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This took me a very long time to finish this project when I was a 2nd yr. student. Like duh, I'm the only one doing this xD The last part of the presentation was downloaded from the internet because …

This took me a very long time to finish this project when I was a 2nd yr. student. Like duh, I'm the only one doing this xD The last part of the presentation was downloaded from the internet because I'm too lazy. Hope this will help the other students and teachers !

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- 1. In a RECTANGLE1. Opposite sides are congruent2. Opposite sides are parallel3. Each diagonal separates the rectangle into two congruent triangles4. Opposite angles are congruent
- 2. 5. Consecutive angles are supplementary6. All angles are right angles.7. Diagonals bisect each other and are congruent.
- 3. In a RHOMBUS1. All the sides are congruent2. Opposite sides are parallel3. Each diagonals bisect the rhombus into two congruent triangles4. Opposite angles are congruent
- 4. 5. Consecutive angles are congruent6. Diagonals bisect each other and are perpendicular
- 5. In a SQUARE1. All the sides are congruent2. All angles are right angles3. Each diagonals separate the square into two congruent triangles4. Opposite angles are congruent and are supplementary
- 6. 5. Consecutive angles are congruent6. Diagonals bisect each other and are perpendicular and are congruent
- 7. PROPERTIES OF PARALLELOGRAM1.Opposite angles are congruent2. Two non-opposite angles of a parallelogram are supplementary3. Opposite sides are congruent
- 8. 4. A diagonal of a parallel divides the parallelogram into congruent angles5. Diagonals bisect each other
- 9. LL = Legs Legs
- 10. HyL = Hypotenuse Legs
- 11. HyA = Hypotenuse Acute
- 12. LA = Legs Acute
- 13. RATIO• Comparisonof twoquantities.
- 14. WAYS OF WRITING• BY COLON 5:6 or a:b• BY FRACTION 5/6 or a/b• BY DIVISION 5 ÷ 6 or a ÷ b
- 15. PROPORTION• An equation in which 2 quantities are set equal a:b = c:d
- 16. aMEANS b EXTREMES c d
- 17. PROPORTIONAL SEGMENTS &BASIC SIMILARITY THEOREM
- 18. ILLUSTRATION Pt. X divides segment AB so that AX to A 12 X 8 B XB is 3 : 2. Pt. Y divides C Y D 6 4 segment CD so that CY to YD is 3 : 2
- 19. THEOREM: PROPORTIONAL SEGMENTS “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A 12 X 8 BSOME PROPORTIONS1. AX CY XB YD C 6 Y 4 D 12 6 8 4 3 3 2 2
- 20. THEOREM: PROPORTIONAL SEGMENTS “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A 12 X 8 BSOME PROPORTIONS2. AB CD C 6 Y 4 D XB YD 20 10 8 4 5 5 2 2
- 21. THEOREM: PROPORTIONAL SEGMENTS “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A 12 X 8 BSOME PROPORTIONS3. AB CD C 6 Y 4 D AX CY 20 10 12 6 5 5 3 3
- 22. THEOREM: PROPORTIONAL SEGMENTS “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A 12 X 8 BSOME PROPORTIONS4. AX +AB CY +CD C 6 Y 4 D AX CY 32 16 12 6 8 8 3 3
- 23. Illustrative Examples
- 24. Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then, A 8 B 12 C1. AB MN M 2 N 3 P BC NP2. AB BC MN NP
- 25. Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then, A 8 B 12 C3. AB MN M 2 N 3 P AC MP4. BC NP AC MP
- 26. Find the unknown parts assuming the segments are divided proportionally.Solution:X : 32 = 6 : 8 X 32Applying the law of proportion 6 88(x) = 6( 32)8x = 192X = 24
- 27. Solution: Let x = the length of the other portion of the second string.x 1560 18x 5 x 6060 6 6x 5(60) 15 18 6x = 300 X = 50, the length of the other portion of the second string
- 28. BASIC PROPORTIONALITY THEOREM If a line intersects two sides of a triangle and is parallel to the third side, then it divides the first two sides proportionally.
- 29. RESTATEMENT OF THE THEOREM If a line (EF) intersects B two sides ( AB & CB) of a triangle ( ABC) and is parallel to the third E F side( AC ), then it divides the first two sides proportionally. A C Thus,
- 30. OTHER PROPORTIONS 1. BE : EA = BF : FC B 2. BE : BA = BF : BC 3. BA : EA = BC : FC E F 4. BE : BF = EA : FC 5. FC : EA = BC : BA 6. EF : AC = BF : BC A C 7. EF : AC = BE : BA
- 31. VERIFYING A PROPORTIONS ( an example)1. BE : EA = BF : FC B 15 : 5 = 12 : 4 By simplifying, 15 123 : 1 = 3 : 1 6 E F 5 4 8 A C
- 32. VERIFYING A PROPORTIONS2. BE : BA = BF : BC B 15 : 20 = 12 : 16 By simplifying, 15 123 : 4 =3:4 E 6 F 5 4 8 A C
- 33. VERIFYING A PROPORTIONS3. BA : EA = BC : FC B 20 : 5 = 16 : 4 15 12 By simplifying, E 6 F4 : 1=4:1 5 4 8 A C
- 34. VERIFYING A PROPORTIONS4. BE : BF = EA : FC B 15 : 12 = 5 : 4 By simplifying, 15 12 E 6 F5 : 4=5:4 5 4 8 A C
- 35. VERIFYING A PROPORTIONS5. FC : EA = BC : BA B 4 : 5 = 16 : 20 By simplifying, 15 12 E 6 F 4: 5=4:5 5 4 8 A C
- 36. VERIFYING A PROPORTIONS6. EF : AC = BF : BC B 6 : 8 = 12 : 16 By simplifying, 15 12 E 6 F3 : 4=3:4 5 4 8 A C
- 37. VERIFYING A PROPORTIONS6. EF : AC = BE : BA B 6 : 8 = 15 : 20 By simplifying, 15 12 E 6 F 3: 4=3:4 5 4 8 A C
- 38. Exercises GIVEN: DE // BC, A AD = 9, AE = 12, DE = 10,DB = 18. 9 12 D E Find, 10BC, AC and CE. 18 B C
- 39. Solution Find BC, BC : DE = BA : DA A BC : 10 = 27 : 9 or 9 12 BC : 10 = 3 : 1 D E 10 Applying principle of proportion 18 BC(1) = 10(3) 30 BC = 30 B C
- 40. Solution Find AC, AC : AE = BA : DA A AC : 12 = 27 : 9 or 9 12 AC : 12 = 3 : 1 D E 10 Applying principle of proportion 18 AC(1) = 12(3)AC = 36 B C
- 41. Solution Find CE, CE : AE = BD : DA A CE : 12 = 18 : 9 or 9 12 CE : 12 = 2 : 1 D E 10 Applying principle of proportion 18 24 CE(1) = 12(2) CE = 24 B C
- 42. Solution Another way to find CE, A CE = AC - AE 9 Hence, AC =36, 12 D then 10 E CE = 36 - 12 24 CE = 24 18 B C
- 43. • If a line divides any two sides of a triangle in the same ratio, then the linemust be parallel to the third side.
