Principles of Mathematics 11 Section, Introduction 01 Introduction, Section Analtic Geometr II As the lesson titles show, this section etends what ou have learned about Analtic Geometr to several related topics, such as applications and the graphing of sstems of inequations (inequalities). The last lesson shows how Analtic Geometr has a lot in common with classical geometr and it s Theorems. Section Outline Lesson 1 Lesson Lesson 3 Lesson Lesson 5 Applications of Sstems Non-Linear Sstems Graphing Linear Inequalities Quadratic, Rational, and Absolute Value Inequalities in Two Variables Verifing and Proving Assertions in Coordinate Geometr Review
0 Section, Introduction Principles of Mathematics 11 Notes
Principles of Mathematics 11 Section, Lesson 1 03 Outcome Lesson 1 Applications of Sstems When ou complete this lesson, ou will be able to solve sstems of linear equations and appl them in problemsolving situations Overview We will be using the techniques of solving sstems of equations and appling them to problem-solving situations. Eample 1 An airplane took hours to fl 190 km when it had a tail wind. Fling back against the same wind, and travelling at the same air speed, the plane took 1 hour longer. Find the speed of the wind and the plane s air speed. Solution Let = air speed of the airplane = speed of the wind distance = rate time ( + ) = 190 (1) 5( ) = 190 () Note: Fling with the wind increases the speed and fling against the wind decreases the speed. Equation (1) : + = 80 (3) Equation () 5: = 38 () Equation (1) + Equation (): = 86 = 3 km/hr Substitute = 3 into Equation 3: 3 + = 80 The air speed of the airplane is 3 km/hr. The speed of the wind is 8 km/hr. = 8 km/hr
0 Section, Lesson 1 Principles of Mathematics 11 Eample The sum of three numbers is 18. The third number is five times the sum of the first two numbers. The sum of the third number, three times the first number, and twice the second number is 17. Write the three equations and find the numbers. Solution Let the three numbers be,, and z. + + z = 18 z = 5( + ) z + 3 + = 17 Rearrange equations: + + z = 18 5 + 5 z = 0 3 + + z = 17 (1) () (3) Equation (1) + Equation (): Equation (1) Equation (3): Equation () 6: Equation (5) + Equation (6): 6 + 6 = 18 = 1 + = 3 = = () (5) (6) Substitute = into Equation (6): + = 3 = 7 Substitute =, = 7 into Equation (1): + 7 + z = 18 z = 15 The three numbers are, 7, and 15 or written as an ordered triple (, 7, 15).
Principles of Mathematics 11 Section, Lesson 1 05 Eample 3 For breakfast, Jean had one cup of oatmeal porridge, one cup of skim milk, and one cup of orange juice for a total of 375 calories. John had two cups of oatmeal porridge, two cups of skim milk, and one cup of orange juice for a total of 60 calories. Joan had one cup of oatmeal porridge, three-quarter cup of skim milk, and one-half cup of orange juice for a total of 90 calories. How man calories are in one cup of oatmeal porridge, one cup of skim milk, and one cup of orange juice? Solution Let = number of calories in a cup of oatmeal Jean: John: Joan: = number of calories in a cup of skim milk z = number of calories in a cup of orange juice + + z = 375 + + z = 60 3 1 + + z = 90 (1) () (3) + + z = 375 (1) + + z = 60 () + 3 + z = 1160 (3) Equation (3) Eqn (1) Eqn () Eqn () Eqn (3) Eqn (5) Substitute = 10 in Eqn (): = 65 + + z = 180 = 10 = 10 () (5) 10 = 65 = 15 = 15 Substitute = 10, = 15 in Eqn (1): 15 + 10 + z = 375 z = 110 Ordered triple: (15, 10, 110). = 15 calories in a cup of oatmeal = 10 calories in a cup of skim milk z = 110 calories in a cup of orange juice
06 Section, Lesson 1 Principles of Mathematics 11 Self-Marking Activit 1. The length of each of the congruent sides of an isosceles triangle is one and one-half times the length of the base. The perimeter of the triangle is 60 cm. Find the length of each side of the triangle.. The sum of the digits of a three-digit number is 13. If the tens and hundreds digits are interchanged, the new number is 90 less than the original, and if the units and hundreds digits are interchanged, the resulting number is 99 less than the original. Find the original number. Hint: Let = hundreds digit = tens digit z = units digit Note: Remember a number like 576 means 100 5 + 7 10 + 6 when ou interpret its place value. Accordingl, the number in this question is 100 + 10 + z. 3. Melissa has loonies and $5 and $10 bills in her purse that are worth $96. If she had one more loonie, she would have just as man loonies as $5 and $10 bills combined. If she has 3 bills and loonies in all, how man of each denomination does she have? Hint: You use variables to represent coin-value in much the same wa as ou did for place-values in the previous question.. A cauliflower, one bunch of celer, and one bag of radishes cost $3.67. Two bunches of celer, a cauliflower, and four bags of radishes cost $8.53. Three bunches of celer, two cauliflowers, and one bag of radishes cost $7.0. Find the cost of each tpe of vegetable.
