UNIT 1B LESSON 2

REVIEW OF LINEAR FUNCTIONS

Equations of Lines

The horizontal line through the point (2, 3) has equation

The vertical line through the point (2, 3) has equation

y = 3

x = 2

The vertical line through the point (a, b)has equation x = a since every x-coordinate

on the line has the same value a.

Similarly, the horizontal line through

(a, b) has equation y = b

(𝟐 ,𝟓 )

(𝟐 ,𝟑 )

(𝟐 ,𝟎 )(𝟐 ,−𝟐 )

(−𝟏 ,𝟑 ) (𝟒 ,𝟑 )(𝟎 ,𝟑 ) (𝟐 ,𝟑 )

Finding Equations of Vertical and Horizontal Lines

Vertical Line is x = – 3

Horizontal Line is y = 8

EXAMPLE 1 Write the equations of the vertical and horizontal lines through the point

(−𝟑 ,𝟖 )

Y1 = 2x + 7

x Y = 2x + 7

y – intercept ( , )

𝟐 (−𝟑 )+𝟕=𝟏𝟐 (𝟎 )+𝟕=𝟕

𝟎 𝟕

Slope y-intercept form y = mx + b

slope y-intercept (0, b)

EXAMPLE 2: Reviewing Slope-Intercept Form of Linear Functions

𝒎=𝒓𝒊𝒔𝒆𝒓𝒖𝒏

=𝒚𝟐− 𝒚𝟏

𝒙𝟐− 𝒙𝟏

𝒎=𝟏−𝟕

−𝟑−𝟎=

−𝟔−𝟑

=𝟐

(𝟎 ,𝟕)

(−𝟑 ,𝟏)

Unit 1B Lesson 2 Page 1EXAMPLESState the slopes and y-intercepts of the given linear functions.

y = 4x slope = m = _______ y -intercept ( , )3.

y = 3x – 5 slope = m = _______ y -intercept ( , )4.

)(xf 23

1x= slope = m = _______ y -intercept ( , )6.

xxf 12

1)(

slope = m = _______ y -intercept ( , )

6.

4

3

⅓

𝒇 (𝒙 )=𝟏𝟐

−𝟏𝟐𝒙

0 , 0

0

0 ,

0 ,

General Linear Equation

Although the general linear form helps in the quick identification of lines, the slope-intercept form is the one to enter into a calculator for graphing.

y = – (A/B) x + C/B

By = – Ax + C

Ax + By = C

Analyzing and Graphing a General Linear Equation

Rearrange for y

Slope is

y-intercept is

Find the slope and y-intercept of the line

−𝟑−𝟑

𝒚=−𝟐−𝟑

𝒙+𝟏𝟓−𝟑

𝒚=𝟐𝟑𝒙−𝟓

𝟐

𝟑𝟒

𝟔

Example 7

Unit 1B Lesson 2 Page 1EXAMPLESState the slopes and y-intercepts of the given linear functions.

x + 2y = 3 slope = m = _______ y -intercept ( , )8.−𝟏𝟐

𝟐 𝒚=−𝒙+𝟑

0 , 3/2

𝒚=−𝟏𝟐𝒙+

𝟑𝟐

slope = m = _______ y -intercept ( , )9.

−𝟑 𝒚=−𝟓 𝒙−𝟒

𝒚=𝟓𝟑𝒙+

𝟒𝟑

𝟓𝟑 0 , 4/3

𝟓=𝟐𝟑

(−𝟑 )+𝒃

EXAMPLE 10Find the equation in slope-intercept form for the line with slope and passes through the point

Step 1: Solve for b using the point

Step 2: Find the equation

(𝟎 ,𝟕)(−𝟑 ,𝟓)

𝒚=𝒎𝒙+𝒃

b = 7

𝟓=−𝟐+𝒃

𝒎=𝟐𝟑

−𝟏=𝟐𝟓

(𝟏𝟎 )+𝒃

EXAMPLE 11Find the equation in slope-intercept form for the line parallel to and through the point (10, -1)

Step 2: Solve for b using the point

Step 3: Find the equation

Step 1: The slope of a parallel line will be

𝒃=−𝟓−𝟏=𝟒+𝒃

𝒚=𝒎𝒙+𝒃

(𝟎 ,−𝟓)

(𝟎 ,𝟐)

EXAMPLE 12Write the equation for the line through the point (– 1 , 2) that is

parallel to the line L: y = 3x – 4

Step 1: Slope of L is 3 so slope of any parallel line is also 3.

Step 2: Find b.Step 3: The equation of the line parallel to L: is

Step 4: Graph on your calculator to check your work. Use a square window. Y1 = 3x – 4 Y2 = 3x + 5

(0, 5)

(0, – 4)

𝒃=𝟓𝟐=−𝟑+𝒃𝟐=𝟑(−𝟏)+𝒃

EXAMPLE 13Write the equation for the line that is perpendicular to and passes through the point (10, – 1 )

Step 2: Solve for b using the point (10, – 1)

Step 3: The equation of the line ┴ to

is

Step 1: The slope of a perpendicular line will be negative reciprocal

Step 4: Graph on your calculator to check your work. Use a square window.

Y1 = Y2 = – x + 24

𝒃=𝟐𝟒−𝟏=−𝟐𝟓+𝒃

−𝟓𝟓

𝟐𝟐

EXAMPLE 14Write the equation for the line through the point (– 1, 2) that is

perpendicular to the line L: y = 3x – 4

Step 1: Slope of L is 3 so slope of any perpendicular line is .

𝟐=−𝟏𝟑

(−𝟏)+𝒃

Step 3: Find the equation of the line perpendicular to L: y = 3x – 4

Step 4: Graph on your calculator to check your work. Use a square window.

Y1 = 3x – 4 Y2

Step 2: Find b.

𝒃=𝟓𝟑𝟐=

𝟏𝟑

+𝒃𝟔𝟑

=𝟏𝟑

+𝒃

−𝟐=−𝟓𝟔

(𝟕)+𝒃

EXAMPLE 15Find the equation in slope-intercept form for the line that passes through the points (7, 2) and (5, 8).

Step 1: Find the slope Step 2: Solve for b using either point

Step 3: Find the equation

𝟖=−𝟓𝟔

(−𝟓)+𝒃𝒎=−𝟐−𝟖𝟕−(−𝟓)

=−𝟏𝟎𝟏𝟐

=−𝟓𝟔

(7, – 2)

(– 5, 8)

𝒃=𝟐𝟑𝟔

−𝟏𝟐𝟔

=−𝟑𝟓𝟔

+𝒃

𝒃=𝟐𝟑𝟔

𝟒𝟖𝟔

=𝟐𝟓𝟔

+𝒃

EXAMPLE 16Write the slope-intercept equation for the line through (– 2, –1) and (5, 4).

Slope = m =

–𝟏=𝟓𝟕

(–𝟐)+𝒃

Equation for the line is

(5, 4)(– 2, – 1)

𝒃=𝟑𝟕

−𝟕𝟕

=−𝟏𝟎𝟕

+𝒃

Finish the 5 questions in Lesson #2