- 44. Segment cut by a transversal corollary
- 45. Definition Inmathematics, a corollary is a statement which follows readily from a previously proven statement, typically a mathematical theorem.
- 46. Segment cut by a transversal Ifthree parallel lines intersect two transversals, then they divide the transversals proportionally.
- 47. If three parallel lines intersect two transversals, then they divide the transversals proportionally Restatement of the A D corollary If AD //EF//BC and E F intersect two transversals (line AB and line DC) , then B C DC : DF = AB : AE BE : EA = CF : FD AE : DF = BE : CF
- 48. Illustrative examplesFind the value of x. Solution:By applying the corollary 10 : x = 12 : x + 2Using the principle of 10 12 proportion 10(x +2) = x ( 12 ) 10x + 20 = 12x x X+2 20 = 12x – 10x 20 = 2x X = 10
- 49. Another solutionBy applying the theoremx : x + 2 = 10 : 12 orx:x+2 =5:6 10 12Using the principle of proportion x X+2 6(x) = 5 (x + 2 ) 6x = 5x + 10 6x – 5x = = 10 X = 10
- 50. Segment cut by angle bisector Theorem
- 51. Segment cut by angle bisector If a ray bisects an angle of a triangle, it divides the opposite side into segments proportional to the other two sides.
- 52. Exploration Construct any triangle. Construct an angle bisector in the triangle and draw the segment along the angle bisector from the vertex to the intersection with the opposite side. Measure the ratio of the adjacent sides . Measure the ratio of the segments cut off by the bisector on the opposite side. Repeat for many triangles .
- 53. ILLUSTRATION The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides.
- 54. ILLUSTRATION C for any triangle ABC, the bisector of the angle at C divides the opposite side into segments of length x and y such that A D B
- 55. ILLUSTRATION Cx ay b orx:y=a:b A D B
- 56. If a ray bisects an angle of a triangle, it divides the opposite side into segments proportional to the other two sides. Restatement of the A theorem If AD bisects angle BAC of triangle ABC, then … BD : DC = AB : AC BD : BC = AB : AB + AC DC : BC = CA : CA + AB B D C
- 57. Illustrative examplesFind the value of x ifa =10, b = 15 and y = 12.
- 58. solutionBy applying the theoremx:y =a:b orx : 12 = 10 : 15Using the principle of proportion 15(x) = 12 (10 ) 15x = 120 x = 120 15 x=8
- 59. another solutionBy applying the theoremx:a =y:b orx : 10 = 12 : 15Using the principle of proportion 15(x) = 12 (10 ) 15x = 120 x = 120 15 x=8
- 60. another solutionBy applying the theoremx : x + y = a : a +b orx : x+12= 10 : 10 +15x : x+12= 10 : 25x : x+12= 2: 5Using the principle of proportion 5(x) = 2 (x +12 ) 5x = 2x + 24 5x – 2x = 24 3x = 24 x=8
- 61. SIMILAR TRIANGLESTwo triangles are similar iftheir corresponding anglesare congruent and themeasures of correspondingtriangles are proportional.~ MEANS SIMILAR
- 62. PYTHAGOREAN THEOREMThe PYTHAGOREAN THEOREM is used to find the length of a missing side in a right triangle.
- 63. 45-45-90• A 45°- 45°- 90° triangle is a special right triangle whose angles are 45°, 45°and 90°. The lengths of the sides of a 45°- 45°- 90° triangle are in the ratio of . Note that a 45°- 45°- 90° triangle is an isosceles right triangle.
- 64. 30-60-90• Another type of special right triangles is the 30°- 60°- 90° triangle. This is right triangle whose angles are 30°, 60°and 90°. The lengths of the sides of a 30°- 60°- 90° triangle are in the ratio of
- 65. • The hypotenuse is always twice the length of the shorter leg (the side facing the 30° angle). The longer leg (the side facing the 60° angle) is times of the shorter leg.
- 66. CENTRAL ANGLE• A central angle is an angle formed by two intersecting radii such that its vertex is at the center of the circle.
- 67. <AOB is a central angle.Its intercepted arc is the minor arc from A to B. m<AOB = 80°
- 68. INSCRIBED ANGLE• An inscribed angle is an angle with its vertex "on" the circle, formed by two intersecting chords.
- 69. <ABC is an inscribed angle. Its intercepted arc is the minor arc from A to C. m<ABC = 50°
- 70. INTERCEPTED ARC• Corresponding to an angle, this is the portion of the circle that lies in the interior of the angle together with the endpoints of the arc.
- 71. ∠ ABC is an inscribed angle andis its intercepted arc.
- 72. THEOREMS• The measure of an inscribed angle in a circle equals half the measure of its intercepted arc.• If two inscribed angles of a circle intercept the same arc or arcs of equal measure, then the inscribed angles have equal measure.• If an inscribed angle intercepts a semicircle, then its measure is 90°.
- 73. X – Axis Y – Axis Cartesian Coordinate SystemQuadrants Ordered Pair Coordinates
- 74. Terminologies: Cartesian Coordinate plane a two-dimensional surface onwhich a coordinate system hasbeen set up; invented by ReneDescartes; also called Cartesianplane, graph, coordinate grid.