Principles of Mathematics 11 Section, Lesson 1 07 5. The sum of the length, width, and height of a rectangular bo is 75 cm. The length is twice the sum of the width and height, and twice the width eceeds the height b 5 cm. Find the dimensions of the bo. 6. Twent-five coins whose value is $.75 are made up of nickels, dimes, and quarters. If the nickels were dimes, the dimes were quarters, and the quarters were nickels, the total value would be $3.75. How man coins of each tpe are there? 7. In a triangle whose perimeter is 5 cm, the length of the longest side is cm less than the sum of the lengths of the other two sides. The length of the longest side is 3 cm more than twice the length of the shortest side. How long is each side of the triangle? Check our answers in the Answer Ke.
08 Section, Lesson 1 Principles of Mathematics 11 Notes
Principles of Mathematics 11 Section, Lesson 09 Outcomes Lesson Non-Linear Sstems When ou complete this lesson, ou will be able to solve quadratic-linear sstems of equations graphicall and algebraicall solve quadratic-quadratic sstems graphicall and algebraicall Overview In the previous lessons, ou investigated sstems of linear equations in two variables graphicall and algebraicall. In this lesson, ou will investigate sstems containing one linear equation and one quadratic equation (second-degree equation). The graph of a second-degree equation is referred to as a conic. You will stud these in Principles of Mathematics 1 but at present ou should be familiar with the different tpes of graphs that could result. The following diagrams are given as a guide to the various graphs that could result from graphing a quadratic equation. Parabolas a) b) c) d)
10 Section, Lesson Principles of Mathematics 11 Circles Ellipses a) b) Hperbolas a) b) c)
Principles of Mathematics 11 Section, Lesson 11 A quadratic-linear sstem of equations includes an equation of one of the conics and a linear equation. The solution of two equations in two variables still corresponds to a point of intersection of their graphs. Draw several different sketches showing intersections of a line and a parabola that result in different numbers of intersection points. How man points of intersection are possible for a line and a parabola? Repeat this process for a line and a circle, a line and an ellipse, and a line and a hperbola. Eample 1 Identif the number of solutions in each sstem. a) b) c) 0 0 0 Solution a) solutions because there are two points of intersection b) 1 solution because there is one point of intersection c) no solution because the graphs do not intersect A quadratic-linear sstem has 0, 1, or real number solutions. As in the case of sstems of linear equations, ou can find approimate solutions b graphing or eact solutions algebraicall.
1 Section, Lesson Principles of Mathematics 11 Eample Solve the following sstem of equations b graphing: = + + = 5 Solution The first equation represents a line with a slope of and a -intercept of. The graph of the second equation is a circle with centre (0, 0) and radius 5. The intersection points appear to be near 5 0 5 1 1, 5 and (3, ). Eample 3 Solve the following sstem of equations algebraicall. = 6 + 9 5 = + Solution Solve the linear equation for. + = 5 = 5 Substitute the epression for in the first equation. = 6 + 9 5 = 6 + 9 0 = 5 + 0 = ( ) ( 1) = or = 1 To find the corresponding -values, substitute the -values into the linear equation and solve for.
Principles of Mathematics 11 Section, Lesson 13 = + = 5 + = 5 = 1 = 1 + = 5 1+ = 5 = The solutions are (, 1) and (1, ), meaning there are two points of intersection on the graphs. Check our answers algebraicall and graphicall. A sstem of two or more quadratic relations or conics is a quadratic-quadratic sstem. A quadratic-quadratic sstem ma have 0, 1,, 3, or real number solutions as illustrated in the following diagrams. No Solution 1 Solution Solutions 3 Solutions Solutions The ma be solved graphicall or algebraicall remembering that the graphs might be approimate solutions but the algebraic solutions are eact.
1 Section, Lesson Principles of Mathematics 11 Eample Solve b (a) graphing, and (b) algebraicall. Solution = 5 + 6 = 0 You can use a table of values or a graphing calculator to get the accompaning graph. 3 5 5 3 =± 5 ± 11 ± 0 0 ± ± 11 ±3. ±3. 6 3 1.5 1 0 1 1.5 3 6 1 3 6 undefined 6 3 1 = 6 0 a) The points of intersection are at ( 3, ) and (3, ).
Principles of Mathematics 11 Section, Lesson 15 b) Algebraicall: = 5 + 6 = 0 Solve the second equation for : Substitute = 6 = 6 = 5 Solve. 5 36 = 0 ( 9) ( + ) = 0 6 for in the first equation. 6 = 5 36 = 5 Multipl b and rearrange. Because + has no real roots, then 9 = 0 ( 3) ( + 3) = 0 = + 3 or 3 Substitute the -values into = 6. If = 3 = 6 3 = If = 3 = 6 3 = Solution is at (3 ) and ( 3, ). Notice that this corresponds to the graph.
