Topic 8 (Writing Equations Of A Straight Lines)

PROPERTIES OF STRAIGHT LINES

A. WRITING EQUATION OF A LINE :

1. GIVEN TWO POINTS.

2. GIVEN THE SLOPE AND A POINT B. PARALLEL AND PERPENDICULAR

LINES.

Equations of lines come in several different forms. Two of those are:

Writing Equations of Lines

1. Slope-intercept form

where m is the slope and b is the y-intercept

y mx b

2. General form 0Ax By C Answers will be written in either of these two

forms only

Writing Equations of Lines

A. Given a Point and a Slope

Find the equation of the line that goes through the point (4, 5) and has a slope of 2.

Solution: m = 2 x1 = 4 y1 = 5

Substitute the above given to point- slope form equation a of line.

1 1( )y y m x x

5 2( 4)y x Simplify

Slope-intercept form General Form

5 2( 4)y x

2 8 5y x

2 3y x

5 2( 4)y x

5 2 8y x

2 3 0x y

2 3 0x y

FINAL ANSWER

Find the equation of the line that goes through the point (-3, 2) and has a slope of -4/5.

Solution: m = -4/5 x1 = -3 y1 = 2

1 1( )y y m x x Substitute the above given to point- slope form equation a of line.

42 ( 3)

5y x

Slope-intercept form General Form

42 [ ( 3)]

5y x

42 ( 3)

5y x

4 122

5 5y x

4 2

5 5y x

4 2

5 5y x

5 4 2y x

4 5 2 0x y

FINAL ANSWER

Writing Equations of Lines

B. Given Two Points

Find the equation of the line that passes through the points (-2, 3) and (1, -6).

Solution: x1 = -2 y1 = 3

Substitute the above given to slope formula to find the slope.

2 1

2 1

y ym

x x

x2 = 1 y2 = -6

6 3

1 2m

9

3m

3m

Slope-intercept form General Form

1 1( )y y m x x 3 3[ ( 2)]y x

3 3( 2)y x

3 3y x

3 3 0x y

FINAL ANSWER

3 3 6y x

3 3y x

Definitions

PerpendicularLines

Two lines that makes a 90° angle.The slopes of perpendicular linesare negative reciprocal of each other .

Parallel Lines Lines that never meet .The slopes of parallel lines arethe same.

Writing Equations of Lines C. Given a point and equation of a line parallel to it.

Solution: x1 = 1 y1 = -5

Rewrite the equation to slope- interceptform to get the slope.

4x - 2y =3.

2m

Find the equation of the line that passes through (1, -5) and is parallel to 4x – 2y =3.

- 2y =-4x +3.-4 3

y = x + .-2 -2

3 y =2x - .

2

A

Slope-intercept form General Form

1 1( )y y m x x ( 5) 2( 1)y x

5 2 2y x

2 7y x

2 7 0x y

FINAL ANSWER

2 2 5y x

2 7y x

Writing Equations of Lines D. Given a point and equation of a line perpendicular to it.

Solution: x1 = 1 y1 = -5

Rewrite the equation to slope- interceptform to get the slope.

4x - 2y =3.

m = -½2m

Find the equation of the line that passes through (1, -5) and is perpendicular to 4x – 2y =3.

- 2y =-4x +3.-4 3

y = x + .-2 -2

3 y =2x - .

2

B

Slope-intercept form General Form

1 1( )y y m x x

1( 5) ( 1)

2y x

1 15

2 2y x

1 9

2 2y x

2 9y x

FINAL ANSWER

1 9

2 2y x

2 9 0x y

LINEAR EQUATION WITH TWO VARIABLES

SOLVING SYSTEM OF EQUATION BY:

1.Graph

2.Substitution

3.Elimination

4.Cramer’s Rule

For two-variable systems, there are then three possible types of solutions:

Solving Systems of Linear Equations

A. Independent system:

one solution andone intersection point

1. two distinct non-parallel lines 2. cross at exactly one point3. "independent" system 4. one solution at (x,y )point.

Properties

Solving Systems of Linear Equations

B. Inconsistent system:

no solution andno intersection point. 1. two distinct parallel lines

2. never cross 3. No point of intersection 4. "inconsistent" system5. no solution.

