The document provides information on properties of straight lines and methods to write equations of straight lines given different conditions. It also discusses solving systems of linear equations using different methods like graphing, substitution, elimination and Cramer's rule. Key points covered include writing equations in slope-intercept and standard form, finding slopes of parallel and perpendicular lines, and properties of consistent, inconsistent and dependent systems of linear equations.

1. 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.

2. Writing Equations of Lines
Equations of lines come in several different forms.
Two of those are:
2. Slope-intercept form y = mx + b
where m is the slope and b is the y-intercept
2. General form
Ax + By + C = 0
Answers will be written in either of these two
forms only

3. 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
y − y1 = m( x − x1 ) Substitute the above given to point-
slope form equation a of line.
y − 5 = 2( x − 4) Simplify

4. Slope-intercept form General Form
y − 5 = 2( x − 4) y − 5 = 2( x − 4)
y = 2x − 8 + 5 y − 5 = 2x − 8
y = 2x − 3 −2 x + y + 3 = 0
2x − y − 3 = 0
FINAL ANSWER

5. 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
y − y1 = m( x − x1 ) Substitute the above given to point-
slope form equation a of line.
4
y − 2 = − ( x − −3)
5

6. Slope-intercept form General Form
4 4 2
y − 2 = − [ x − (−3)] y = − x−
5 5 5
4 5 y = −4 x − 2
y − 2 = − ( x + 3)
5
4 12 4x + 5 y + 2 = 0
y−2= − x−
5 5
4 2
y = − x−
5 5
FINAL ANSWER

7. 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 x2 = 1 y2 = -6
y2 − y1 Substitute the above given to slope
m= formula to find the slope.
x2 − x1
−6 − 3 −9
m= m= m = −3
1+ 2 3

8. Slope-intercept form General Form
y − y1 = m( x − x1 ) y = −3x − 3
y − 3 = −3[ x − (−2)]
y − 3 = −3( x + 2) 3x + y + 3 = 0
y − 3 = −3x − 6
y = −3x − 3
FINAL ANSWER

9. Definitions
Perpendicular Two lines that makes a 90° angle.
Lines The slopes of perpendicular lines
are negative reciprocal of each
other .
Parallel Lines Lines that never meet .
The slopes of parallel lines are
the same.

10. Writing Equations of Lines
• Given a point and equation of a line parallel
to it.
A Find the equation of the line that passes
through (1, -5) and is parallel to 4x – 2y =3.
Solution: x1 = 1 y1 = -5
4x - 2y =3. Rewrite the equation to slope- intercept
form to get the slope.
- 2y =-4x +3.
-4 3 3
y= x+ . y =2x - . m=2
-2 -2 2

11. Slope-intercept form General Form
y − y1 = m( x − x1 ) y = 2x − 7
y − (−5) = 2( x − 1)
y + 5 = 2x − 2 2x − y − 7 = 0
y = 2x − 2 − 5
y = 2x − 7
FINAL ANSWER

12. Writing Equations of Lines
• Given a point and equation of a line
perpendicular to it.
B Find the equation of the line that passes
through (1, -5) and is perpendicular to 4x – 2y =3.
Solution: x1 = 1 y1 = -5
4x - 2y =3. Rewrite the equation to slope- intercept
form to get the slope.
- 2y =-4x +3.
-4 3 3
y= x+ . y =2x - . m=2
-2 -2 2
m = -½

13. Slope-intercept form General Form
y − y1 = m( x − x1 ) 1 9
y = − x−
2 2
1
y − (−5) = − ( x − 1)
2 2y = −x − 9
1 1 x + 2y + 9 = 0
y+5= − x+
2 2
1 9
y = − x−
2 2 FINAL ANSWER

14. LINEAR EQUATION WITH TWO
VARIABLES
SOLVING SYSTEM OF EQUATION BY:
2.Graph
3.Substitution
4.Elimination
5.Cramer’s Rule

15. Solving Systems of Linear Equations
For two-variable systems, there are then three
possible types of solutions:
Properties
A. Independent system:
one solution and 1. two distinct non-parallel lines
one intersection point
2. cross at exactly one point
3. "independent" system
4. one solution at (x,y )point.

