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Analytic Geometry Opt me In entering the Database of... Analytic Geometry® is a PowerPoint Presentation by The Fantastic Four©. All rights Reserved XD
Table of Contents ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Circle Equations
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Circle Equations Ex. Diagram
Distance Equations Finding distance can be tricky, but this section will show you just how to solve them.
Distance Equations HOW TO SOLVE Example: Line1  0 = 4x - 2y - 8 y = 2x - 4  Line2  4y = 8x + 8 y = 2x + 2  Find the horizontal distance for Line1. Go to ‘x’ axis if    you can determine the distance between the two intercepts. 3 6 Line2 Line1 Find the vertical distance for Line2. Go to ‘y’ intercepts and find the difference  between them. (1, 4) (1, -2) The shortest distance distance formula   |Ax + By + C| A + B 2 2 * Continued on next page
Distance Equations Continued .. When you use the  formula,   use the ‘x’ and ‘y’ of the  perpendicular line. Line1  0 = 4x - 2y - 8 A = 4   B = -2 C = -8  L =  |4(1) + -2(4) + -8| 4 + -2 1 2 2 =  |4 – 8 – 8| 16 + 4 2 2 =  |-12| 20 20 20 =  12  20 20 Line2  0 = 2x - y + 2 A = 2 B = -1 C = 2 L =  |2(1) + -1(-2) +2| 2 + -1 2 2 2 =  |2 + 2 + 2| 5 =  |6| 5 5 5  =  6  5 5
Two Variable Systems In this section, we will show you how to solve two variable systems.
Two Variable Equations * Remember that if there are 2 variables, it requires 2 equations Three ways  to solve .. ,[object Object],[object Object],[object Object],2) Elimination Example: 3x - 2y = 6   -2y + x = 4 Step 1: Place both equations in same form so variables align 3x - 2y = 6  (the ‘x’ and ‘y’ align)    x - 2y = 4 Step 2: Multiply both equations by an appropriate value to eliminate one of the variables 3x - 2y = 6 (the ‘y’ can eliminate)   x - 2y = 4 Continued on next page
Two Variable Equations Continued .. Step 3: Either add or subtract to eliminate 3x - 2y = 6 -  x - 2y =  4 2x  0  = 2 Algebraically solve for ‘x’ 2x = 2   x = 1  Step 4: Substitute the value back into the original equation to solve for  second variable 3x - 2y = 6 3(1) - 2y = 6 3 - 2y = 6 -2y = 3 y = -3/2 * Solution is (1, -3/2)
Three Variable Systems A repeat of two variable systems, but an extra equation is added. A different solution is presented.
3 Variable Systems …  Oh, three variables, three equations…    LOVELY! HOW TO SOLVE; 1) -9x + 5y – 8z = -61 2) -4x + y – 2z = -7 3) 7x – 5y + 11z = 96 Now, to solve this, we must make two more equations ; so using equations 1&2, we can make equation four like so ; -9x + 5y – 8z = -61 5(-4x + y – 2z = -7)    11x  + 2z = -26 <- multiplied the equation by 5 so that at least one of the variables have the same value [in this case ‘’y’’]  so that we can remove one of the variables to find one of the remaining two. Using equations 1&3, we can make equation 5 ; **Note ; When making equations 4 & 5 be sure that you are eliminating the same variable. -9x + 5y – 8z = -61 +7x – 5y + 11z = 96    -2x + 3z = 35 RULE ; 3 Variables, 3 Equations 4) 11x  + 2z = -26 5) -2x + 3z = 35 Now that we have these two, we can now substitute.   3(11x  + 2z = -26)  33x + 6z =-78  37x  = - 148 2(-2x + 3z = 35)  - (-4x + 6z = 70)   37  37  Therefore  X = -4 Next, we can find the other variables by subbing in ‘’x’’, for either equations 4 or 5. 11(-4) + 2z = -26    -44 + 44 + 2z = -26 +44     2z  =  18     Z = 9 ……… . 2  2 So far, we have variables Z and X. Now we can go back to one of our original equations and sub in these values to find the final variable and solve the system. -9(-4) = 5y – 8(9) = -61    36 - 36 + 5y – 72 + 72  = -61 + 72 - 36    5y = -25    y = -5 Thus, we have solved the 3 variable system with a solution of (-4, -5, 9) Not hard when you think about it… just time consuming.