- 75. Coordinates the numbers in an orderedpair that locate a point in thecoordinate plane. Ordered pair a pair of numbers, written as(x, y), that represents a point on a coordinate grid.
- 76. X-axisthe horizontal number line on a coordinate grid. Y-axis the vertical number line on a coordinate grid
- 77. QuadrantsThe four parts made by X and Y axis Origin Intersection of X and Y axis
- 78. y Quadrant 2 Quadrant 1 3 (-,+) (+,+) 2 1 Origin-x x -3 -2 -1 1 2 3 -1 -2 (-,-) (+,-) -3 Quadrant 3 Quadrant 4 -y
- 79. yPlot the point: 31. ( 1, 2) 2 1 -x x -3 -2 -1 1 2 3 -1 -2 -3 -y
- 80. yPlot the point: 32. ( 2, -3) 2 1 -x x -3 -2 -1 1 2 3 -1 -2 -3 -y
- 81. yPlot the point: 32. ( -2, 3) 2 1 -x x -3 -2 -1 1 2 3 -1 -2 -3 -y
- 82. yPlot the point: 32. (-2,-3) 2 1 -x x -3 -2 -1 1 2 3 -1 -2 -3 -y
- 83. LINEAR EQUATIONS PART I 1. Basic Coordinate Plane Info Assignments 2. Review on Plotting Points 3. Finding Slopes 4. x and y intercepts 5. Slope-Intercept Form of a Line 6. Graphing Lines7. Determine the equation of a line given two points, slope and one point, or a graph.
- 84. COORDINATE PLANE Y-axis Parts of a plane 1. X-axis 2. Y-axis QUAD II QUAD I 3. Origin 4. Quadrants I-IV Origin ( 0 , 0 ) X-axisQUAD III QUAD IV
- 85. PLOTTING POINTS Remember when plotting points you always start at B the origin. Next you go left C (if x-coordinate is negative) or right (if x-coordinate is positive. Then you go up (if y-coordinate is positive) or down (if y-coordinate is negative) AD Plot these 4 points A (3, -4), B (5, 6), C (-4, 5) and D (-7, -5)
- 86. SLOPESlope is the ratio of the vertical rise to the horizontalrun between any two points on a line. Usuallyreferred to as the rise over run. Run is 6 Slope triangle between two because we points. Notice that the slope went to the triangle can be drawn twoRise is 10 right different ways.because we Rise is -10went up because we went down 10 5 The slope in this case is Run is -6 6 3 because we went to the left 10 5 The slope in this case is 6 3 Another way to find slope
- 87. FORMULA FOR FINDING SLOPE The formula is used when you know two points of a line.They look like A( X 1 , Y1 ) and B( X 2 , Y2 ) RISE X 2 X1 X1 X 2SLOPE RUN Y2 Y1 Y1 Y2 EXAMPLE
- 88. Find the slope of the line between the two points (-4, 8) and (10, -4) If it helps label the points. X 1 Y1 X2 Y2 Then use the formula X 2 X1 (10 ) ( 4) Y2 Y1 SUBSTITUTE INTO FORMULA ( 4) (8) (10 ) ( 4) 10 4 14 7 Then Simplify ( 4) (8) 4 ( 8) 12 6
- 89. X AND Y INTERCEPTSThe x-intercept is the x-coordinate of a pointwhere the graph crosses the x-axis.The y-intercept is the y-coordinate of a pointwhere the graph crosses the y-axis. The x-intercept would be 4 and is located at the point (4, 0). The y-intercept is 3 and is located at the point (0, 3).
- 90. SLOPE-INTERCEPT FORM OF A LINE The slope intercept form of a line is y = mx + b, where “m” represents the slope of the line and “b” represents the y-intercept. When an equation is in slope-intercept form the “y” is always on one side by itself. It can not be more than one y either. If a line is not in slope-intercept form, then we must solve for “y” to get it there. Examples
- 91. IN SLOPE-INTERCEPT NOT IN SLOPE-INTERCEPT y = 3x – 5 y – x = 10 y = -2x + 10 2y – 8 = 6x y = -.5x – 2 y + 4 = 2xPut y – x = 10 into slope-intercept form Add x to both sides and would get y = x + 10Put 2y – 8 = 6x into slope-intercept form. Add 8 to both sides then divide by 2 and would get y = 3x + 4Put y + 4 = 2x into slope-intercept form. Subtract 4 from both sides and would get y = 2x – 4.
- 92. GRAPHING LINES BY MAKING A TABLE OR USING THE SLOPE-INTERCEPT FORM I could refer to the table method by input-output table or x-y table. For now I want you to include three values in your table. A negative number, zero, and a positive number. Graph y = 3x + 2 INPUT (X) OUTPUT (Y) -2 -4 0 2 1 5By making a table it gives me three points, in this case (-2, -4) (0, 2) and (1, 5) to plotand draw the line. See the graph.
- 93. Plot (-2, -4), (0, 2) and (1, 5)Then draw the line. Make sure yourline covers the graph and hasarrows on both ends. Be sure touse a ruler. Slope-intercept graphing
- 94. Slope-intercept graphingSteps1. Make sure the equation is in slope-intercept form.2. Identify the slope and y-intercept.3. Plot the y-intercept.4. From the y-intercept use the slope to get another point to draw the line. 1. y = 3x + 2 2. Slope = 3 (note that this means the fraction or rise over run could be (3/1) or (-3/-1). The y-intercept is 2. 3. Plot (0, 2) 4. From the y-intercept, we are going rise 3 and run 1 since the slope was 3/1.
- 95. FIND EQUATION OF A LINE GIVEN 2 POINTS Find the equation of the line between (2, 5) and (-2, -3).1. Find the slope between the two points. 1. Slope is 2.2. Plug in the slope in the slope- 2. y = 2x + b intercept form. 3. Picked (2, 5) so3. Pick one of the given points and plug (5) = 2(2) + b in numbers for x and y. 4. b = 14. Solve and find b. 5. y = 2x + 15. Rewrite final form. Two other ways
- 96. Steps if given the slope and If given a graph there are three ways.a point on the line.1. Substitute the slope into One way is to find two points on the slope-intercept the line and use the first method we talked about. form.2. Use the point to plug in Another would be to find the for x and y. slope and pick a point and use the second method.3. Find b.4. Rewrite equation. The third method would be to find the slope and y-intercept and plug it directly into y = mx + b.
