Slope-Intercept Form and Point-Slope Form

(1)

Slope-Intercept Form and Point-Slope Form

In this section we will be discussing Slope-Intercept Form and the Point-Slope Form of a line. We will also discuss how to graph using the Slope-Intercept Form.

Slope-intercept form:

When an equation is in slope-intercept form we can directly read the slope and the y-intercept from the equation. We can also graph immediately using this information.

Slope-intercept form: y = mx + b where m is the slope and (0,b) is the y-intercept.

Example: y = 2x + 1

We have slope of 2 and y-intercept (0,1).

Note that the purple square is the y-intercept (0,1) and red points are found by using the slope of 2.

Remember that we are seeing a visual of all the solutions to the equation (every point on the line is a solution).

-5 -4 -3 -2 -1 1 2 3 4 5

x y

(2)

Point-Slope form:

When an equation is in point-slope form, we can directly read the slope and a point that the line passes through from the equation. Usually we don’t graph from this form because we don’t often see problems written this way, but we can.

Point-slope form: y

−

y₁

=

m x

( −

x₁

)

or y

=

m x

( −

x₁

) +

y₁ where m is the slope and

1 1

( ,x y )is a point on the line.

Example: y – 3 = 2(x – 1) or y = 2(x – 1) +3

Here the slope is 2 and the point that the line passes through is (1,3). Note that when given in the first form y – 3 = 2(x – 1) the subtraction signs do not affect the coordinates of the points.

Note that the purple square is the point given by the equation (1,3) and red points are found by using the slope of 2.

You might have noticed that the graphs are the same. If we solve the equation given in point-slope form for y, we end up with the equation in slope-intercept form:

( )

3 2 1

3 2 2 2 1

y x

− = −

= +

Standard form:

Standard form is not a very helpful form as far as graphing, but some books ask students to write their equations in standard form. Your instructor may want you to write your work in slope-intercept

-5 -4 -3 -2 -1 1 2 3 4 5

x y

(3)

Standard Form: Ax + By = C where A,B and C are real numbers and A and B can not both equal 0 at the same time.

We have seen the graph of y = 2x + 1 above. In standard form, this equation would look like -2x + y = 1. However, this is not the only way we could write it. Often books will have A positive.

We can do this if we multiply both sides of the equation by -1: 2x – 1 = -1. There are multiply ways to write an equation in Standard Form and it still be correct. So, if your book writes the answers in Standard Form, you may need to manipulate their equation to see if it matches what you found.

Since there are multiply ways to write Standard Form, many instructors prefer students write their lines in slope-intercept form.

Notes about slope-intercept form:

1. The slope is a number, so don’t write a variable when you are identifying m.

2. Write the y-intercept as a point. Since it is a point on the y-axis, the x-coordinate will be 0.

3. You MUST solve for y to find the slope and y-intercept. 2y = 3x+1 is not slope-intercept form, and the slope is NOT 3 and the y-intercept is NOT (0,1).

4. 4

2 y x+

= is not in slope-intercept form. We must rewrite with the 2 under each of the

terms in the numerator: 4 2 2

y = x+ , which simplifies to 2 2

y = x+ ; So, 1

m =2 and the y- intercept is (0,2).

5. When graphing using the slope intercept form, first plot the y-intercept. Then use the slope

to find additional points. Note that a slope of 2

3 means we would “rise” up in the positive direction 2 units and “run” to the right in the positive direction 3 units. We can repeat this as

often as we want. Also, since 2 2 is the same as

3 3

−

− , we could also “rise” down in the negative direction 2 units and “run” to the left in the negative direction 3 units and be on the same line. Similarly for slopes such as -5, we can associate the negative with the

(4)

numerator and denominator to get additional points going both “directions” on the line:

5 5

5 1 1

− = − =

− ^.

EXAMPLE 1: Put -2x + 4y = 12 into slope-intercept form, find the slope and the y-intercept (write as a point). Graph the equation using the y- intercept and the slope. Plot the y-intercept point and three additional points using the slope.

Solution:

First we need to put −2x+4y=12 slope-intercept form by solving for y:

The slope is 1

m = 2 and the y-intercept is (0,3)

Note the we have labels on the axes, the scale on both the axes, and the arrows on the line. Also note how the slope gives us points going both “ways” on the line from the y-intercept and that the line rises from left to right as it should with a positive slope.

2 4 12

4 2 12

2 12

4 4

2 3

x y

y x

y x y x

− + =

= +

-5 -4 -3 -2 -1 1 2 3 4 5

x y

(5)

Finding equations of lines given various information:

To find the equation of a non-vertical line, we need only 2 pieces of information:

1. The slope.

2. A point on the line.

You may use either the slope-intercept form or the point-slope form to find equations of lines. The example below shows both forms. Pick your favorite and practice using it.

Note: If finding the equation for a horizontal or vertical line, it may be easiest to graph it to find the equation!

EXAMPLE 2: Find the equation of a line in slope-intercept form (if possible) for the line that passes through the points (3, 3) and (6, 5) and write it as a function (if possible)

Solution:

We need to find the slope since it is not given:

5 3 2 6 3 3

m −

= =

− ^.

Now we need to find b, the y-intercept. We can do this two ways:

1. Using the slope-intercept form:

2 ( )

5 6 (1)

3 5 2(2) (2)

5 4 (3)

1 (4)

So we have 2 1 (5) 3

y mx b b b b b

y x

= +

=

= +

(1): Fill-in one of the points into the variables x and y. It does not matter which point you pick. Also fill-in the slope.

