FINDS THE EQUATION
OF A LINE GIVEN TWO
POINTS AND THE
SLOPE AND A POINT
TWO-POINT FORM
of equation of a line is
𝑦2 −𝑦1
𝑦 − 𝑦1 =
𝑥 −𝑥1 , where
𝑥2− 𝑥1
(𝑥1 , 𝑦1 ) is the coordinate of 𝑃1
Point) and (𝑥2 , 𝑦2 ) is the coordinate
nd
of 𝑃2 (2 Point), respectively.
st
(1
𝑃1 (𝑥1 , 𝑦1 )
𝑃2 (𝑥2 , 𝑦2 )
𝒚𝟐 − 𝒚𝟏
𝒚 − 𝒚𝟏 =
𝒙 −𝒙𝟏
𝒙𝟐− 𝒙𝟏
𝑃1
( 2 ,3 )
𝑃2
( 4 ,5 )
𝒚𝟐 − 𝒚𝟏
𝒚 − 𝒚𝟏 =
𝒙 −𝒙𝟏
𝒙𝟐− 𝒙𝟏
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑥1 = 𝟐
𝒚𝟐 − 𝒚𝟏
𝒚 − 𝒚𝟏 =
𝒙 −𝒙𝟏
𝒙𝟐− 𝒙𝟏
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑥1 =
𝒚𝟐 − 𝒚𝟏
𝒚 − 𝒚𝟏 =
𝒙−𝟐
𝒙𝟐 − 𝟐
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑦1 = 𝟑
𝒚𝟐 − 𝒚𝟏
𝒚 − 𝒚𝟏 =
𝒙−𝟐
𝒙𝟐 − 𝟐
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑥2 = 𝟒
𝒚𝟐 − 𝟑
𝒚−𝟑=
𝒙−𝟐
𝒙𝟐 − 𝟐
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑦2 = 𝟓
𝒚𝟐 − 𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
(
4
,
5
)
𝑃2 𝑥2, 𝑦2
𝑦2 =
𝟓−𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟓−𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟓−𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟓−𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟐
𝟓−𝟑
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟐
𝟐
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟐
𝒚−𝟑=
𝒙−𝟐
𝟒−𝟐
𝟐
𝒚−𝟑= 𝒙−𝟐
𝟐
𝟐
𝒚−𝟑= 𝒙−𝟐
𝟐
𝒚−𝟑=𝟏 𝒙−𝟐
𝒚−𝟑=𝟏 𝒙−𝟐
𝒙
𝒚−𝟑=𝟏 𝒙−𝟐
𝒚−𝟑=𝒙
𝒚−𝟑=𝟏 𝒙−𝟐
−𝟐
𝒚−𝟑=𝒙
𝒚−𝟑=𝟏 𝒙−𝟐
𝒚−𝟑=𝒙−𝟐
𝒚−𝟑=𝒙−𝟐
𝒚−𝟑=𝒙−𝟐
𝒚−𝟑+𝟑=𝒙−𝟐+𝟑
𝒚−𝟑+𝟑=𝒙−𝟐+𝟑
𝟎
𝒚+𝟎=𝒙−𝟐+𝟑
𝒚=𝒙−𝟐+𝟑
1
𝒚=𝒙+𝟏
ACTIVITY: 1
Find the equation of a line with the given two points
1. (4, 6) and (3, -5)
2. (5,1) and (2, 4)
3. (-1, 0) and (2,3)
4. (0, 0) and (-1, 4)
POINT-SLOPE FORM
of equation of a line is
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
where m is the slope and x1 and
y1 are coordinates of the fixed
point.
𝑃1
( 2 ,3 )
𝑥1 , 𝑦1
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑚=2
𝑥1 = 2
𝑦1 = 3
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑚=
𝑥1 = 2
𝑦1 = 3
𝑦 − 𝑦1 = 2(𝑥 − 𝑥1 )
𝑚=
𝑥1 =
𝑦1 = 3
𝑦 − 𝑦1 = 2(𝑥 − 2)
𝑚=
𝑥1 =
𝑦1 =
𝑦 − 3 = 2(𝑥 − 2)
𝑦 − 3 = 2(𝑥 − 2)
𝑦−3=
2𝑥
𝑦 − 3 = 2(𝑥 − 2)
𝑦 − 3 = 2𝑥
𝑦 − 3 = 2(𝑥 − 2)
𝑦 − 3 = 2𝑥
−4
𝑦 − 3 = 2(𝑥 − 2)
𝑦 − 3 = 2𝑥 − 4
𝑦 − 3 = 2𝑥 − 4
𝑦 = 2𝑥 − 4 − 3
𝑦 = 2𝑥 − 4 + 3
𝑦 = 2𝑥 − 4 + 3
−1
𝑦 = 2𝑥 − 1
𝑦 = 2𝑥 − 1
ACTIVITY: 2
Find the equation of the line of the form 𝑦 = 𝑚𝑥 + 𝑏
given the slope and a point by using the linear equation
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ).
1.
2.
3.
4.
5.
𝑚 = 2 𝑎𝑛𝑑 0,4
𝑚 = 1 𝑎𝑛𝑑 5, −2
𝑚 = −5 𝑎𝑛𝑑 −3,9
𝑚 = −7 𝑎𝑛𝑑 4, −1
7
𝑚 = − 𝑎𝑛𝑑 (−4,3)
2