16 Section, Lesson Principles of Mathematics 11 Eample 5 Solve: + = 9 3 = 1 This eample can be done algebraicall b substitution or b addition-subtraction. Solution 1: Substitution Solve the first equation for : = 9 Substitute 9 for in the second equation and solve. ( 9 ) 3 = 1 98 8 3 = 1 11 = 110 = 10 = 10 or 10 Substitute -values. 10 or 10 for in the first equation to find the ( + 10 = 9 ) + 0 = 9 = 9 = ± 3 The solution is made up of ( ) ( ) ( ) ( ) 3, 10, 3, 10, 3, 10, 3, 10.
Principles of Mathematics 11 Section, Lesson 17 Solution : Addition Subtraction Multipl first equation + = 9 b and subtract 3 = 1 + 8 = 98 3 = 1 11 = 110 = 10 = ± 10 The net steps are a repeat of Solution 1. This sstem is graphed below. 0
18 Section, Lesson Principles of Mathematics 11 Eample 6 Solve: = = Solution Graph = = Algebraic solution: = (1) = () Solve Eqn () for in terms of and substitute that epression into Eqn (1). = = = ( ) = 16 + 16 = 0 This quadratic equation does not factor so use the discriminant to determine whether it has solutions. a = 1 b = c = 16 b ac = ( ) ( 1)( 16) = 8 Because b ac < 0, there is no solution. This means the graphs of the two equations do not intersect.
Principles of Mathematics 11 Section, Lesson 19 Self-Marking Activit 1. Solve each sstem both graphicall and algebraicall. a) = b) ( ) + ( 3) = c) = 1 = 8 + + 3 = 0 + = d) + = 18 e) = + = 16 = 10. The line = intersects the parabola = at the origin and the point Q. Let P be the verte of the parabola. a) Find the coordinates of Q. b) Find the coordinates of P. c) Find the length of OQ. d) Find the distance from P to OQ. e) Find the area of OQP. P Q O 3. Solve algebraicall. a) b) + = 18 = c) 3 = + = 0 + + = 9 = 0 d) + = e) + = 9 f) + = 16 = + = 3 + = 0
0 Section, Lesson Principles of Mathematics 11. Find the dimensions of a rectangle if its diagonal is 17 cm and the perimeter is 6 cm. 5. The perimeter of a rectangle is 6 cm. Its area is 1 cm. Find the dimensions of the rectangle. 6. The sum of two numbers is 7. Their product is 16. What are the two numbers? 7. The sum of two numbers is 3. The sum of the squares of these numbers is 17. What are the two numbers? 8. The sum of the squares of two positive numbers is 65. The difference of their squares is 33. What are the two numbers? 9. The area of a rectangle is 60 cm. The measure of each of its diagonals is 13 cm. Determine the length and width of the rectangle. Check our answers in the Answer Ke.
Principles of Mathematics 11 Section, Lesson 3 1 Outcomes Lesson 3 Graphing Linear Inequalities In Two Variables When ou complete this lesson, ou will be able to graph the solution of a linear inequalit in two variables find the solution to a sstem of inequalities Overview You have learned several was of solving sstems of linear equations. You will now investigate the idea of a linear inequalit because much of the mathematics that ou use in dail life does not come in the form of an equation. Statements such as, I want to find a job that pas more than the minimum wage. or I have to spend less than $10 this week. are eamples of statements of inequalities. The graph of a line separates the plane into three distinct regions: two half planes and the line itself. The line itself is the boundar line of each half plane. It divides the coordinate plane into two half planes. Half- Half- Plane Boundar Line Plane A linear inequalit has a boundar line that can be epressed in the form = m + b. The solution to a linear inequalit is the set of all ordered pairs that make the inequalit true. When the inequalit is (read less than or equal to ) or (read greater than or equal to ), the solution includes the points on the boundar line and the graph has a solid boundar line.
Section, Lesson 3 Principles of Mathematics 11 When the inequalit is < or > the solution does not include the points on the boundar line and the graph will have a dashed boundar line. Eample 1 Graph + <. Solution The boundar line is + =. You can graph the line b transferring it into = +. You can use the slope intercept form, with b = and the slope. Use a dashed line to show that the points on the line are not solutions to + <.
Principles of Mathematics 11 Section, Lesson 3 3 8 8 8 8 To determine which half plane is the solution, select a point from each half plane and see which ordered pair makes the inequalit true. A good test point is the origin (0, 0) as long as the boundar line does not go through that point. (0, 0) on the left side (, 0) on the right side < + < + 0 < (0) + 0 < () + 0 < True 0 < False Because the test point (0, 0) makes the inequalit true, shade the half plane in which this point is located. 8 8 8 8 Note: Some graphing calculators, like the TI-83Plus, can graph inequalities and shade one side for ou. You ma use that feature if ou have it on assignments and tests in this course, but show the working for a test point if the question asks for it.
