Finding the equation of a line algebraically - Purdue University

16-week Lesson 16 (8-week Lesson 12)

Finding the equation of a line algebraically

The slope formula is an equation with a fraction ( = ). To eliminate

the fraction, we can multiply both sides of the equation by the

denominator.

=

()

=

()

=

This means the vertical change () of a line is equal to its slope times its horizontal change (). This leads us to what is known as point-slope form, which is a way to find the equation of a line using any point that the line passes through and the slope of the line.

Point-slope form: - = or - 1 = ( - 1) - a formula used to find the equation of a line that is passing through a point (1, 1) and that has a slope o plug in the -coordinate of the point for 1, plug in the coordinate of the point for 1, plug in the slope of the line for , and then simplify completely removing parentheses and

combining like terms

- when using this formula we will always leave and as and and

only enter values for 1, 1, and

1

16-week Lesson 16 (8-week Lesson 12)

Finding the equation of a line algebraically

Example 1: Find the equation of a line that passes through the point (1,7), and has a slope of - 34. Isolate the variable in the equation.

We have a point that the line passes through (1,7) and the slope of the line ( = - 34), so we can use point-slope form to find the equation of the line.

- 1 = ( - 1)

3 - 7 = - 4 ( - 1)

33 - 7 = - 4 + 4

33 = - 4 + 4 + 7

= - +

2

16-week Lesson 16 (8-week Lesson 12)

Finding the equation of a line algebraically

The equation we found in Example 1 was written as in terms of

(

=

-

3 4

+

341).

That means that the variable is dependent on the

variable. This is what is known as slope-intercept form, where the

coefficient of the variable is the slope and the constant term is the -

intercept. Looking at the equation from Example 1, the slope of that line

is

-

3 4

and

the

-intercept

is

341.

Slope-intercept form: - = +

- a common way of writing the equation of a line that identifies the slope and the -intercept (0, )

- the result of solving a linear equation for the variable o this means expressing in terms of

Example 2: Find the equation of a line that passes through the point (-5, -2), and has a slope of 52. Enter exact answers only (no approximations), and write the equation in slope-intercept form

( = + ).

We have a point that the line passes through (-5, -2) and the slope of the line ( = 52), so we can use point-slope form to find the equation of the line.

- 1 = ( - 1)

5 - (-2) = 2 ( - (-5))

+

2

=

5 2

(

+

5)

5 25 + 2 = 2 + 2

= +

3

16-week Lesson 16 (8-week Lesson 12)

Finding the equation of a line algebraically

Example 3: Find the equation of the line that passes through the point (3, -2) and has a slope 5. Enter exact answers only (no approximations), and write the equation in slope-intercept form ( = + ).

We have a point that the line passes through (3, -2) and the slope of the line ( = 5), so we can use point-slope form to find the equation of the line.

- 1 = ( - 1)

- (-2) = 5( - 3)

+ 2 = 5 - 15

= -

Example

4:

Find

the

equation

of

the

line

that

has

a

-intercept

of

-

2 5

and

has a slope 17. Enter exact answers only (no approximations), and write

the equation in slope-intercept form ( = + ).

- :

(, )

- = ( - )

( , )

- = ( - )

4

16-week Lesson 16 (8-week Lesson 12)

Finding the equation of a line algebraically

Example 5: Find the equation of the line that has an -intercept of - 2

and a slope of . Enter exact answers only (no approximations), and

write the equation in slope-intercept form ( = + ).

- :

(, )

- = ( - )

(- , )

-

= ( -

- )

To find the equation of a line, you need to have one point that the line passes through and the slope of the line; this is how we found the equations of the lines on Examples 1 ? 5. If you do not have the slope of the line, you will need two points that the line passes through so you can find the slope; this is what we will do on Examples 6 and 7.

Once you have the slope of the line, and at least one point that the line passes through, you can use point-slope form ( - 1 = ( - 1)) to find the equation of the line. Point-slope form was used (or will be used) on each example except Example 4, when slope-intercept form was used since we had the slope of the line and the -intercept.

5

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