[2.4.1] Point-slope form for lines: y-y₁= m (x-x₁)

When given two-points:

1. First, you must find the slope (m) by using the “two-points” formula by plugging in your two given points into the following equation:

m= y₂-y₁

      x₂-x₁

2. Now that you have found slope (m) and were given two points on the line, choose one point given and use the point-slope form to find the equation of the line.

Here is an example:

(-2, -1) (3, 4) ßtwo given points

m= y₂-y₁ à  m= 4-(-1) à  m= 5 à  m =1         finding the slope (m)

x₂-x₁                    3-(-2)             5

y-y₁= m(x-x₁)                                                       point-slope form

y-(-1)= 1[x-(-2)]                                                  substitute (m) with your answer

                                                                            substitute x and y with the points given

y + 1= x + 2                                                        simplify

y= x + 1

When given the slope and one point:

1. Because you know the slope (m), you do not need to use the “two points” formula to find the slope (m).

2. When given a slope and one point on the line, you should use the point-slope form to write an equation of the line.

3. Once you have used the slope-point form to find the equation of the line, you can simplify the result by using the “slope-intercept form” and distributing.

Here is an example:

½ ,(3,4) ß given slope and point on the line

y-y₁= m (x-x₁)                                          point-slope form

y-4= ½ (x-3) substitute (m) with the given slope

…now simplify…

y-4= ½ (x-3)                                        your point-slope form

y-4= ½ x – ³_/2 distributive property

y= ½ x + 2.5                        converted into slope-intercept form

Equation of perpendicular lines:

For a line to be perpendicular to each other, the lines must have reciprocal slopes.

Here is an example to understand the concept:

1. Write an equation of the line that passes through (6,4) and is perpendicular to the line y= -6x + 4

2. The given line (y= 6x + 4 ) has a slope (m) of -6. A line that is perpendicular to this line must be the negative reciprocal of (m₂) = -1 = 1/6

m₁

3. Because you know the slope and a point on the line, use the point-slope form

and replace (x₁,y₁) with the given points to find an equation of the line.

y-y₁= m₂(x-x₁) point-slope form

y-4= 1/6(x-(-6)) substitute (m₂) with the reciprocal

substitute x₁ and y₁ with the given point

y-4= 1/6x + 1 distributive property

y=1/6x + 4 converted into slope-intercept form

Equation of parallel lines:

For a line to be parallel to each other, the lines must have the exact same slope.

Here is an example to understand the concept:

1. For parallel lines, obviously they must have the same slope in order to be parallel to each other, making slope (m₁₎ = slope (m₂₎ true.

2. Use the point-slope form to find the equation of parallel lines

3, (2,1) ß given slope and point

y-y₁= m₂(x-x₁) point-slope form

y-1= 3(x-2) substitute (m₂) with the given slope

substitute x₁ and y₁ with the given point

y-1= 3x -6 distributive property

y= 3x -5 converted into slope-intercept form

!!!!!!!NOW YOU TRY!!!!!!!!!!

1. Write an equation of the line that passes through the given point and has the given slope.

a) (0,4), m= 2

b) (-6, 5), m= 0

c) (3, -2), m= -4/3

2. Write an equation of the line that passes through (1,-1) and is perpendicular to the line y= -1/2x + 6

3. Write an equation of the line that passes though

(2, -7) and is parallel to the line x= 5