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