Section, Lesson 3 Principles of Mathematics 11 Etensions of this would be: a) + becomes + The boundar line would be solid and shaded below the line. 8 8 8 8 b) + > becomes > + The boundar line would be dashed to show that the points on the line are not solutions of > +. Use the same test points as in the initial questions. (0, 0) on the left side (, 0) on the right side > + > + 0 > (0) + 0 > () + 0 > False 0 > True The coordinates of (, 0) make the inequalit true. 8 8 8 8
Principles of Mathematics 11 Section, Lesson 3 5 c) + The boundar line is solid and the shading is then the same as in b). 8 8 8 8 A linear inequalit in and is an inequalit that can be written in one of the following forms: a + b < c, a + b c, a + b > c, or a + b c. An ordered pair (, ) is a solution of a linear inequalit in and if the inequalit is true when a and b are substituted for and, respectivel. For instance (, 0) is a solution to > + because 0 is greater than () + =. To sketch the graph of a linear inequalit in two variables, first sketch the line given b the corresponding equation. Use a dashed line for inequalities with < or >. This line separates the coordinate plane into two half planes. In one of the half planes are points that are solutions of the inequalit. In the other half plane no point is a solution. You can decide whether the points in an entire half plane satisf the inequalit b testing one point in the half plane.
6 Section, Lesson 3 Principles of Mathematics 11 Eample Sketch the graph of the inequalit 3. 3 Eample 3 Graph the solid horizontal line = 3. Net, test a point (0, 0). 3 0 3 False Sketch the graph of the inequalit <. Solution 1 3 1 Sketch the dashed vertical line <. 1 3 The origin is not a solution and it is below the line. Therefore, the graph is all points on or above the line = 3. 3 1 3 1 1 3 1 3 Net, choose a test point (0, 0). < 0 < True This is a solution which represents all points to the left of. Two or more linear inequalities on the same coordinate plane make up a sstem of linear inequalities. The following process ma help ou sketch the graph of a sstem of linear inequalities. 1. Sketch the line that corresponds to each inequalit. It is useful to put the line in the form = m + b. Use a dashed line for inequalities with < or > and a solid line for inequalities with or.
Principles of Mathematics 11 Section, Lesson 3 7. Lightl shade the half plane that is the graph of each linear inequalit. Using a different-coloured pencil or different shading to shade each inequalit ma help ou distinguish the different half planes. 3. The graph of the sstem is the intersection of each of the half planes. If ou used coloured pencils or different shading it is the region that has been shaded with each colour. Eample Given: + Solution < 1 Graph +, or, showing the boundar as a solid line. Note: < 1 < + 1 > 1 when ou divide b ( 1) the sign of the inequalit changes. In the same coordinate plane, graph < 1, or > 1, showing the boundar as a dashed line. The graph of the solution set of the sstem is the doubl shaded region above and on the graph of + = and also above the graph of = 1. = 1 + = + and < The word and means the intersection of the two sets of points. The solution is the dark shaded region. Algebraicall it can be written as {( ) } ( ) { }, +, < 1
8 Section, Lesson 3 Principles of Mathematics 11 + or 1 The word or means the union of the two sets of points. The solution is all the shaded area in the graph. Algebraicall it can be written as {( ) } ( ) { }, +, < 1 Self-Marking Activit 1. Sketch the graph of the sstem of linear inequalities. 1 + 0 0. Match the sstem of inequalities with its graph. a) b) > 1 > 3 + 1 c) < d) > 3 0 3 < 1 + 1 i) ii) 1 3 1 1 1 6 3 iii) iv)
Principles of Mathematics 11 Section, Lesson 3 9 3. Write a sstem of linear inequalities for the region shown below. 8 6 l B (9, 3) l A (0, 3) l D (, 1) l 6 C (5, 8 1). Sketch the following sstems defined b a) > b) 5 < 1 c) + >3 + < d) + < 5 5 3 5. Graph the sstem + > 6 and + 3 < 5. 6. Graph the sstem < 5 or + > 3. 7. Sketch the region defined b 5 6 ³ 6, 3 +, and ³ 3. 8. a) Draw the region defined b ³ 0, ³ 0, 3 +, and > 1. b) Name the coordinates of the vertices in part (a). 9. Find the area of the region determined b the graph of the solutions of, 6, and ³ 0. Check our answers in the Answer Ke.