- 97. Systems of Linear Equations Using a Graph to Solve
- 98. All the slides in this presentation are timed. You do not need to click the mouse or press any keys onthe keyboard for the presentation on each slide to continue. However, in order to make sure the presentation does notgo too quickly, you will need to click the mouse or press a keyon the keyboard to advance to the next slide. You will know when the slide is finished when you see asmall icon in the bottom left corner of the slide. Click the mouse button to advance the slide when you see this icon.
- 99. What is a System of Linear Equations?A system of linear equations is simply two or more linear equationsusing the same variables.We will only be dealing with systems of two equations using twovariables, x and y.If the system of linear equations is going to have a solution, thenthe solution will be an ordered pair (x , y) where x and y makeboth equations true at the same time.If the lines are parallel, there will be no solutions.If the lines are the same, there will be an infinite number of solutions.We will be working with the graphs of linear systems and how to findtheir solutions graphically.
- 100. How to Use Graphs to Solve Linear Systems yConsider the following system: x – y = –1 x + 2y = 5Using the graph to the right, we cansee that any of these ordered pairs willmake the first equation true since they xlie on the line. (1 , 2)We can also see that any of thesepoints will make the second equationtrue.However, there is ONE coordinate thatmakes both true at the same time… The point where they intersect makes both equations true at the same time.
- 101. •If the lines cross once, there will be one solution.•If the lines are parallel, there will be no solutions.•If the lines are the same, there will be an infinite number of solutions.
- 102. How to Use Graphs to Solve Linear Systems yConsider the following system: x – y = –1 x + 2y = 5We must ALWAYS verify that yourcoordinates actually satisfy bothequations. x (1 , 2)To do this, we substitute thecoordinate (1 , 2) into bothequations. x – y = –1 x + 2y = 5(1) – (2) = –1 (1) + 2(2) = Since (1 , 2) makes both equations 1+4=5 true, then (1 , 2) is the solution to the system of linear equations.
- 103. Graphing to Solve a Linear SystemSolve the following system by graphing: 3x + 6y = 15 Start with 3x + 6y = 15 –2x + 3y = –3 Subtracting 3x from both sides yields 6y = –3x + 15 While there are many different Dividing everything by 6 gives us… ways to graph these equations, we 1 5 will be using the slope - intercept y= - 2 x+ 2 form. Similarly, we can add 2x to both sides and then divide everything by To put the equations in slope 3 in the second equation to get intercept form, we must solve both equations for y. y = 2 x- 1 3 Now, we must graph these two equations.
- 104. Graphing to Solve a Linear SystemSolve the following system by graphing: y 3x + 6y = 15 –2x + 3y = –3Using the slope intercept form of theseequations, we can graph them carefully xon graph paper. (3 , 1) 5 y = - 1 x+ 2 2 y = 2 x- 1 3 Label theStart at the y - intercept, then use the slope. solution!Lastly, we need to verify our solution is correct, by substituting (3 , 1).Since 3(3)+ 6(1)= 15 and - 2(3)+ 3(1)= - 3, then our solution is correct!
- 105. Graphing to Solve a Linear System Lets summarize! There are 4 steps to solving a linear system using a graph.Step 1: Put both equations in slope - Solve both equations for y, so thatintercept form. each equation looks like y = mx + b.Step 2: Graph both equations on the Use the slope and y - intercept forsame coordinate plane. each equation in step 1. Be sure to use a ruler and graph paper!Step 3: Estimate where the graphs This is the solution! LABEL theintersect. solution!Step 4: Check to make sure your Substitute the x and y values into bothsolution makes both equations true. equations to verify the point is a solution to both equations.
- 106. Graphing to Solve a Linear System Lets do ONE more…Solve the following system of equations by graphing. 2x + 2y = 3 y x – 4y = -1 LABEL the solution!Step 1: Put both equations in slope -intercept form. (1 , 1 ) 2 y = - x+ 3 2 y = 1 x+ 1 4 4 xStep 2: Graph both equations on thesame coordinate plane.Step 3: Estimate where the graphsintersect. LABEL the solution! 2(1)+ 2(1 )= 2 + 1 = 3Step 4: Check to make sure your 2solution makes both equations true. 1- 4 ( 1 ) = 1- 2 = - 1 2
- 107. 3-2: Solving Systems of Equations using Substitution
- 108. Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y= ; x= ;a=) 2. Substitute the expression from step one intothe other equation. 3. Simplify and solve the equation. 4. Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of thesystem.
- 109. Example #1: y = 4x 3x + y = -21Step 1: Solve one equation for one variable. y = 4x (This equation is already solved for y.)Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21Step 3: Simplify and solve the equation. 7x = -21 x = -3
- 110. y = 4x 3x + y = -21Step 4: Substitute back into either original equation to find the value of the other variable. 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12).
- 111. y = 4x 3x + y = -21Step 5: Check the solution in both equations. Solution to the system is (-3,-12). y = 4x 3x + y = -21 -12 = 4(-3) 3(-3) + (-12) = -21 -12 = -12 -9 + (-12) = -21 -21= -21
- 112. Example #2: x + y = 10 5x – y = 2Step 1: Solve one equation for one variable. x + y = 10 y = -x +10Step 2: Substitute the expression from step one into the other equation. 5x - y = 2 5x -(-x +10) = 2
- 113. x + y = 10 5x – y = 2Step 3: Simplify and solve the equation. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x=2
- 114. x + y = 10 5x – y = 2Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y=8 Solution to the system is (2,8).
