Roda Frogs
Transcript of Roda Frogs
QUIZZES#11. Find the directed distance:
A B C D E F G H I J K L M N O P Q
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
1. ❑AH
=x2−x1 2. ❑PB
=x2−x1 3. ❑EK
=x2−x1
¿1+8 ¿ (−7 )+ (−7 ) ¿2+4¿7 ¿−14 ¿6
4. ❑LD
=x2−x1 5.❑CO
=x2−x1
¿−5+(−3) ¿6+6¿−8 ¿12
#21. Find the perimeter of the with vertices A (4, -1), B (3,1) C (-4,-2)
❑AB
=√¿¿¿
¿√(3−4 )2+(1+1)2
¿√1+4
¿√5∨2.2units
❑BC
=√¿¿¿
¿√(−4−3)2+(−2−1)2
¿√ (−7 )2+(−3 )2
¿√49+9
¿√58∨7.6units
❑AC
=√¿¿¿
¿√(−4−4)2+(−2−1)2
¿√64+1
¿√65∨8.1units
P=S1+S2+S3
¿2.2+7.6+8.1
¿17.9Units
Therefore the P of the ABC is 17.9 units
#3Find the length of the medians of the with the vertices:A(2,-4) B(8,4) C(0,6)
❑AB
=12
(2+8 )
X=5
¿ 12
(−4+4 )
Y=0 (5,0 )→ ❑AB
❑AC
=12(2+0)
X=1
¿ 12(−4+6)
¿ 12
(2 )
Y=1 (1,1 )→ ❑AC
❑BC
=12(8+0)
X=4
¿ 12(4+6)
Y=5 ( 4,5 )→ ❑BC
(1,1) (8,4)
¿√ (8−1 )2+( 4−1 )2
¿√58
¿7.6
(4,5) (2,-4)
¿√ (−2−4 )2+(−4−5 )2
¿√(−2)2+(−9)2
¿√85
¿9.2
(5,0) (0,6)
¿√(0−5)2+(6−0)2
¿√(−5)2+(6)2
¿√61
¿7.8
#4Prove that the triangle with vertices A (2 , 1) B (6 , 9) and C (-2 , 3) is a right triangle and find it’s area.
d=√¿¿¿
¿√ (6−2 )2+ (9−1 )2
¿√ (4 )2+ (8 )2
¿√80
❑BA
=8.9
m1=(2 ,1 )(6 ,9)
¿y2− y1
x2−x1
¿ 9−16−2
¿ 84
¿ 21
m2=(−2 ,3 ) (2 ,1 )
¿y2− y1
x2−x1
¿ 1−32+2
¿ −12
A=12bh
¿ 12
(4.5 ) (8.9 )
¿ 12
(40.05 )
Therefore the with vertices A (2 , 1) B (6 , 9) C (-2 , 3) is a right triangle because BA is ┴ to CA and ∟A is 90° and the area of the is 20.03 units.
#5Find the area of a whose vertices are:
(2,3) (5,7) (4,-2)
12
X Y 12 3 15 7 17 -2 1
A.
27 1
-35 1
+15 7
-2 1 4 1 4 -2
B.
2 7(1)+2(1) -3 5(1)-(1) +1 5(2)-4(7)
C.
2 7 + 2 -3 5-4 +1 -10 - 28
D.
2 9 -3 1 +1 -38
E. 12
-23
F.12
23
G. 11.5 units Area = 11.5 Units
FINALS
1. Give the center and radius of the circle determined by its equation. Sketch the graph.
a. ¿ ( x−5 )2+( y−1 )2=10
c= (5 ,1 )
r=√10
b. (x−12 )
2
+( y+ 14 )
2
= 494
c=( 12,−14 )
r=72
c. 5(x+3)2+5( y−2)2=45
c= (3 ,−2 )
r=?