30 Section, Lesson 3 Principles of Mathematics 11 Notes
Principles of Mathematics 11 Section, Lesson 31 Lesson Quadratic, Rational, and Absolute Value Inequalities Outcomes When ou complete this lesson, ou will be able to sketch the graph of a quadratic inequalit in two variables solve a quadratic inequalit in one variable solve a rational inequalit in one variable solve an absolute value inequalit in one variable Overview The graphs of linear and quadratic inequalities in two variables are closel related to the graphs of lines and parabolas. If a line has the equation = m + b, an point (, ) above the line satisfies the inequalit > m + b. An point (, ) below the line = m + b satisfies the inequalit < m + b. > m + b = m + b < m + b If the quadratic has the equation = a + b + c, an point (, ) above the parabola satisfies the inequalit > a + b + c. An point (, ) below the parabola = a + b + c satisfies the inequalit < a + b + c. > a + b + c = a + b + c < a + b + c
3 Section, Lesson Principles of Mathematics 11 In this lesson, ou will investigate four tpes of quadratic inequalities: < a + b + c a + b + c > a + b + c a + b + c The steps used to sketch the graph of a quadratic inequalit are as follows: 1. Sketch the graph of the parabola = a + b + c. Use a dashed parabola for inequalities with < or > and a solid parabola for inequalities with or.. Test a point inside and one outside the parabola. 3. Onl one of the test points will be a solution. Shade the region that contains the test point. Eample 1 Sketch the graph of 8. Solution b The parabola has its verte at = ( = ) = 1. a ( 1) Substitute = 1 for = 8 = 1 ( 1) 8 = 1 8 = 9 Verte: (1, 9) Zeros: 8 = 0 ( )( + ) = 0 = or = Draw the sketch using a solid line for the parabola. 8 = 8
Principles of Mathematics 11 Section, Lesson 33 Choose one point inside region and one outside region and check. (0, 0) (5, 0) 8 8 0 0 (0) 8 0 5 (5) 8 0 8 true 0 7 false The region containing (0, 0) must be shaded as the graph of the inequalit. 8 = 8 At this stage ou have graphed quadratic inequalities in two variables. Because there were two variables, the solution set was represented graphicall b a two-dimensional area which ou shaded. To solve a quadratic inequalit which has onl one variable, ou can use a number line or a sign graph of the function. The steps are as follows: 1. Determine the zeros of the function and plot them on a number line.. Use open circles for inequalities with < or >. 3. Use solid circles for inequalities with or.. Test one value of from each one of the intervals determined b the zeros to determine the sign of each factor. 5. Determine the intervals for which the values are true.
3 Section, Lesson Principles of Mathematics 11 Eample Solve 8 0 using a sign graph. Solution Find and plot the zeros: 8 = 0 ( )( + ) = 0 The zeros are and. 0 Determine the signs b testing one value of in each interval. Interval: or (, ]. Test = 3. f ( ) = ( + )( ) f ( 3) = ( 3 + )( 3 ) ( ) ( ) = + signs of product signs of factors + 0 Interval: or [, ]. Test = 0. f ( ) = ( + )( ) f( 0) = ( 0 + )( 0 ) ( + ) ( ) = + 0 signs of factors signs of product Interval: or [, ). Test = 5. f ( ) = ( + )( ) f( 5) = ( 5 + ) ( 5 ) ( + ) ( + ) = + signs of factors signs of product + + 0 The quadratic ( + )( ) is positive or zero in the interval (, ] and in the interval [, ). Therefore, the solution is all real numbers in the interval (, ] or [, ). This can also be written as or.
Principles of Mathematics 11 Section, Lesson 35 When f() > 0 the graph is above the -ais. When f() < 0 the graph is below the -ais. The above method uses the fact that a polnomial can change sign onl at its zeros (that is, the -values that make the polnomial zero). Between consecutive zeros a polnomial must be entirel positive or negative. When the real zeros of a polnomial are put in order, the divide the number line into intervals in which the polnomial has no sign changes. These zeros are referred to as the critical numbers of the inequalit and the resulting intervals are the test intervals for the inequalit. Not all polnomials change sign at a zero. Eample 3 Solve ( 1) 0. Solution Find zeros and plot. Critical number: 1 = 0, = 1 + + 0 1 Determine signs of factors in the interval. Interval: [1, ), = ( 1)( 1) 0 ( + ) ( + ) = + signs of factors Interval: [,1], = 0 ( 1)( 1) 0 ( ) ( ) = + signs of factors sign of product sign of product The signs are positive on each side of the zero. A polnomial will not change sign at a zero if a corresponds to the squared factor ( a).