16. Solving Systems of Linear Equations
B. Inconsistent system: Properties
no solution and
no intersection point. 1. two distinct parallel lines
2. never cross
3. No point of intersection
4. "inconsistent" system
5. no solution.

17. Solving Systems of Linear Equations
Properties
C. Dependent system:
infinitely many solution
1. only one line.
2. same line drawn twice.
3. "intersect" at every point
4. "dependent" system,
5. Infinitely many solutions.

18. Methods of Solving
Systems
of Linear Equations

19. A. Systems of Linear Equations:
Solving by Graphing
Solve the following system by graphing.
2x – 3y = –2
4x + y = 24
Solve for y for each equation
Equation 1 Equation 2
2x – 3y = –2
2x + 2 = 3y 4x + y = 24
y = (2/3)x + (2/3) y = –4x + 24

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

21. Using the table of values we can now graph
and look for the intersection:
y=
(2/3
)x +
(2/3
24
x+ solution: (x, y) = (5, 4)
)
–4
y=

22. B. Systems of Linear Equations:
Solving by Substitution
Solve the following system by substitution.
2x – 3y = –2
4x + y = 24
Solution:
4x + y = 24 solve the second equation for y:
y = –4x + 24
2x – 3(–4x + 24) = –2 substitute it for "y" in
2x + 12x – 72 = –2 the first equation
14x = 70
x=5 solve for x

23. Equation 1 Equation 2
x=5 x=5
4x + y = 24 plug this x-value back 2x – 3y = –2
into either equation,
4( 5 ) + y = 24 and solve for y 2( 5 ) – 3y = –2
20 + y = 24 10 – 3y = –2
y = 24 - 20 - 3y = –2 - 10
- 3y = - 12
y=4
y=4
Then the solution is ( x, y ) = (5, 4).

24. C. Systems of Linear Equations:
Solving by Elimination
Solve the following system using elimination.
2x + y = 9
3x – y = 16
Solution:
2x + y = 9 add down, the y's will cancel out
3x – y = 16
5x = 25
x=5 divide through to solve for x

25. Equation 1 Equation 2
x=5 x=5
2x + y = 9 using either of the 3x – y = 16
original equations, to
2( 5 ) + y = 9 find the value 3( 5 ) – y = 16
10 + y = 9 of y 15 – y = 16
y = 9 - 10 - y = 16 - 15
-y= 1
y = -1
y = -1
Then the solution is ( x, y ) = (5, -1).

26. D. Systems of Linear Equations:
Solving by Cramer’s Rule
Solve the following system using cramer’s rule.
2x – 3y = –2
4x + y = 24
Solution: Nx Ny
x= ,y=
D D
D - determinant of the coefficient of the variables
Nx - determinant taken from D replacing the coefficient of x
Ny - and y by their corresponding constant terms leaving all
other terms unchanged

27. 2 −3
D= D = (2) −( −12) =14
4 1
−2 −3 N x = (−2) − (−72) = 70
Nx = N x 70
24 1 x= = =5
D 14
N y = (48) − (−8) = 56
2 −2
Ny = Ny 56
4 24 y= = =4
D 14
FINAL ( 5, 4 )
ANSWER

28. ANSWER THE FOLLOWING PROBLEMS
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
• The diagram shows the graph of 2y = x - 2.
The values of a and b are respectively.
2 and -1

29. • The graphs of x - 2y - 3 = 0 and 6 + 4y - 2x = 0
are identical lines
• 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

30. Find the solution for each system of equation
using any method:
1. 5x – 2y = 0 3. y=x+3
4x + y = 13 5y + 6x = 15
Solution : Solution :
( x , y) = ( 2, 5 ) ( x , y) = ( 0, 3 )
2. 5y = 6x – 3 4. 1 1
x + y =1
2y = x – 4 3 3
1
Solution : y =1 − x
( x , y) = ( -2, -3 ) 2
Solution :
( x , y) = ( -1, 4 )