- 115. x + y = 10 5x – y = 2Step 5: Check the solution in both equations. Solution to the system is (2, 8). x + y =10 5x – y = 2 2 + 8 =10 5(2) - (8) = 2 10 =10 10 – 8 = 2 2=2
- 116. Solving Systems of EquationsThe Elimination Method
- 117. Objectives• Learn the procedure of the Elimination Method using addition• Learn the procedure of the Elimination Method using multiplication• Solving systems of equations using the Elimination Method
- 118. Elimination using Addition Consider the system x - 2y = 5 Lets add both equations 2x + 2y = 7 to each otherREMEMBER: We are trying to find the Point of Intersection. (x, y)
- 119. Elimination using Addition Consider the system x - 2y = 5 Lets add both equations+ 2x + 2y = 7 to each otherNOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 120. Elimination using Addition Consider the system x - 2y = 5 Lets add both equations+ 2x + 2y = 7 to each other 3x = 12 x=4 ANS: (4, y)NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 121. Elimination using Addition Consider the system x - 2y = 5 Lets substitute x = 4 into this equation. 2x + 2y = 7 4 - 2y = 5 Solve for y - 2y = 1 y= 1 2 ANS: (4, y)NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 122. Elimination using Addition Consider the system x - 2y = 5 Lets substitute x = 4 into this equation. 2x + 2y = 7 4 - 2y = 5 Solve for y - 2y = 1 y= 1 2 1 ANS: (4, 2 )NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 123. Elimination using Addition Consider the system 3x + y = 14 4x - y = 7NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 124. Elimination using Addition Consider the system 3x + y = 14+ 4x - y = 7 7x = 21 x=3 ANS: (3, y)
- 125. Elimination using Addition Consider the system 3x + y = 14 Substitute x = 3 into this equation 4x - y = 7 3(3) + y = 14 9 + y = 14 y=5ANS: (3, 5 )NOTE: We use the Elimination Method, if we can immediately cancel out two like terms.
- 126. Examples…1. 2. 2x + y = 5 6y + x = 11 3x y = 15 2y x = 5 ANS: (4, -3) ANS: (-1, 2)
- 127. Elimination using MultiplicationConsider the system6x + 11y = -56x + 9y = -3
- 128. Elimination using Multiplication Consider the system 6x + 11y = -5+ 6x + 9y = -3 12x + 20y = -8 When we add equations together, nothing cancels out
- 129. Elimination using MultiplicationConsider the system6x + 11y = -56x + 9y = -3
- 130. Elimination using Multiplication Consider the system-1 (6x + 11y = -5 ) 6x + 9y = -3
- 131. Elimination using Multiplication Consider the system - 6x - 11y = 5+ 6x + 9y = -3 -2y = 2 y = -1 ANS: (x, -1 )
- 132. Elimination using MultiplicationConsider the system6x + 11y = -56x + 9y = -3 Lets substitute y = -1 into this equationy = -16x + 9(-1) = -36x + -9 = -3 +9 +9 6x = 6 x=1 ANS: (x, -1 )
- 133. Elimination using MultiplicationConsider the system6x + 11y = -56x + 9y = -3 Lets substitute y = -1 into this equationy = -16x + 9(-1) = -36x + -9 = -3 +9 +9 6x = 6 x=1 ANS: ( 1, -1 )
- 134. Elimination using MultiplicationConsider the system x + 2y = 6 Multiply by -3 to eliminate the x term3x + 3y = -6
- 135. Elimination using Multiplication Consider the system-3 ( x + 2y = 6 ) 3x + 3y = -6
- 136. Elimination using Multiplication Consider the system -3x + -6y = -18+ 3x + 3y = -6 -3y = -24 y=8 ANS: (x, 8)
- 137. Elimination using MultiplicationConsider the system x + 2y = 6 Substitute y =14 into equation3x + 3y = -6 y =8 x + 2(8) = 6 x + 16 = 6 x = -10 ANS: (x, 8)
- 138. Elimination using MultiplicationConsider the system x + 2y = 6 Substitute y =14 into equation3x + 3y = -6 y =8 x + 2(8) = 6 x + 16 = 6 x = -10 ANS: (-10 , 8)
- 139. Examples1. 2. x + 2y = 5 x + 2y = 4 2x + 6y = 12 x - 4y = 16 ANS: (3, 1) ANS: (8, -2)
- 140. More complex ProblemsConsider the system 3x + 4y = -25 Multiply by 2 2x - 3y = 6 Multiply by -3
- 141. More complex ProblemsConsider the system2( 3x + 4y = -25 )-3( 2x - 3y = 6)
- 142. More complex Problems Consider the system 6x + 8y = -50+ -6x + 9y = -18 17y = -68 y = -4 ANS: (x, -4)
- 143. More complex ProblemsConsider the system 3x + 4y = -25 2x - 3y = 6 Substitute y = -42x - 3(-4) = 6 2x - -12 = 6 2x + 12 = 6 2x = -6 x = -3 ANS: (x, -4)
- 144. More complex ProblemsConsider the system 3x + 4y = -25 2x - 3y = 6 Substitute y = -42x - 3(-4) = 6 2x - -12 = 6 2x + 12 = 6 2x = -6 x = -3 ANS: ( -3 , -4)
- 145. Examples…1. 2. 4x + y = 9 2x + 3y = 1 3x + 2y = 8 5x + 7y = 3 ANS: (2, 1) ANS: (2, -1)
- 146. Parallel and PerpendicularLines
- 147. Parallel Lines Two lines with the same slope are said to be parallel lines. If you graph them they will never intersect. We can decide algebraically if two lines are parallel by finding the slope of each line and seeing if the slopes are equal to each other. We can find the equation of a line parallel to a given line and going through a given point by: a.) first finding the slope m of the given line; b.) finding the equation of the line through the given point with slope m.
- 148. Testing if Lines are ParallelAre the lines 12 x 3 y 9 and -8x 2 y 14 parallel?Find the slope of 12 x 3 y 9 The slope m = -4 3y 12 x 9 y 4x 3 Find the slope of 8 x 2 y 14 The slope m = -4 2 y 8 x 14 y 4x 7 Since the slopes are equal the lines are parallel.