36 Section, Lesson Principles of Mathematics 11 ( 1) 0 is alwas true, but if the inequalit is given as ( 1) > 0 it is true for all values ecept = 1. The concept of critical numbers and test intervals can be etended to inequalities involving rational epressions. This uses the fact that the value of a rational epression can change sign onl at its zeros (-value for which its numerator is zero) and its undefined values (-values at which the denominator is zero). These are the two tpes of numbers that make up the critical numbers of a rational inequalit. Eample Solve: 1 < 0 ( )( + 3) Solution Determine critical numbers and plot. The critical numbers are 1 = 0, = 1 (the numerator is zero when = 1) and ( )( + 3) = 0, = and 3 (the denominator is zero when = and 3. Critical numbers are 3, 1,. Open dots are used because f() is < 0. Interval (, 3) Test point: ( + 3) ( 1) ( ) ( + 3) ( 1) ( ) ( ) ( ) ( ) = signs of factors sign of product 3 + + 1 0 1 3 Note: In the original equation, two of these factors were in the denominator. But the sign is what we want, and it is the same if factors are all multiplied together or if some are divided. So we simpl multipl them that s easier to do and take the sign of the product. Interval ( 3, 1) Test point: 0 ( + 3) ( 1) ( ) ( 0+ 3) ( 0 1) ( 0 ) ( + ) ( ) ( ) = + signs of factors sign of product
Principles of Mathematics 11 Section, Lesson 37 Interval (1, ) Test point: = 3 ( + 3) ( 1) ( ) 3 + 3 3 1 3 ( + ) ( + ) ( ) = signs of factors sign of product Interval (, ) Test point: = 3 ( + 3) ( 1) ( ) ( 3+ 3) ( 3 1) ( 3 ) ( + ) ( + ) ( + ) = + signs of factors sign of product The solution is the intervals for which the sign of the quotient is < 0 (negative), (, 3), or (1, ). Eample 5 Solve: ( 1)( + 3) 0 Solution Zeros are at + 1 = 0, 3 = 0, and = 0. = 1, 3, Critical numbers are 1, 3, Because f() can equal 0, use solid dots for 1 and 3 and an open dot for as the denominator can never be zero. Interval (, 1]. Test point: =. ( + 1) ( 3) ( 3) ( ) ( ) ( ) ( ) ( ) = + signs of factors sign of product + + 3 1 0 1 3 5
38 Section, Lesson Principles of Mathematics 11 Interval [ 1, 3]. Test point: = 0. ( + 1) ( 3) ( 3) ( ) ( + ) ( ) ( ) ( ) = signs of factors sign of product Interval [3, ) Test point: 7 ( + 1) ( 3) ( 3) ( ) ( + ) ( + ) ( + ) ( ) = Interval (, ) Test point: = 5 ( + 1) ( 3) ( 3) ( ) ( + ) ( + ) ( + ) ( + ) = + The solution interval is the intervals where the quotient is negative or zero [ 1, ) or, 1 <. Eample 6 Solve: Solution signs of factors signs of factors > 1 3 + 1 + > 0 3 ( + ) ( 3) > 0 ( 3) ( + ) + + 3 ( 3) ( + ) > 0 + + 3 ( 3) ( + ) > 0 sign of product sign of product For + + 3, the discriminant b ac = 1 (1)(3) = 11 which means that the numerator has no real zeros. 3 = 0 and + = 0 = +3 and = are the critical numbers
Principles of Mathematics 11 Section, Lesson 39 Interval (, ) Test point: = ( ) + + + 3 ( ) ( 3) ( + ) ( ) ( ) = + signs of factors sign of product + + 0 3 Interval (, 3) Test point: = 0 ( ) + + + 3 ( ) ( 3) ( + ) ( + ) ( ) = signs of factors sign of product Interval (3, ) Test point: = ( ) + + + 3 ( ) ( 3) ( + ) ( + ) ( + ) = + signs of factors sign of product The solution intervals where the quotient is > 0 is (, ) or (3, ). The net inequalit to investigate is the absolute value inequalit. The absolute value of a number,, denoted b can be interpreted geometricall as the distance from to zero in either direction on the number line. Eamine the following chart involving this interpretation. Statement Geometric Interpretation Graph Solution = 3 Values of that lie 3 units from 0 3 3 3 0 3 = 3 and = 3 < 3 Values of that lie less than 3 units from 0 3 3 3 1 0 1 3 3 < < 3* ( 3, 3) > 3 Values of that lie more than 3 units from 0 3 3 3 0 3 < 3 or > 3 (, 3) (3, ) * Remember 3 < < 3 means 3 < and < 3.
0 Section, Lesson Principles of Mathematics 11 The following represents the meaning and the solution of absolute value equations and inequalities involving and c, where c represents a distance with c 0. Statement Geometric Interpretation Graph Solution = c The distance from to 0 is eactl c units. c 0 c = c or = c < c The distance from to 0 is less than c units. c 0 c c < < c or ( c, c) > c The distance from to 0 is more than c units. c 0 c < c or > c (, c) (c, ) These statements are valid if is used instead of < and replaces >. Equations and inequalities involving a, where a is a constant, can be solved b interpreting a as the distance from to a specific point a on the number line. The net two tables eplain a. The first table uses a specific value of 1 for a, and the second table eplains the universal case for an value of a. Statement Geometric Interpretation Graph Solution 1 = 3 The distance from to 1 is 3. 3 3 1 = or = 1 < 3 The distance from to 1 is less than 3. 3 3 1 < < 1 > 3 The distance from to 1 is greater than 3. 1 < or >
Principles of Mathematics 11 Section, Lesson 1 Statement Geometric Interpretation Graph Solution a = c The distance from to a is c. = a c or = a + c a < c The distance from to a is less than c. c < a < c ie. a c < < a < c a > c The distance from to a is greater than c. a > c or a < c ie. > a + c or < a c Eample 7 Solve 1 > 7 (a) geometricall and (b) algebraicall. Solution a) Geometricall 1 > 7 means the distance from to 1 is greater than 7 units 7 7 6 1 8 b) Algebraicall Translated it means that 1 < 7 or 1 > 7. i.e., < 6 or > 8.
