In mathematics, the slope-intercept form and the point-slope form are two well-known techniques used to calculate the equation of a straight line. The equation of the straight line must be linear that is why it is also called the linear equation of the line.
Both techniques are helpful in calculating
the linear equation of the line in different manners. In this post, we will
cover the basics of the slope-intercept form and the point-slope form along
with their definitions, equations, and solved examples.
What is the point-slope form?
In algebra, a linear equation of the line
that is evaluated with the help of the slope (steepness) of the line and any
point that exists on the line is known as the point-slope form. This technique
is frequently used when the slope and the point on the line are given.
The expression that is helpful in
determining the point-slope form is:
y – y1 = m (x – x1)
·
m = the slope (inclination or steepness) of the line.
·
x1 & y1 = the coordinate points of the line.
·
x & y = the fixed points of the line.
How to calculate the point-slope form?
The equation of the point-slope form can be
calculated easily by following the below steps.
·
If the slope is not given
calculate it (in the case of two points technique)
·
Take any point that exists on
the line such as (x1, y1).
·
Take the general expression of
the point-slope form and substitute the slope and point to that expression.
·
The result must be the point-slope
form of the line.
Here are a few solved examples of this
technique of line equation to learn how to calculate it.
Example 1: When two points are given
Determine the point-slope form of the linear equation of the line by using two
points method. If the existing points of the line are:
(x1,
y1) = (5, 6) & (x2, y2) = (15, 4)
Solution
Step 1: Write the given existing
points on the line.
x1 = 5, x2 = 15, y1
= 6, y2 = 4
Step 2: First of all, calculate
the slope of the line by substituting the given points for the its general
expression.
Formula for the slope of the line
Slope of the line = m = [y2 – y1]
/ [x2 – x1]
Put the given values
Slope = m = [4 – 6] / [15 – 5]
Slope = m = [-2] / [10]
m = -2/10
m = -1/5 = -0.2
Step 3: Now take the general
form of the point-slope form.
(y – y1) = m (x – x1)
Step 4:
Now substitute the calculated slope of the line and any
point that existed on the line to calculate the point-slope form of the line.
(y –
y1) = m * (x – x1)
(y – 6) = -0.2 * (x – 5)
(y – 6) = -0.2 * x + 0.2 * 5
y - 6 = -0.2x + 1
y - 6 + 0.2x – 1 = 0
y + 0.2x - 7 = 0
0.2x + y - 7 = 0
Example 2: When slope and point are
given
Determine the point-slope form of the linear equation of the line. If the slope of
the line is 2 and existed point on the line is:
(x1,
y1) = (15, -7)
Solution
Step 1: Write the given slope
and existing point on the line.
x1 = 15
y1 = -7
m = 2
Step 2: Now take the general
form of the point-slope form.
(y – y1) = m (x – x1)
Step 3:
Now substitute the given slope of the line and any
point that existed on the line to calculate the point-slope form of the
line.
(y –
y1) = m * (x – x1)
(y – (-7)) = 2 * (x – 15)
(y + 7) = 2 * (x – 15)
(y + 7) = 2 * x – 2 * 15
y + 7 = 2x – 30
y + 7 – 2x + 30 = 0
y – 2x + 37 = 0
2x – y – 37 = 0
What is the slope-intercept form?
In algebra, a linear equation of the line
that is evaluated with the help of the slope (steepness) of the line and the y-intercept
of the line is known as the slope-intercept form. This technique is frequently
used when the slope and the y-intercept of the line are given.
The expression that is helpful in
determining the slope-intercept form is:
y =
mx + b
·
m = the slope (inclination or steepness) of the line.
·
b = the y-intercept (y value) of the line.
·
x & y = the fixed points of the line.
How to calculate the slope-intercept form?
The equation of the slope-intercept form can be calculated easily either by using a slope intercept calculator or by following the below steps.
·
If the slope is not given
calculate it (in the case of two points technique)
·
After that calculate the
y-intercept of the line if it is not given.
·
Take the general expression of
the slope-intercept form and substitute the slope and y-intercept to that
expression.
·
The result must be the slope-intercept
form of the line.
Here are a few solved examples of this
technique of line equation to learn how to calculate it.
Example 1: When two points are given
Determine the slope-intercept form of the linear equation of the line by using two
points method. If the existing points of the line are:
(x1,
y1) = (12, -6) & (x2, y2) = (18, 12)
Solution
Step 1: Write the given existing
points on the line.
x1 = 12, x2 = 18, y1
= -6, y2 = 12
Step 2: First of all,
calculate the slope of the line by substituting the given points for its
general expression.
Formula for the slope of the line
Slope of the line = m = [y2 – y1]
/ [x2 – x1]
Put the given values
Slope = m = [12 – (-6)] / [18 – 12]
Slope = m = [12 + 6] / [18 – 12]
Slope = m = [18] / [6]
Slope = m = 18/6
Slope = m = 3
Step 3: Now take the general
form of the slope-intercept form and evaluate the y-intercept of the line.
y =
mx + b
-6 =
3(12) + b
-6 = 36 +
b
- 6 – 36
= b
-42 = b
Step 4:
Now substitute the calculated slope of the line and y-intercept
of the line to calculate the slope-intercept form of the line.
y =
mx + b
y =
3x + (-42)
y = 3x – 42
y = 3(x - 12)
Example 2: When the slope and point are
given
Determine the slope-intercept form of the linear equation of the line If the slope
of the line is -6 and existed point on the line is:
(x, y) =
(4, 3)
Solution
Step 1: Write the given slope
and the points on the line.
x1 = 4, x2 = 3
m = -6
Step 2: Now take the general
form of the slope-intercept form and evaluate the y-intercept of the line.
y =
mx + b
3 = -6(4)
+ b
3 = -24 +
b
3 + 24 =
b
27 = b
Step 3:
Now substitute the given slope of the line and
y-intercept of the line to calculate the slope-intercept form of the line.
y =
mx + b
y =
-6x + (27)
y = -6x + 27
y = -3(2x – 9)
Final words
In this lesson, we have covered all the
basics of slope-intercept and point-slope forms along with solved examples. Now
you can solve any problem of these techniques to determine the linear equation
of the line by learning the above lesson.