- 149. Graphs of Parallel Lines The red line is the graph of y = – 4x – 3 and the blue line is the graph of y = – 4x – 7
- 150. Practice Testing if Lines are ParallelAre the lines 6 x 3 y 5 and 2 y 4 x 4 parallel? (click mouse for answer) 6x 3 y 5 2y 4x 4 Since the slopes are different 3y 6x 5 y 2x 2 the lines are not parallel. y 2x 5 m 2 3 m 2Are the lines x 2 y 4 and 2 x 4 y 12 parallel? (click mouse for answer) x 2y 4 2 x 4 y 12 Since the slopes are equal 2y x 4 4y 2 x 12 the lines are parallel. y 1 x 3 y 1 x 2 2 2 m 1 m 1 2 2
- 151. Constructing Parallel LinesFind the equation of a line going through the point (3, -5) andparallel to y 2 x 8 3Using the point-slope equation where the slope m = -2/3andthe point is (3, -5) we get y 5 2 x 3 3 y 5 2 x 2 3 y 2 x 3 3
- 152. Practice Constructing Parallel LinesFind the equation of the line going through the point (4,1) andparallel to y 3x 7 (click mouse for answer) y 1 3 x 4 y 1 3 x 12 y 3 x 13Find the equation of the line going through the point (-2,7) andparallel to 2 x y 8 (click mouse for answer) y 7 2 x 2 y 7 2 x 2 y 7 2x 4 y 2x 3
- 153. Perpendicular Lines Perpendicular lines are lines that intersect in a right angle. We can decide algebraically if two lines are perpendicular by finding the slope of each line and seeing if the slopes are negative reciprocals of each other. This is equivalent to multiplying the two slopes together and seeing if their product is –1. We can find the equation of a line perpendicular to a given line and going through a given point by: a.) first finding the slope m of the given line; b.) finding the equation of the line through the given point with slope = –1 /m.
- 154. Testing if Lines Are Perpendicular 1Are the lines 2 x y 5 and y x 4 perpendicular? 2Find the slope of 2 x y 5 m 2 y 2x 5 1 1Find the slope of y x 4 m 2 2 Since the slopes are negative reciprocals of each other the lines are perpendicular. 1 2 1 2
- 155. Graphs of Perpendicular Lines The red line is the graph of y = – 2x + 5 and the blue line is the graph of y = – 1/2 x +4
- 156. Practice Testing if Lines Are PerpendicularAre the lines 6 x 3 y 5 and 2 y 4 x 4 perpendicular?6x 3 y 5 2y 4x 4 Since the slopes are not 3y 6x 5 y 2x 2 negative reciprocals of y 2x 5 m 2 each other (their product 3 is not -1) the lines are m 2 not perpendicularAre the lines x 2 y 4 and 4 x 2 y 6 perpendicular? 4x 2 y 6 Since the slopes arex 2y 4 2y x 4 2y 4x 6 negative reciprocals of y 2x 3 each other (their y 1 x 2 2 m 2 product is -1) the lines m 1 are perpendicular. 2
- 157. Constructing Perpendicular LinesFind the equation of a line going through the point (3, -5) andperpendicular to y 2 x 8 3The slope of the perpendicular line will be m = 3/2 Usingthe point-slope equation where the slope m = 3/2 andthe point is (3, -5) we get 3 y 5 x 3 2 y 5 3 x 9 2 2 y 3 x 19 2 2
- 158. Practice Constructing Perpendicular LinesFind the equation of the line going through the point (4,1) andperpendicular to y 3x 7 (click mouse for answer) y 1 1 x 4 3 y 1 1 x 4 3 3 y 1 x 1 3 3Find the equation of the line going through the point (-2,7) andperpendicular to 2 x y 8 (click mouse for answer) y 7 1 x 2 2 y 7 1 x 2 2 y 7 1 x 1 2 y 1 x 8 2
- 159. The Distance and MidpointFormulas Topic 7.1
- 160. Distance Formula Used to find the distance between two points 2 2 distance ( x2 x1 ) ( y2 y1 )
- 161. Example Find the distance between A(4,8) and B(1,12) A (4, 8) B (1, 12) 2 2 distance ( x2 x1 ) ( y2 y1 ) 2 2 distance (1 4) (12 8) 2 2 distance ( 3) (4) distance 9 16 25 5
- 162. YOU TRY!! Find the distance between: A. (2, 7) and (11, 9) 2 2 (9) (2) 85 B. (-5, 8) and (2, - 4) 2 2 (7) ( 12) 193
- 163. Midpoint Formula Used to find the center of a line segment x2 x1 y2 y1midpoint , 2 2
- 164. Example Find the midpoint between A(4,8) and B(1,12) A (4, 8) B (1, 12) x2 x1 y2 y1 midpoint , 2 2 1 4 12 8 midpoint , 2 2 midpoint 5 ,10 2
- 165. YOU TRY!! Find the midpoint between: A) (2, 7) and (14, 9) midpoint = 8,8 B) (-5, 8) and (2, - 4) -3 midpoint = ,2 2
- 166. 4-7 Introduction to Coordinate Proof 4-7 Introduction to Coordinate Proof Warm Up Lesson Presentation Lesson Quiz Holt GeometryHolt Geometry
- 167. 4-7 Introduction to Coordinate Proof Warm Up Evaluate. 1. Find the midpoint between (0, 2x) and (2y, 2z). (y, x + z) 2. One leg of a right triangle has length 12, and the hypotenuse has length 13. What is the length of the other leg? 5 3. Find the distance between (0, a) and (0, b), where b > a. b – aHolt Geometry
- 168. 4-7 Introduction to Coordinate Proof Objectives Position figures in the coordinate plane for use in coordinate proofs. Prove geometric concepts by using coordinate proof.Holt Geometry
- 169. 4-7 Introduction to Coordinate Proof Vocabulary coordinate proofHolt Geometry
- 170. 4-7 Introduction to Coordinate Proof You have used coordinate geometry to find the midpoint of a line segment and to find the distance between two points. Coordinate geometry can also be used to prove conjectures. A coordinate proof is a style of proof that uses coordinate geometry and algebra. The first step of a coordinate proof is to position the given figure in the plane. You can use any position, but some strategies can make the steps of the proof simpler.Holt Geometry
- 171. 4-7 Introduction to Coordinate ProofHolt Geometry
- 172. 4-7 Introduction to Coordinate Proof Example 1: Positioning a Figure in the Coordinate Plane Position a square with a side length of 6 units in the coordinate plane. You can put one corner of the square at the origin.Holt Geometry