Section, Lesson Principles of Mathematics 11 Eample 8 Solve + 1 7 (a) geometricall and (b) algebraicall. Solution a) Geometricall Make use of the fact that ab = a b. +1 7 + 1 7 + 1 7 + 1 7 + 1 7 1 7 Remember + 1 = 1 This means that the distance from to 1 is less than or equal to 7. Thus, 3 or [, 3]. b) Algebraicall + 1 7 This means 7 + 1 7 8 6 3 Rewritten Subtract 1 from each part Divide b
Principles of Mathematics 11 Section, Lesson 3 Self-Marking Activit 1. Determine whether the ordered pair is a solution of the inequalit. a) > + + (1, 6) b) + 1 ( 1, 5). Match the inequalit with its graph. a) > 6 b) + 3 5 c) + 3 + 5 d) > 6 i) ii) 6 6 6 iii) iv) 6 6 6 8 3. Sketch the graph of the inequalities: a) b) c) + 1 > + 3 + < 3 6
Section, Lesson Principles of Mathematics 11. Find the solution set for the following. Epress our answer in set notation and interval notation. a) b) c) d) + 3 0 + 3 5 > 0 0 < 0 + 3 18 5. Solve the following inequalities. Check our solutions. a) + 3 d) + < + 5 + 1 0 1 + 1 1 b) + 1 e) 0 1 0 5 + c) 0 f) 6. Solve the following inequalities geometricall and algebraicall. a) 5 8 d) + 8 > b) + 1 5 e) 3 + > 10 c) b g 3 > 9 5 3 0 3 + 5 Check our answers in the Answer Ke.
Principles of Mathematics 11 Section, Lesson 5 5 Lesson 5 Verifing and Proving Assertions in Coordinate Geometr Outcome When ou complete this lesson, ou will be able to use the properties of slope, midpoint and distance to verif and prove assertions in coordinate geometr Overview We will be using the slope, midpoint, distance formulas to verif or prove assertions in coordinate geometr, pertaining to triangles, quadrilaterals, and circles. You ma find the following formulas useful in doing our proofs or verifications. Slope ( m)= 1 1 Midpoint = 1 +, 1 + Distance ( d)= ( 1 ) + ( 1 ) Distance ( d)= A + B 1 + C A + B If two lines are parallel their slopes are equal. If l 1 l then m 1 m = -1 Also included is a review of the properties of quadrilaterals. Parallelograms Opposite sides are parallel. > > > Opposite sides are equal. >
6 Section, Lesson 5 Principles of Mathematics 11 Opposite angles are equal. Adjacent angles are supplementar. + = 180 Diagonals bisect each other. Rectangle All angles are right angles. Diagonals are equal. A B C D AD = BC
Principles of Mathematics 11 Section, Lesson 5 7 Rhombus All sides are equal; opposite sides are parallel. Diagonals are perpendicular to each other and bisect each other. Square (rectangle and rhombus) All angles are right angles. All sides are equal. Diagonals are equal and perpendicular to each other. A B AD = BC C D
8 Section, Lesson 5 Principles of Mathematics 11 Eample 1 Show that A(, 1), B(1, 3), C(6, 5), and D(7, 1) are the vertices of a parallelogram. Solution There are several was to solve this problem. 1. You could show that the opposite sides have the same slope and, therefore, are parallel. Using the slope formula: B(1, 3) A(, 1) C(6, 5) D(7, 1) m = 1 1 3 mab = ( 1 ) = 1 3 mbc = 5 = 1 6 5 5 mcd = 1 6 7 = m 1 1 = 7 = DA 5 Because mab = mcd = and mbc = mda =, opposite 5 sides are parallel and the given ordered pairs are vertices of a parallelogram.. You could show that the opposite sides are equal in length b determining their lengths. d = ( 1 ) + ( 1 ) AB = BC = CD = ( 1) + ( 1 3) = 1+16 = 17 ( 1 6) + ( 3 5) = 5 + = 9 ( 7 6) + ( 1 5) = 1+16 = 17 AD = ( 7 ) + 1 ( 1) ( ) = 5+ = 9 Because AB = CD = 17 and BC = AD = 9, the figure is a parallelogram.
Principles of Mathematics 11 Section, Lesson 5 9 Eample Decide whether ABCD is a rectangle, rhombus, or a square given these vertices: A(3, 9), B(1, ), C( 9, 7), and D( 7, ). Solution In order to be a rectangle or square, the figure must have four right angles. A(3, 9) D( 7, ) B(1, ) C( 9, 7) If DC CB, then C would be 90. DC CB if m DC m BC = 1. When l 1 l, then m 1 m = 1. m m DC BC = ( 7 ) ( ) = 11 = 11 7 9 = 7 ( ) = 5 9 1 10 = 5 10 11 Because mdc mbc = C is not 5 1, 10 90. quadrilateral ABCD is not a rectangle or a square
50 Section, Lesson 5 Principles of Mathematics 11 If it is a rhombus, then all four sides must be congruent. d = ( ) ( ) 1 1 ( ) = + = AB = ( 3 1) + 9 ( ) 11 15 = 5 5 ( ) + ( ) = + = = BC = 1 ( 9) ( 7) 100 5 15 5 5 ( ) + CD = 9 ( 7) ( 7 ) = + 11 = 15 = 5 5 AD = ( 7 3) + ( 9) = 100 + 5 = 15 = 5 5 Because all the lengths are equal, the figure must be a rhombus. The following proof involves using coordinate geometr to prove a theorem in plane geometr. The connection between these two geometries is a ver important one in mathematics. Eample 3 Prove that the segment determined b the midpoint of two sides of a triangle is: parallel to the third side half the length of the third side Solution A(a, b) E(a, b) D(a + 1, b) B (0, 0) C (, 0) Let B be at the origin and BC lie along the -ais so that C has coordinates (, 0). Arbitraril choose A to have coordinates (a, b).