Properties

Solving Systems of Linear Equations

C. Dependent system:

infinitely many solution1. only one line. 2. same line drawn twice. 3. "intersect" at every point4. "dependent" system, 5. Infinitely many solutions.

Properties

Methods of Solving Systems

of Linear Equations

Solve the following system by graphing. 2x – 3y = –24x + y = 24

2x – 3y = –22x + 2 = 3y 4x + y = 24

Solve for y for each equation

y = (2/3)x + (2/3) y = –4x + 24

Systems of Linear Equations: Solving by Graphing

A.

Equation 1 Equation 2

x y = (2/3)x + (2/3) y = –4x + 24

–4 –8/3 + 2/3 = –6/3 = –2 16 + 24 = 40

–1 –2/3 + 2/3 = 0 4 + 24 = 28

2 4/3 + 2/3 = 6/3 = 2 –8 + 24 = 16

5 10/3 + 2/3 = 12/3 = 4 –20 + 24 = 4

8 16/3 + 2/3 = 18/3 = 6 –32 + 24 = –8

Get the ( x, y) values for both equation to facilitate easy graphing. The table below shows it

y = –4x + 24

y = (2/3)x + (2/3)

solution:  (x, y) = (5, 4)

Using the table of values we can now graphand look for the intersection:

Systems of Linear Equations: Solving by Substitution

B.

Solve the following system by substitution. 2x – 3y = –24x + y = 24

4x + y = 24y = –4x + 24

Solution:

substitute it for "y" in the first equation

solve the second equation for y:

solve for x

2x – 3(–4x + 24) = –2

x = 5

2x + 12x – 72 = –214x = 70

plug this x-value back into either equation,

and solve for y

4x + y = 24

x = 5

4( 5 ) + y = 24

2x – 3y = –2

x = 5

2( 5 ) – 3y = –2

20 + y = 24 y = 24 - 20

y = 4

10 – 3y = –2- 3y = –2 - 10- 3y = - 12

y = 4

Equation 1 Equation 2

Then the solution is ( x, y ) = (5, 4).

Systems of Linear Equations: Solving by Elimination

C.

Solve the following system using elimination. 2x + y = 93x – y = 16

2x + y = 93x – y = 165x = 25

Solution:

add down, the y's will cancel out

divide through to solve for xx = 5

using either of the original equations, to

find the value of y

x = 5

2( 5 ) + y = 9

x = 5

3( 5 ) – y = 16

10 + y = 9 y = 9 - 10

y = -1

15 – y = 16- y = 16 - 15- y = 1

y = -1

Equation 1 Equation 2

Then the solution is ( x, y ) = (5, -1).

2x + y = 9 3x – y = 16

D. Systems of Linear Equations: Solving by Cramer’s Rule

2x – 3y = –24x + y = 24

Solve the following system using cramer’s rule.

, yxNN

x yD D

Solution:

D - determinant of the coefficient of the variablesNx - determinant taken from D replacing the coefficient of x

and y by their corresponding constant terms leaving all other terms unchanged

Ny -

2 3

4 1D

(2) ( 12) 14D

2 3

24 1xN

2 2

4 24yN

( 2) ( 72) 70xN 70

514

xNx

D

(48) ( 8) 56yN 56

414

yNy

D

FINAL ANSWER

( 5, 4 )

1. (a) Explain why the simultaneous equations 8x – 4y = 20 and y =2x – 3 have no solution . What can you say about the straight lines representing these two equations?

They are parallel

2. The diagram shows the graph of 2y = x - 2. The values of a and b are respectively.

2 and -1

ANSWER THE FOLLOWING PROBLEMS

3. The graphs of x - 2y - 3 = 0 and 6 + 4y - 2x = 0 are identical lines 4. Find the graph of y = -2x - 1?

5. The diagram shows the graph of y = ax + b. Find the values of a and b.

a = 2, b = 2

Find the solution for each system of equationusing any method:

5x – 2y = 0 4x + y = 13

1.

2. 4.

3. y = x + 3 5y + 6x = 15

5y = 6x – 32y = x – 4

1 11

3 3x y

11

2y x

Solution : ( x , y) = ( 2, 5 )

Solution : ( x , y) = ( 0, 3 )

Solution : ( x , y) = ( -2, -3 )

Solution : ( x , y) = ( -1, 4 )