MATHLECT1LECTUREFFFFFFFFFFFFFFFFFFHJ.pdf

lecture 1
Prepared by:
Assist. Lect. Hiba Abdul –Kareem Assist. Lect. Azam Isam

A SLOPE OF A LINE
 In Mathematics, a slope of a line is the change in y coordinate
with respect to the change in x coordinate.
 The change in y-coordinate is represented by Δy and the
change in x-coordinate is represented by Δx.
 Hence, the change in y-coordinate with respect to the change
in x-coordinate is given by:
 m = change in y/change in x = Δy/Δx
 Where “m” is the slope of a line.

A SLOPE OF A LINE
m = (y2 – y1)/(x2 – x1)
Example 1)What is the slope of a line passes through points
(4,6) and (3,4)?
Solution:
m = (y2 – y1)/(x2 – x1)
m = (4 – 6)/(3 – 4)
m = -2/-1
m=2

A SLOPE OF A LINE
Example 2)Determine the Slope that passes through the given
pair of points:
1) (3, 2), (6, 6)
• Solution:
m = (y2 – y1)/(x2 – x1)
m = (6 – 2)/(6 – 3)
m = 4/3

A SLOPE OF A LINE
3) (-4, 2), (-5, 4)
Solution:
m = (y2 – y1)/(x2 – x1)
m = (4 – 2)/(-5 –(- 4))
m = 2/-1
m=-2
2) (-9, 6), (-10, 3)
Solution:
m = (y2 – y1)/(x2 – x1)
m = (3 – 6)/(-10 –(- 9))
m = -3/-1
m=3

TYPES OF SLOPES
The slope of a line can be of 4 types depending on the change in the x
and y-coordinates.
1)Positive Slope:
 A line has a positive slope when both the x and y-coordinates
successively increase or decrease.
 This means that if the x-coordinate increases, so do the y-coordinate
and vice versa.

TYPES OF SLOPES
2)Negative Slope:
 The relation between the x and y-coordinates reverses in a negative
slope.
 This goes to say that when either one of the coordinates increases,
the other will decrease. Thus, if the x-coordinate increases, the y-
coordinate will decrease and vice versa.

TYPES OF SLOPES
3)Zero Slope
 A perfectly horizontal line has zero slope. This type of line will be
parallel to the x-axis.

TYPES OF SLOPES
4) Undefined Slope or Infinite Slope
an infinite slope is perfectly vertical. Such a line will be parallel to the
y-axis.

TYPES OF SLOPES
Example 3) What is the type of Slope that passes through the points
(-1,-1) and (2,-2) ?
Solution:
m = (y2 – y1)/(x2 – x1)
m = (-2 – (-1))/(2 –(- 1))
m = -1/3
negative slope
(2,0) and (0,-3) ?
Solution:
m = (y2 – y1)/(x2 – x1)
m = ((-3)-0))/(0 -2)
m = 3/2
Positive Slope

TYPES OF SLOPES
(1,2) and (3, 2) ?
Solution:
m = (y2 – y1)/(x2 – x1)
m = (2 – 2)/(3 –2)
m = 0/2
m=0
Zero Slope
(1,2) and (1,5) ?
Solution:
m = (y2 – y1)/(x2 – x1)
m = (5-1))/(2 -1)
m = 3/0 error!
Undefined Slope

classwork
Determine the Slope of the following linear
functions that passes through the given pair of
points:

LINEAR EQUATION
A linear equation is the equation of a line
The standard form of a linear equation is
Ax + By = C
Examples of linear equations:
2x + 4y =8 The equation is in the standard form
6y = 3 – x x + 6y = 3
4x-y=-21

LINEAR EQUATION
• The following equations are NOT in the standard
form of Ax + By =C:

LINEAR EQUATION
• Determine whether the equation is a linear
equation, if so write it in standard form.