- 173. 4-7 Introduction to Coordinate Proof Check It Out! Example 1 Position a right triangle with leg lengths of 2 and 4 units in the coordinate plane. (Hint: Use the origin as the vertex of the right angle.)Holt Geometry
- 174. 4-7 Introduction to Coordinate Proof Once the figure is placed in the coordinate plane, you can use slope, the coordinates of the vertices, the Distance Formula, or the Midpoint Formula to prove statements about the figure.Holt Geometry
- 175. 4-7 Introduction to Coordinate Proof Example 2: Writing a Proof Using Coordinate Geometry Write a coordinate proof. Given: Rectangle ABCD with A(0, 0), B(4, 0), C(4, 10), and D(0, 10) Prove: The diagonals bisect each other.Holt Geometry
- 176. 4-7 Introduction to Coordinate Proof Example 2 Continued By the Midpoint Formula, 0 4 0 10 mdpt. of AC , (2,5) 2 2 mdpt. of The midpoints coincide, therefore the diagonals bisect each other.Holt Geometry
- 177. 4-7 Introduction to Coordinate Proof Check It Out! Example 2 Use the information in Example 2 (p. 268) to write a coordinate proof showing that the area of ∆ADB is one half the area of ∆ABC. Proof: ∆ABC is a right triangle with height AB and base BC. 1 area of ∆ABC = bh 2 1 = (4)(6) = 12 square units 2Holt Geometry
- 178. 4-7 Introduction to Coordinate Proof Check It Out! Example 2 Continued By the Midpoint Formula, the coordinates of D = 0+4 , 6+0 = (2, 3). 2 2 The x-coordinate of D is the height of ∆ADB, and the base is 6 units. 1 The area of ∆ADB = bh 2 = 1 (6)(2) = 6 square units 2 Since 6 = 1 (12), the area of ∆ADB is one half the 2 area of ∆ABC.Holt Geometry
- 179. 4-7 Introduction to Coordinate Proof A coordinate proof can also be used to prove that a certain relationship is always true. You can prove that a statement is true for all right triangles without knowing the side lengths. To do this, assign variables as the coordinates of the vertices.Holt Geometry
- 180. 4-7 Introduction to Coordinate Proof Example 3A: Assigning Coordinates to Vertices Position each figure in the coordinate plane and give the coordinates of each vertex. rectangle with width m and length twice the widthHolt Geometry
- 181. 4-7 Introduction to Coordinate Proof Example 3B: Assigning Coordinates to Vertices Position each figure in the coordinate plane and give the coordinates of each vertex. right triangle with legs of lengths s and tHolt Geometry
- 182. 4-7 Introduction to Coordinate Proof Caution! Do not use both axes when positioning a figure unless you know the figure has a right angle.Holt Geometry
- 183. 4-7 Introduction to Coordinate Proof Check It Out! Example 3 Position a square with side length 4p in the coordinate plane and give the coordinates of each vertex.Holt Geometry
- 184. 4-7 Introduction to Coordinate Proof If a coordinate proof requires calculations with fractions, choose coordinates that make the calculations simpler. For example, use multiples of 2 when you are to find coordinates of a midpoint. Once you have assigned the coordinates of the vertices, the procedure for the proof is the same, except that your calculations will involve variables.Holt Geometry
- 185. 4-7 Introduction to Coordinate Proof Remember! Because the x- and y-axes intersect at right angles, they can be used to form the sides of a right triangle.Holt Geometry
- 186. 4-7 Introduction to Coordinate Proof Example 4: Writing a Coordinate Proof Given: Rectangle PQRS Prove: The diagonals are . Step 1 Assign coordinates to each vertex. The coordinates of P are (0, b), the coordinates of Q are (a, b), the coordinates of R are (a, 0), and the coordinates of S are (0, 0). Step 2 Position the figure in the coordinate plane.Holt Geometry
- 187. 4-7 Introduction to Coordinate Proof Example 4 Continued Given: Rectangle PQRS Prove: The diagonals are . Step 3 Write a coordinate proof. By the distance formula, PR = √ a2 + b2, and QS = √a2 + b2 . Thus the diagonals are .Holt Geometry
- 188. 4-7 Introduction to Coordinate Proof Check It Out! Example 4 Use the information in Example 4 to write a coordinate proof showing that the area of ∆ADB is one half the area of ∆ABC. Step 1 Assign coordinates to each vertex. The coordinates of A are (0, 2j), the coordinates of B are (0, 0), and the coordinates of C are (2n, 0). Step 2 Position the figure in the coordinate plane.Holt Geometry
- 189. 4-7 Introduction to Coordinate Proof Check It Out! Example 4 Continued Step 3 Write a coordinate proof.Holt Geometry
- 190. 4-7 Introduction to Coordinate Proof Check It Out! Example 4 Continued Proof: ∆ABC is a right triangle with height 2j and base 2n. 1 The area of ∆ABC = bh 2 1 = (2n)(2j) 2 = 2nj square units By the Midpoint Formula, the coordinates of D = 0 + 2n, 2j + 0 = (n, j). 2 2Holt Geometry
- 191. 4-7 Introduction to Coordinate Proof Check It Out! Example 4 Continued The height of ∆ADB is j units, and the base is 2n units. 1 area of ∆ADB = bh 2 1 = (2n)(j) 2 = nj square units Since nj = 1 (2nj), the area of ∆ADB is one half the 2 area of ∆ABC.Holt Geometry
- 192. 4-7 Introduction to Coordinate Proof Lesson Quiz: Part I Position each figure in the coordinate plane. Possible answers: 1. rectangle with a length of 6 units and a width of 3 units 2. square with side lengths of 5a unitsHolt Geometry
- 193. 4-7 Introduction to Coordinate Proof Lesson Quiz: Part II 3. Given: Rectangle ABCD with coordinates A(0, 0), B(0, 8), C(5, 8), and D(5, 0). E is mdpt. of BC, and F is mdpt. of AD. Prove: EF = AB By the Midpoint Formula, the coordinates of E are 5 ,8 . 2 5 and F are ,0 . Then EF = 8, and AB = 8. 2 Thus EF = AB.Holt Geometry
