Slope Point Form Of A Linear Equation

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Slope point form of a linear equation article deals with way to write the equation of the line when the slope and the point on line are given and the model problems related to slope point form of a linear equation

Slope point form of a linear equation:

The nature of the line depends on the nature slope of the line and the point on the line will satisfy the condition of the line. With the help of the slope and one point on line, we can write the slope point form of a linear equation.
Formula for Slope Point Form of a Linear Equation:

When the slope and the point on the line are given, we can find the slope point form of a linear equation using the formula:

The slope point form of linear equation is

y-y_1=m(x-x_1)

m is the slope of the line

(x1, y1) is the point on the line
Model Problem to Slope Point Form of a Linear Equation:

Problem 1: find the slope point form of a linear equation using slope and one point of the line slope -1 and point is (5, 3)

Solution:

The slope point form of a linear equation is

y- y_1= m(x-x_1)

Here

m ...
... is the slope of the line, m=-1

(x_1,y_1) is the point on the line, (x_1, y_1) = (5, 3)

Plug in the values

y- 3 = -1(x-5)

y-3 = -x+5

y+x=5+3

x+y=7

The standard equation of the line is x+y-7=0

Problem 2: find the standard equation of the line using slope and one point formula containing the two points (3, 1) and (1, 5)(using the slope point form of a linear equation)

Solution:

Since two points are given (x_1, y_1) = (3, 1)

Slope, m= ((y_2-y_1)/(x_2-x_1)) (x_2, y_2) = (1, 5)

m= ((5-1)/ (1-3))

m= (4/-2)

m= -2

We have the slope and the point on the line.

The slope point form of a linear equation is

y- y_1= m(x-x_1)

Here

m is the slope of the line

(x_1,y_1) is the point on the given line. (x_1, y_1) = (3, 1)

y-1 = -2(x-3)

y-1 = -2x+6

1+6= 2x+y

7= 2x-y

The equation of the line is 2x-y-7=0.

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Problem 3: find standard equation of the line (using point slope form of a linear equation) whose perpendicular line slope is -3 and the point on the line is (4, 7).

Solution:

The slope of the line = (1/ (-"slope of the perpendicular line"))

= (1/ (-(-3)))

= (1 /3)

The slope point form of a linear equation

y- y_1= m(x-x_1)

Here

m is the slope of the given line, m= (1/3)

(x_1,y_1) is the point on the given line, (x_1, y_1) = (8, 9)

Plug in the values

y- 7 = (1/3) (x-4)

3y-21= x-4

-21+4= x-3y

-17 = x-3y

x-3y+17=0

The equation of the line is x-3y+17=0

Linear equations lesson plans is the method of solving equations and finding the values of each variable. Linear equations are resolved through eliminating the unknown values. Linear equations lesson plans of systems of equations can be done by two methods as linear equations systems of equations using elimination and linear equations systems of equations using substitution. Let us see sample problems for linear equations lesson plans.
Linear Equations Lesson Plans:

Linear equations using elimination by addition and subtraction.

Do solving by elimination using addition or subtraction. Elimination is method of getting rid of something, in this case, remove a variable. Again, our idea is to have only one variable to solve for.

Steps to Linear equations using elimination.

Example:

x + y = 64 Since the y values have coefficients as positive &
+ x - y = 12 negative one, add them together to eliminate y.
-------------------
2x = 76

` (2x)/2` = `76/2`

x = 38

Using x = 38, substitute this new value for x into either of the original equations.

38 + y = 64
-38 -38
-------------------------
y = 26

Therefore, the solution is x=38 and y=26.

Given: a+b = 16 Since the a values have coefficients of positive one,

- a+2b = 4 subtract to eliminate a.
--------------------
-1b = 20

`(-1b)/(-1)` = `20/(-1)`

b = -20

Using b= -20, substitute this new value for b into either of the original equations.

a+(- 20) = 16
+20 +20
----------------------
a = 36
Linear Equations Lesson Plans:

Steps Linear equations lesson plans using substitution.

Example:

Solve the pair of simultaneous equations: 4x + y = 14 and x + 2y = 1.

Solution:
4x + y = 14 (1)
and
x + 2y = 1 (2)

Rewrite equation (1) as:

y = 14 − 4x. (3)

Substitute (3) in (2):

x + 2(14 − 4x) = −7x+28 = 1.

So, −7x = −27, that is x = 3.8.

Now let us substitute x = 3.8 in equation (1):

4(3.8) + y = 15.2 + y = 14.

So, y = −1.2.

Check through substituting x = 3.8 and y = −1.2 into (2)

3.8 + 2(−1.2) = 1

3.8 - 2.4 = 1 as required.Linear Equations Lesson Plans

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