EQUATIONS OF THE STRAIGHT LINES
Horizontal lines
The standard form of equation
in Horizontal lines is:
where b is the y-intercept. Note:
m = 0.
Y=b
Vertical lines
The standard form of equation
in vertical lines is:
where a is the x-intercept. Note:
m is undefined
x=a

Neither horizontal nor Vertical
a)point-slop equation
The general form of point-slope equation of the point (x1,y1)
with the slope m is :
Example 7): Write the point-slope form for the equation of the line through the
point (-2, 5) with a slope of 3.
y-y1= m(x-x1)
Use the point-slope form, y – y1 = m(x – x1), with m = 3 and
(x1, y1) = (-2, 5).
y – 5 = 3(x – (-2))
y – 5 = 3(x + 2)
y= 3x+11

Example 8) Write the point-slope form of the equation of the
line passing through (-1,3)with a slope of 4.
Solution We use the point-slope equation of a line with m = 4,
x1= -1, and y1 = 3.
y-y1= m(x-x1)
y– 3 = 4[x – (-1)] Substitute the given values. Simply.
We now have the point-slope form of the equation for the
given line.
y – 3 = 4(x + 1)
y – 3 = 4x + 4
y = 4x + 7

Steps:
1. Find slope 2. Place a point and
the slope in into point
slope form
3. Point-Slope form
Example 9) Given two points, write the equation of a line in
point-slope form (2,-3) and (4,-2)
2  3
4  2

2  3
2

1
2
y  y1  m(x  x1)
y  3 
1
2
(x  2)
y  3 
1
2
(x  2)
)
2
(
2
1
3 

 x
y
4
2
1

 x
y

20
Example 10): Write the slope-intercept form for the equation
of the line through the points (4, 3) and (-2, 5).
y – y1 = m(x – x1) Point-slope form
Slope-intercept form
y = - x + 13
3
1
3
2 1
5 – 3
-2 – 4
= -
6
= -
3
Calculate the slope.
m =
Use m = - and the point (4, 3).
y – 3 = - (x – 4)
1
3 3
1

Example 11: Find the equation of a straight line that passes
through the points (1, 3) and (-2, 4).
Solution: To determine the equation of the line, we will use the
formula point-slope form.
y-y1= m(x-x1)
For this, we first need to find the slope of the line (m).
m = (4-3)/(-2-1) = -1/3
Therefore, the equation of the line passing through (1, 3) and (-2, 4)
is
y - 4 = (-1/3) (x + 2)
y - 4 = -x/3 - 2/3
y + x/3 = 4 - 2/3
x + 3y = 10

classwork
Give the equation of the linear function y
with the given slope and passing through
given points

Neither horizontal nor Vertical
b)Slope intercept equation
The general form of Slope intercept equation of a line L
with y-intercept b and the slope m is :
Y=mx + b

Example13: Find the equation of the line that has a slope of 2/3 and a
y intercept of (0, 4)

Example14) Find the equation of the line passing
through the points with coordinates (2, 4) and (8, 7).

classwork
Give the equation of the linear function y in
slope intercept form given its slope and y-
intercept

Distance Between Two Points
• Distance between two points is the length of the line
segment that connects the two points in a plane.
• The formula to find the distance between the two points is
usually given by
• d= ((x2 – x1)² + (y2 – y1)²).
• This formula is used to find the distance between any two
points on a coordinate plane or x-y plane

Example 15) What is the distance between two points
A and B whose coordinates are (3, 2) and (9, 7),
respectively?
• Solution: Given, A (3,2) and B(9,7) are the two points
in a plane.
distance = ((x2 – x1)² + (y2 – y1)²).
Here, x1 = 3, x2 = 9, y1 = 2 and y2 = 7.
Thus, putting all the values of x and y in the formula,
we get;
• d= ((9−3)² + (7−2)²)
• d = 61 unit.

Example 16) : Find the distance of a point P(4, 3)
from the origin.
Solution: Given, P(4, 3) is a point at a distance from the
origin.
The coordinates of a point at the origin will be (0,0).
Using the distance between formula for two points, we
know;
distance = ((x2 – x1)² + (y2 – y1)²).
Here, x1 = 4, x2 = 0, y1 = 3 and y2 = 0.
Thus, putting all the values of x and y in the formula,
we get;
d = ((0−4)² + (0−3)²)
d = d= 25
d = 5 unit

classwork
Find the distance of a point P(4, 3) from
the origin.
1. A(6,7) B(-1,7)
2. E (-1,7) F(12,0)

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