- 194. Equations of Circles
- 195. Equation of a CircleThe center of a circle is givenby (h, k)The radius of a circle is givenby rThe equation of a circle instandard form is(x – h)2 + (y – k)2 = r2
- 196. Finding the Equation of a CircleCircle AThe center is (16, 10)The radius is 10The equation is(x – 16)2 + (y – 10)2 = 100
- 197. Finding the Equation of a CircleCircle BThe center is (4, 20)The radius is 10The equation is(x – 4)2 + (y – 20)2 = 100
- 198. Finding the Equation of a CircleCircle OThe center is (0, 0)The radius is 12The equation isx 2 + y 2 = 144
- 199. Graphing Circles(x – 3)2 + (y – 2)2 = 9Center (3, 2)Radius of 3
- 200. Graphing Circles(x + 4)2 + (y – 1)2 = 25Center (-4, 1)Radius of 5
- 201. Graphing Circles(x – 5)2 + y2 = 36Center (5, 0)Radius of 6
- 202. Writing Equations of CirclesWrite the standard equation of the circle:Center (4, 7) Radius of 5 (x – 4)2 + (y – 7)2 = 25
- 203. Writing Equations of CirclesWrite the standard equation of the circle:Center (-3, 8) Radius of 6.2 (x + 3)2 + (y – 8)2 = 38.44
- 204. Writing Equations of CirclesWrite the standard equation of the circle:Center (2, -9) Radius of 11 (x – 2)2 + (y + 9)2 = 11
- 205. Writing Equations of CirclesWrite the standard equation of the circle:Center (0, 6) Radius of 7 x 2 + (y – 6)2 = 7
- 206. Writing Equations of CirclesWrite the standard equation of the circle:Center (-1.9, 8.7) Radius of 3 (x + 1.9)2 + (y – 8.7)2 = 9
- 207. The standard form of the equation of a circle with its center at the origin is 2 2 2 x y r r is the radius of the circle so if we take the square root of the right hand side, well know how big the radius is.Notice that both the x and y terms are squared. Linearequations don’t have either the x or y terms squared.Parabolas have only the x term was squared (or only they term, but NOT both).
- 208. 2 2 Lets look at the equation x y 9 This is r2 so r = 3 The center of the circle is at the origin and the radius is 3. Lets graph this circle.Count out 3 in all Center at (0, 0)directions sincethat is the radius -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
- 209. If the center of the circle is NOT at the origin then the equation for the standard form of a circle looks like this: 2 2 2 x h y k r The center of the circle is at (h, k). This is r2 so r = 4 2 2 x 3 y 1 16 Find the center and radius and graph this circle.The center of the circle is at (h, k) which is (3,1).The radius is 4 - - - - - - - 12345678 765432 1 0
- 210. If you take the equation of a circle in standard form for example: 2 2 x 2 y 4 4 This is r2 so r = 2 (x - (-2)) Remember center is at (h, k) with (x - h) and (y - k) since the x is plus something and not minus, (x + 2) can be written as (x - (-2)) You can find the center and radius easily. The center is at (-2, 4) and the radius is 2.But what if it was not in standard form but multiplied out (FOILED) 2 2 x 4x 4 y 8 y 16 4Moving everything to one side in descending order andcombining like terms wed have: 2 2 x y 4 x 8 y 16 0
- 211. 2 2 x y 4 x 8 y 16 0 If wed have started with it like this, wed have to complete the square on both the xs and ys to get in standard form. Move constantGroup x terms and a place Group y terms and a place to the other side to complete the square to complete the square 2 2 x 4 4 x ____ y 8 y 16 ____ 4 16 16 ___ ___ Complete the square Write factored and wahlah! back in standard form. 2 2 x 2 y 4 4
- 212. Now lets work some examples:Find an equation of the circle with center at (0, 0) and radius 7. Lets sub in center and radius values in the standard form 2 2 2 x 0 h y k 0 r 7 2 2 x y 49
- 213. Find an equation of the circle with center at (0, 0) that passesthrough the point (-1, -4).Since the center is at (0, 0) well have 2 2 2 x y rThe point (-1, -4) is on the circle so should work whenwe plug it in the equation: 2 2 2 1 4 r 1 16 17 Subbing this in for r2 we have: 2 2 x y 17
- 214. Find an equation of the circle with center at (-2, 5) and radius 6 Subbing in the values in standard form we have: 2 2 2 x -2 h y k 5 r 6 2 2 x 2 y 5 36
- 215. Find an equation of the circle with center at (8, 2) and passesthrough the point (8, 0). Subbing in the center values in standard form we have: 2 2 2 x 8 h y k 2 r Since it passes through the point (8, 0) we can plug this point in for x and y to find r2. 2 2 2 8 8 0 2 r 4 2 2 x 8 y 2 4
- 216. Identify the center and radius and sketch the graph: 2 2 9x 9y 64 9 9 9 To get in standard form we dont want coefficients on the squared terms so lets divide everything by 9. Remember to square 2 2 64 root this to get the x y radius. 9So the center is at (0, 0) and the radius is 8/3. - - - - - - - 12345678 765432 1 0
- 217. Identify the center and radius and sketch the graph: 2 2 x 4 y 3 25 Remember the center values end up being the opposite sign of what is with the x and y and the right hand side is the radius squared. So the center is at (-4, 3) and the radius is 5. - - - - - - - 01 234 56 78 765432 1
- 218. Find the center and radius of the circle: 2 2 x y 6x 4 y 3 0 We have to complete the square on both the xs and ys to get in standard form. Move constantGroup x terms and a place Group y terms and a place to the other side to complete the square to complete the square 2 2 x 9 6 x ____ y 4 4 y ____ 9 4 3 ___ ___ Write factored for standard form. 2 2 x 3 y 2 16 So the center is at (-3, 2) and the radius is 4.

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