


















































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
pre calculus basic study notes for students
Typology: Study notes
1 / 58
This page cannot be seen from the preview
Don't miss anything!



















































By words; If a line segment has two- dimensional endpoints at ( ๐ฅ 1 , ๐ฆ 1 ) and ( ๐ฅ 2 , ๐ฆ 2 ), then the midpoint of the segment has coordinates of ( ๐ฅ 1 +๐ฅ 2 2
๐ฆ 1 +๐ฆ 2 2
By symbols; Midpoint = ๐ฅ 1 +๐ฅ 2 2
๐ฆ 1 +๐ฆ 2 2
Find the midpoint of the line segment who has (- 5 , 6 ) and ( 1 , 7 ). Let say that (- 5 , 6 ) is the coordinates for (๐ฅ 1 , ๐ฆ 1 ) and ( 1 , 7 ) is the coordinates for (๐ฅ 2 , ๐ฆ 2 ). Solution, M = ๐ฅ 1 +๐ฅ 2 2 , ๐ฆ 1 +๐ฆ 2 2 M = โ 5 + 1 2 , 6 + 7 2 midpoint = (โ 2 , 13 2 )
Find the coordinates of the midpoint of the segment with endpoints of ( 6 , - 5 , 1 ) and (- 2 , 4 , 0 ). Let say that ( 6 , - 5 , 1 ) is the coordinates for (๐ฅ 1 , ๐ฆ 1 , ๐ง 1 ) and (- 2 , 4 , 0 ) is the coordinates for (๐ฅ 2 , ๐ฆ 2 , ๐ง 2 ). Solution, M = ๐ฅ 1 +๐ฅ 2 2 , ๐ฆ 1 +๐ฆ 2 2 , ๐ง 1 +๐ง 2 2 M = 6 +(โ 2 ) 2 , โ 5 + 4 2 , 1 + 0 2 midpoint = ( 2 , โ 1 2 , 1 2 )
By words; The distance between two points with coordinates (๐ฅ 1 , ๐ฆ 1 ) and (๐ฅ 2 , ๐ฆ 2 ) is given. By symbols; ๐๐๐ ๐ก๐๐๐๐ = (๐ฅ 2 โ ๐ฅ 1 ) 2 +(๐ฆ 2 + ๐ฆ 1 ) 2
From the Pythagorean Theorem; ๐ 2 = ๐ 2
Given a general quadratic equation of the form ๐๐ฅ 2
2 โ 4๐๐ 2๐
By definition, the conic sections are the classes of non- degenerate curves generated by the intersections of a plane with one or two nappes of a cone. A conic section can also be realized as the zero set of a quadratic equation in two variables. General Equation of Conic Sections ๐ด๐ฅ 2
Solve for the system: ๐ฆ^2 + 2 ๐ฅ = 7 ๐ฅ โ 2 ๐ฆ = โ 7 Through substitution; ๐ฅ โ 2 ๐ฆ = โ 7 ๐ฅ = 2 ๐ฆ โ 7 ( 1 ) Substitute equation 1 to ๐ฆ^2 + 2 ๐ฅ = 7 ๐ฆ^2 + 2 ( 2 ๐ฆ โ 7 ) = 7 ๐ฆ^2 + 4 ๐ฆ โ 14 โ 7 = 0 ๐ฆ^2 + 4 ๐ฆ โ 21 = 0 ๐ฆ = โ 7 & ๐ฆ = 3 Substitute the values of y to the equation ( 1 ) If ๐ฆ = โ 7 ๐ฅ = 2 ๐ฆ โ 7 ๐ฅ = 2 (โ 7 ) โ 7 ๐ฅ = โ 21 If ๐ฆ = 3 ๐ฅ = 2 ๐ฆ โ 7 ๐ฅ = 2 ( 3 ) โ 7 ๐ฅ = โ 1 Then the solutions are (- 21 , 7 ) & (- 1 , 3 )