This the slope-intercept form of a line.

0x

by

Example: Find the equation of the line with slope -1/2 and y-intercept 4.

Solution: Let m = -1/2 and b = 4. Substitute to the y = mx + b,

we get 4x21

y .

Simplifying, x + 2y – 8 = 0 is the equation of the line.

x + 2y – 8 = 0

b = 4

2. What are the slope and y-intercept of the line whose equation is 5x – 3y = -9.

Solution: Transform the given equation to slope-intercept form, we have

3y = 5x + 9 and so, y = 3x35

. Thus, the slope m = 35

and the y-int is b = 3.

5x – 3y = -9

b = 3

m = 35

2. Point-Slope Form

Let P(x1, y1) be a point on the line and let m be the slope of the line.

P (x1, y1)

(x, y)

The slope of the line is

m = 1

1

xx

yy

.

Simplifying, y – y1 = m(x – x1). This is the Point-Slope Form of a line.

Example:What is the equation of the line that passes

through the points (-1, 4) and (3, -6)?

(3, -6)

(-1, 4)

The slope of the line is

m = 25

3164

.

Take the point (-1, 4), we get

)1x(25

4y which is

5x + 2y – 3 = 0.

3. Intercept Form

Let a and b be the x and y intercept of the line respectively.

(0, b)

(a, 0)

The slope of the line is

m = ab

a00b

.

Using point-intercept form,

xab

by ,

ay – ab = -bx, bx + ay = ab. Divide both sides by ab,

1by

ax

.

This is the I ntercept Form of a line.

Directed Distance of a Line to a Point Theorem 2.3: The directed distance from the slant

line Ax+By+C=0 to the point P1(x1,y1) is given by the formula

Where the denominator is given the sign of B. The distance is positive if the point P is above the line, and negative if P is below the line.

22

11

BA

CByAxd

Example1: Find the distance from the line 5x=12y+26 to

the points P1(3,-5), P2(-4,1) and P3(9,0).

Example2: Find the distance between the parallel lines

15x+8y+68=0 and 15x+8y-51=0.

Example3: Find the equation of the bisector of the pair of

acute angles formed by the line x-2y+1=0 and x+3y-3=0.

CIRCLES

Definition 3.1A circle is a set of all points in a plane equidistant from a fixed

point called the center. The constant equal distance is called the radius.

Circle with Center at the Origin:

If the center of the circle is at the origin and if the point (x, y) is any point on the circle, then by distance formula,

r)y()x( 22 00 , which is 222 ryx

This is the equation of the circle with center at the origin. If the radius is 1, it is called a unit circle.

(x, y)r

Circle with Center at C(h, k)Let the point (h, k) be the center of a circle and let r be its

radius. By distance formula,

(x – h)2 + (y – k)2 = r2.

This is the center-radius form of an equation of a circle. This is also called the standard form of the equation.

P(x, y)

C(h, k)r

Example: 1. Find an equation of the circle where the endpoints of the diameter are at (-2, 4) and (5, 2).

Solution: Solving for the distance between the two points will give

d = 532452 22 )()( , which is the diameter of the circle.

So, the radius is r = 2

53. To find the center of the circle, we need to get the

midpoint of the diameter. Thus, the center is at (3/2, 3). If follows that the equation is (x – 3/2)2 + (y – 3)2 = 53/4.

Simplifying,x2 + y2 – 3x – 6y – 2 = 0

(-2, 4)(3/2, 3)

(5, 2)

2. Find an equation of the circle with center at (-3,-4) and tangent to the y-axis.

Solution:Since the center is already given, we must be able to find

the radius. Note that the circle is tangent to the y-axis. It means that

the center is 3 units from the left of the y-axis. Thus, the radius is 3.

Hence, the equation is (x + 3)2 + (y + 4)2 = 9.

Simplifying, x2 + y2 + 6x + 8y + 16 = 0

(-3, -4)

3. Find an equation of the circle with center at (4, -1) and tangent to the line x – 2y = 2.

Solution: The radius of the circle is the distance of the line x – 2y = 2 from the point (4, -1). It follows that

d = 5

4

21

21241

22

)(

)()( and so r =

5

4.

The equation of the circle is (x - 4)2 + (y + 1)2 = 16/5

(4, 1)

which is also x2 + y2 – 8x + 2y + 69/5 = 0.

General Form of an Equation of a Circlex2 + y2 + dx + ey + f = 0 where d, e, and f are constants.

Example:

Reduce the equation to center-radius form and draw the circle.4x2 + 4y2 + 24x – 4y + 1 = 0

.

First, divide both sides by 4. We get,

041

y - 6x + y + x 22 .

Transpose the constant to the right side and by completing the squares,

41

41

- 41

y -y + 9 6x + x 22 9

This is equal to

9

2

221

-y 3x where the center is

at

2

13, and radius is 3.

(-3, ½)

3. Find an equation of the circle passing through the points (-1,2), (1,1) and (3,2).