Linear Inequalities In  mathematics , an  inequality  is a statement about the relative size or order of two objects.
4 Types of Inequality ,[object Object],[object Object],[object Object],[object Object],In inequalities, the four main types of inequality you’ll see  are the above. Greater than, less than, greater than or equal to, and lesser than or equal to equations. The difference between them determines what kind of sign diagram you would use for solving the equation, as well as what kind of line you draw. If the equation is either < or > then the line is dashed. If it is  ≤ or ≥ then the line you draw on the graph is solid. linear inequalities
Solving Inequalities ,[object Object],[object Object],[object Object],Now on the Left hand side we have the equation graphed in a y =mx+b format, and from the example equation, we get the line illustrated. As mentioned before, the next step is to test one point anywhere on the graph. Usually this point is the origin (0,0) for the sake of ease. y > x + 2 0 > 2 As you can see, the point ( 0, 0 ) makes the equation false (because 0 is not greater then 2) so the solution to the inequality is the area, on the side of the boundary line, opposite to where ( 0, 0 ) is located. This is seen on the graph to the right. *Note* when putting an equation into y = mx+b format, you may have to divide by -1. When you do this however, you must reverse the symbol ( < becomes >, and > becomes < ) linear inequalities
Intersections of Inequalities ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],The line marked as  red  designates the line as y < x -2, while the line marked at  yellow  designates the line as y > 5/2x -5. Now, following the same procedure as before, we’ll test the point ( 0, 0 ), this time with both equations. Y < x – 2  Y > 5/2x - 5  (0) < (0) – 2  (0) > 5/2 (0) - 5 0 < 2  0 > - 5 Now that we’ve plugged in the coordinates of the origin, we can look for the solution. With the first inequality the point ( 0, 0) makes the statement true (because 0 is indeed, less than 2). So the first part of the solution is on the same side of the line y = x – 2 as the origin. Next we look at the second equation.  Since the origin made the equation true once again we know that the second part of the solution is on the same side of the line y = 5/2x -5 as the origin. Now, since we’re looking for the area covered by both, it’s only where the two  overlap , that we will shade, as the solution. This is shown in the graph on the left. linear inequalities
Things to remember… ,[object Object],[object Object],[object Object],[object Object]
Rational Inequalities
Rational Inequalities We are given this  ;  X – 2  (x -4)(x+6) < 0 How to solve this you ask? Four Simple Steps. FIRST  We factor the Numerator and Denominator SECOND , Determine The  CRITICAL  Numbers THIRD , Place Critical Numbers on a Number Line LASTLY,  Test each Interval **Note ;  CRITICAL  means the restricted values & the values that will make the numerator  zero . Critical Values in this inequality are 2, 4 and -6 With those done, we can test values. *Test any value from each section    Testing -11    we get values of   -     -   = - Area /  Testing 0     -      -   =  +  Area - - + - + - Testing 3     +      +   =  -  Area / Testing 11     +   =  +  Area - + - + + **Note that the  <  in the equation means we’re looking for a negative area & vice versa. If something like   X   >  1   appears, just rearrange it to  X   -  1  > 0 (X-4)  (X+3) (X-4)  (X+3) Follow through Normally     X(X+3) – (X-4)     X ² +3X – X – 4     X ² + 2X – 4   (X-4) (X+3)  (X-4) (X+3)  (X-4) (X+3) Then, Just plot them on a number line and Test the area’s. So our solution is at  (  - ∞  , -3)(4,  ∞  ) So our solution is at  (  - ∞  , -6)(2,   4   )
Absolute Value Inequalities
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Absolute Value Inequalities

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Analytic Geometry Period 1

  • 1. Analytic Geometry Opt me In entering the Database of... Analytic Geometry® is a PowerPoint Presentation by The Fantastic Four©. All rights Reserved XD
  • 2.
  • 4.
  • 5. Distance Equations Finding distance can be tricky, but this section will show you just how to solve them.