Principles of Mathematics 11 Section, Lesson 5 51 Now, find the midpoints of AB and AC at E and D respectivel. Midpoint of AB Midpoint of AC E D 1+ 1+ 1 + 1 +,, 0+ a + b + a + b E D, 0, 0 E( ab, ) D( 1+ ab, ) Determine the slope of DE and BC m m DE BC b b 0 = = = 0 1 + a a 1 0 0 0 = = = 0 0 Because m DE = m BC, then DE BC. Length of DE = ( a + 1 a) + ( b b) = 1= 1 Length of BC = ( 0) + ( 0 0) = = 1 DE BC The line segment determined b the midpoints of two sides of a triangle is parallel to the third side and half its length. Note: The proof could also be done b choosing A to be (a,b) and C to be (c,0).
5 Section, Lesson 5 Principles of Mathematics 11 Self-Marking Activit 1. Three vertices of a rectangle ABCD are A( 9, 0), B(5, ), and C(7, 3). a) Find the coordinates of the th verte of the rectangle. b) Find the perimeter of the rectangle. c) Find the area of the rectangle.. A triangle has vertices at A(, ), B(, 8), and C(, 6). Is this a right triangle? Verif our answer. 3. Show that the quadrilateral with vertices A( 5, ), B(1, 1), C(, ), and D(, 3) is a parallelogram.. Line l 1 contains the points (, 3) and (, 1). Line l 1 line l. Line l contains the points (5, ) and (1, ). Find the value of. Eplain our rationale. 5. Line l 3 contains the points (r, 3) and (, 1). Line l 3 is parallel to line l which contains the points (5, ) and (1, ). Find the value of r. Describe the procedure used. 6. Prove that the quadrilateral defined b the following lines is a square: l 1 : = 3 6, l : + 3 = 33, l 3 : = 3 + 19, and l : + 3 8 = 0. 7. A triangle is given b the vertices P( 5, ), Q(1, 8), and R( 1, ). Is the median from P to RQ perpendicular to RQ? Eplain our answer. 8. A quadrilateral PQRS has vertices at P(5, 6), Q(3, 0), R( 1, ), and S( 5, ). Verif that the midpoints of the sides of this quadrilateral form the vertices of a parallelogram. Check our answers in the Answer Ke. Review Section before attempting the review questions beginning on the net page. These questions should help ou consolidate our knowledge as ou prepare for the Section Assignment 3..
Principles of Mathematics 11 Section, Review 53 Review 1. Solve each sstem for and. a) + = 10 + = b) + = 5 = 7 c) 3 + = 3 3 + 3 = 1 d) + 3 = 6 3+ = 5. Graph the following sstems of inequalities. State the solution algebraicall. a) > and + < 6 b) + 6 or 3. Graph: < 0.. Solve the following inequalities: a) b) c) d) + 3 0 + 1 0 + > 7 3 + 1 8 5. A triangle is given b the vertices P( 5, ), Q(1, 8), and R( 1, ). Is the median from P to QR perpendicular to QR? Eplain our answer. Check our answers in the Answer Ke. Now do the Section Assignment which follow this section.
5 Section, Review Principles of Mathematics 11 Notes
Principles of Mathematics 11 Section Assignment 3. 55 PRINCIPLES OF MATHEMATICS 11 Section Assignment 3.
56 Section Assignment 3. Principles of Mathematics 11
Principles of Mathematics 11 Section Assignment 3. 57 Total Value: 0 marks (Mark values in brackets) Section Assignment 3. Analtical Geometr II () Scientific and/or graphing calculator is permitted. 1. Find the points where the line = + 3 intersects the parabola = + 3. (). Graph the inequalities. a) 3
58 Section Assignment 3. Principles of Mathematics 11 b) + < 8 (3) 3. Graph the sstems of inequalities. Clearl mark the solution set. a) + > < ()
Principles of Mathematics 11 Section Assignment 3. 59 () b) + 3 6 1 1 (3). Graph the following inequalit. Clearl mark the solution set. > + + 3
60 Section Assignment 3. Principles of Mathematics 11 5. Solve the sstem algebraicall. = 11 + = 16 () 6. Give the solution set for the inequalit, geometricall and algebraicall. () + 6 <
Principles of Mathematics 11 Section Assignment 3. 61 (3) 7. State the solution set for each of the following inequalities. a) 3 + > () b) ( + 1) ( + )( 3) < 0
6 Section Assignment 3. Principles of Mathematics 11 8. Give the solution set for the following. Epress our answer in set notation and interval notation. (3) 3 10 0 9. Given right triangle B (0,b) D C A (a,0) ABC with D the midpoint of BA.. Show that D is equidistant from B and C. () (Total: 0)