(2)-(4): Solve for b.

(5): Fill-in the slope and the y- intercept, but do NOT fill-in numbers for x and y. The x and y are our variables and MUST stay variables as they represent every point on the line as ordered pairs (x,y)!

(6)

2. Using the point-slope form:

( ) ( )

( )

1 1

where

1

,

1

is a point on the line.

3 2 3 (1)

3 3 2 2 (2)

3 2 1 3

3

y y m x x x y

y x

y x y x

− = −

= +

Note that we have the same equation for both methods!

EXAMPLE 3: Find the equation of a line in slope-intercept form (if possible) for a line that has undefined slope and passes through the point (4, 6).

Solution:

We already have a point and a slope. But, the slope isn’t one we can “fill-in” to either formula because it isn’t a number! Recall that vertical lines have undefined slopes. So, we want a vertical line that passes through the point (4,6). Visualize this by making a graph:

The equation of this line is x = 4. We can not put it in slope-intercept form because the slope is undefined and there is no y-intercept.

(1): Fill-in one of the points into the

1and 1

x y . It does not matter which point you pick. Also fill-in the slope.

(2)-(3): Solve for y.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

x y

(7)

EXAMPLE 4: Find the equation of a line in slope-intercept form (if possible) that passes through the point (-1,-2) and is parallel to the line

1 2

y=2x+ . Solution:

This problem is often misunderstood by students. Often students think that the line 1 2 2 y= x+ passes through the point (-1, -2). It does not. We can verify this by filling in the coordinates of the point into the line:

?

( )

?

2 1 1 2 2

2 1 2 2 2 3

2 − = − +

− = − +

− ≠

It may be helpful to graph what we are given to understand what we want to find…

We want to find the equation of a line that is parallel to the given line and passes through the point (-1, -2). There are infinitely many lines that are parallel to the given line, but only one that passes through the point (-1, -2). See the graph – the pink line is the one we want the equation for. The nose-mucus green lines are just examples of other parallel lines to the original (blue) line.

-4 -3 -2 -1 1 2 3 4

x y

(-1,-2)

y=(1/2)x + 2

-4 -3 -2 -1 1 2 3 4

x y

(-1,-2)

y=(1/2)x + 2

(8)

Recall that we only need two pieces of information to find the equation of a line: (1) the slope and (2) a point on the line.

We know that the pink line has slope of 1

2 because it is parallel to the given line 1 2 2

y= x+ that has slope of 1

2. We also know a point that our pink line will pass through: (-1,-2). We can use the slope-intercept form or the point-slope form to find the equation:

Slope-intercept form: Point-slope form:

1

( )

2 1

2 2 1

2 3 2

y mx b b b b b

= +

− = − +

− =− +

− + =

− =

( )

( ) ( ( ) )

( )

1 1

2 1 1

2

2 1 1

2 2 1

2 2 3 2 2 y y m x x

y x

y x y x

− = −

− − = − −

+ = +

= −

So the equation of the pink line is 1 3

2 2

y= x− .

EXAMPLE 5: Let’s change the prior example slightly: Find the equation of a line in slope-intercept form (if possible) that passes through the point

(-1,-2) and is perpendicular to the line 1 2 2 y= x+ .

Solution:

The only change to this problem and what we had in the prior example is now we want to find a line that is perpendicular instead of parallel. Again, it might help to think of this visually:

(9)

-4 -3 -2 -1 1 2 3 4

x y

(-1,-2)

y=(1/2)x + 2

As before the nose-mucus green lines are just random perpendicular lines to the original blue line that we are given. We want to find the equation of the pink line that is perpendicular, but also passes through the point (-1, -2).

Recall that we only need two pieces of information to find the equation of a line: (1) the slope and (2) a point on the line.

We know that the pink line has slope of -2 because it is perpendicular to the given line 1 2 2 y= x+ that has slope of 1

2 (recall that perpendicular lines have slopes that are negative reciprocals of each other). We also know a point that our pink line will pass through: (-1,-2). We can use the slope-intercept form or the point-slope form to find the equation:

Slope-intercept form: Point-slope form:

( )

2 2 1

2 2 4

y mx b b b b

= +

− = − − +

− = +

− =

( )

( ) ( ( ) )

( )

1 1

2 2 1

2 2 2

2 4

y y m x x

y x

− = −

− − = − − − + = − + + = − −

= − −

So the equation of the pink line is y= −2x−4. We could have also found this one pretty easily by

“back-tracking” from the graph.

Vertical lines and the slope-intecept form (or the point-slope form):

Vertical lines are the only type of line that we can’t put into slope-intercept form (or point-slope form). This is because the slope is undefined. We can’t write the word “undefined” for m in y = mx

(10)

+ b! Equations of vertical line will be “x = the number where the line crosses the x-axis.” Example:

x = 4 would be the equation of a vertical line that crosses the x-axis at 4.

Final comments about lines:

1. Horizontal lines will have equations of the form “y = b” where b is the y-intercept (where the line crosses the y-axis). Note that y = b is slope-intercept form with the slope as 0: y = 0x+b

2. A line of the form y = mx is also in slope-intercept form, where the y-intercept is (0,0).

Pros and Cons for using Point-Slope Form, y

−

y1

=

m x

( −

x1

)

to find equations of lines:

Pro: Many students find it easier and don’t make the mistake of filling-in numbers into the variables (x and y) for the final equation that sometimes happens when using the slope- intercept form.

Con: You will need to memorize another formula and know when and how to apply it.