  • 6. Distance Equations HOW TO SOLVE Example: Line1 0 = 4x - 2y - 8 y = 2x - 4 Line2 4y = 8x + 8 y = 2x + 2 Find the horizontal distance for Line1. Go to ‘x’ axis if you can determine the distance between the two intercepts. 3 6 Line2 Line1 Find the vertical distance for Line2. Go to ‘y’ intercepts and find the difference between them. (1, 4) (1, -2) The shortest distance distance formula |Ax + By + C| A + B 2 2 * Continued on next page
  • 7. Distance Equations Continued .. When you use the formula, use the ‘x’ and ‘y’ of the perpendicular line. Line1 0 = 4x - 2y - 8 A = 4 B = -2 C = -8 L = |4(1) + -2(4) + -8| 4 + -2 1 2 2 = |4 – 8 – 8| 16 + 4 2 2 = |-12| 20 20 20 = 12 20 20 Line2 0 = 2x - y + 2 A = 2 B = -1 C = 2 L = |2(1) + -1(-2) +2| 2 + -1 2 2 2 = |2 + 2 + 2| 5 = |6| 5 5 5 = 6 5 5
  • 8. Two Variable Systems In this section, we will show you how to solve two variable systems.
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  • 10. Two Variable Equations Continued .. Step 3: Either add or subtract to eliminate 3x - 2y = 6 - x - 2y = 4 2x 0 = 2 Algebraically solve for ‘x’ 2x = 2 x = 1 Step 4: Substitute the value back into the original equation to solve for second variable 3x - 2y = 6 3(1) - 2y = 6 3 - 2y = 6 -2y = 3 y = -3/2 * Solution is (1, -3/2)
  • 11. Three Variable Systems A repeat of two variable systems, but an extra equation is added. A different solution is presented.
  • 12. 3 Variable Systems … Oh, three variables, three equations… LOVELY! HOW TO SOLVE; 1) -9x + 5y – 8z = -61 2) -4x + y – 2z = -7 3) 7x – 5y + 11z = 96 Now, to solve this, we must make two more equations ; so using equations 1&2, we can make equation four like so ; -9x + 5y – 8z = -61 5(-4x + y – 2z = -7)  11x + 2z = -26 <- multiplied the equation by 5 so that at least one of the variables have the same value [in this case ‘’y’’] so that we can remove one of the variables to find one of the remaining two. Using equations 1&3, we can make equation 5 ; **Note ; When making equations 4 & 5 be sure that you are eliminating the same variable. -9x + 5y – 8z = -61 +7x – 5y + 11z = 96  -2x + 3z = 35 RULE ; 3 Variables, 3 Equations 4) 11x + 2z = -26 5) -2x + 3z = 35 Now that we have these two, we can now substitute.  3(11x + 2z = -26) 33x + 6z =-78 37x = - 148 2(-2x + 3z = 35) - (-4x + 6z = 70)   37 37 Therefore X = -4 Next, we can find the other variables by subbing in ‘’x’’, for either equations 4 or 5. 11(-4) + 2z = -26  -44 + 44 + 2z = -26 +44  2z = 18  Z = 9 ……… . 2 2 So far, we have variables Z and X. Now we can go back to one of our original equations and sub in these values to find the final variable and solve the system. -9(-4) = 5y – 8(9) = -61  36 - 36 + 5y – 72 + 72 = -61 + 72 - 36  5y = -25  y = -5 Thus, we have solved the 3 variable system with a solution of (-4, -5, 9) Not hard when you think about it… just time consuming.
  • 13. Linear Inequalities In mathematics , an inequality is a statement about the relative size or order of two objects.
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  • 19. Rational Inequalities We are given this ; X – 2 (x -4)(x+6) < 0 How to solve this you ask? Four Simple Steps. FIRST We factor the Numerator and Denominator SECOND , Determine The CRITICAL Numbers THIRD , Place Critical Numbers on a Number Line LASTLY, Test each Interval **Note ; CRITICAL means the restricted values & the values that will make the numerator zero . Critical Values in this inequality are 2, 4 and -6 With those done, we can test values. *Test any value from each section  Testing -11  we get values of -  - = - Area / Testing 0  -  - = + Area - - + - + - Testing 3  +  + = - Area / Testing 11  + = + Area - + - + + **Note that the < in the equation means we’re looking for a negative area & vice versa. If something like X > 1 appears, just rearrange it to X - 1 > 0 (X-4) (X+3) (X-4) (X+3) Follow through Normally  X(X+3) – (X-4)  X ² +3X – X – 4  X ² + 2X – 4 (X-4) (X+3) (X-4) (X+3) (X-4) (X+3) Then, Just plot them on a number line and Test the area’s. So our solution is at ( - ∞ , -3)(4, ∞ ) So our solution is at ( - ∞ , -6)(2, 4 )
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