2. Find the equation of the circle.
a. c= (0 ,3 )r=2
( x−0 )2+ ( y−3 )2=22
( x−0 )2+ ( y−3 )2=4
x2+ ( y−3 )2=4
b. c= (0 ,1 )∧passes through(0 ,212 )
d=√(2.5+1 )2
¿√ (3.5 )2
¿√12.25
r=3.5∨72
x2+ ( y+1 )2=12.25
c. c=( 14,1)d=1
23
(x−14 )
2
+( y−1 )2=53
(x−13 )
2
+( y−1 )=0.70
SEATWORKSMIDTERM
1. The Point (0,1) is the midpoint of a line segment joining the (Y,4) and (-3,Y)
O=12
( X1+(−3)) 1=12(Y 2+4)
O=X1+(−3)
21=1
2(Y 2+4)
O=X1−3
21=
Y 2+4
2
O=X1−3 2=Y 2+4
O+3=X1 2−4=Y 2
X1=3 Y 2=−2
Therefore (3,4) Therefore (-3, 2)
#2
(2 , 5) (4 , 2)
¿ 2−54−2
m=−32
m=23
#3
The Slope of the ┴ to the line through (-3 , -2) (x , 2) is −74
m2=47= 4X+3
4 X+12=28
4 X=28−12
4 X=164
X=4
#4A= (5 , 4) B= (-5 , 8) C = (-7 , 3) D = (3 , 1)
(5 , 4) (3 , -1) (-5 , 8) (-7 , 3) Therefore, AB is ║ to CD because they
have the same slope which is 52
and AD is ║to BC
m1=Y 2−Y 1
X2−X1
m2=Y 2−Y 1
X2−X1
¿ −1−43−5
¿ 3−8−7+5
¿ 52
¿ 52
#5 A = (5 , 2) B = (-5 , 1) C = (-2 , -1)
(5 , 2) (-5 , 1)
m1=Y 2−Y 1
X2−X1
¿ 1−2−5−5
❑AB
= 110
tan A=m2−m1
1+m2m1
¿
37− 1
10
1+37+( 1
10 )
¿
30−770
1+3
70
¿
23707370
(-5 , 1) (-2 , 1)
m2=Y 2−Y 1
X2−X1
¿ −1−1−2+5
❑BC
=−23
tanB=m2−m1
1+m2m1
¿
110
−(−23 )
1+ 110 (−2
3 )
¿
3+2030
1+(−230 )
¿
23302830
(5 , 2) (-2 , -1)
m3=Y 2−Y 1
X2−X1
¿ −1−2−2−5
❑AC
=37
tanC=m2−m1
1+m2m1
¿
−23
−37
1+−23 ( 3
7 )
¿
−14−921
1+(−621 )
¿
−2321
21−621
¿ 2373
∨17.49 ° ¿ 2328
∨39.40 ° ¿ −2315
∨123.11°
#6
Given: ( 12,−14 ) (3 ,4 ), find the equation of a line.
→ y+14=
4+ 14
3−12
(x−12 )
→ y+ 14=16+1
46−1
2
(x−12 )
→ y+ 14=17
10 (x−12 )
→ y+14=
17 x−172
10
→4 y+1
4=34 x−17
20
→80 y+20=136 x−68
→O=136 x−68−80 y−20
→O=136 x−80 y−88
O=17 x−10 y−11
#7A. Reduce the equation to slope – intercept form, find the m and b.1. 2 x=7 y
y=mx+b
y=2x+0
m=2
b=0
2. x4+ y
2=1
y=mx+b
21 ( y2 =−x
4+1) 2
1
y=−x2
+2
m=−12
b=2
B. Find the equation of line if the line passes through: (23
, 0) (0 , −34
)
xa+ yb=1
x23
+ y−34
=1
3x2
−4 y3
=1
9 x−8 y6
=1
9 x−8 y−6=0
Seatwork #1
Draw the circle and find the standard form.
1. Center (2 , 1) r = 3
a=¿2. Center ( 1
2 ) ,1 d=212
r2=(x−b)2+( y−k )2
2516
=(x−12 )
2
+ ( y−1 )2
3. Center (0 , -2) passes through (3 , 1)
d=√(x2−x1)2+( y¿¿2− y1)
2 ¿
¿√ (3−0 )2+ (1+2 )2
¿√9+9
¿√18
r2=(x−h)2+( y−k )2
(√18 )2=(x−h)2+( y−k )2
18=(x )2+( y+2)2
#2
1. x2+ y2+6 x−7=0
→x2+ y2+6 x−7=0
(x¿¿2+6 x+a)+( y¿¿2+0)=7¿¿
(x+3)2+( y+0)2=7+9
¿
c=(−3 ,0)
r=4
2. 4 x2+4 y2+8 x−16 y−29=0
4 x2+8 x+4 y2−16 y=294
(x¿¿2+2 x+1)+( y¿¿2−4 y+4)=294
+1+4 ¿¿
(x+1)2+( y−2)2=494
c= (−1 ,2 )
r=72
#3 (6 , 7) (1 , 8) r = 5
(−6 ,7 )→¿
(36+12h+h¿¿2)+(49−14k+k¿¿2)=25¿¿
h2+k2+12h−14k+36+49=25
h2+k2+12h−14k=25−85
h2+k2+12h−14k=−60
(1 ,8 )→ (1−h )2+ (8−k )2=52
(1−2h+h2 )+(64−16k+k¿¿2)=25¿
h2+k2−2h−16 k+1+64=25
h2+k2−2h−16 k=25−65
h2+k2−2h−16 k=−40
[h2+k2+12h−14k=−60 ]−1
h2+k2−2h−16 k=−40
→h2−k 2−12h+14k=60h2+k2−12h−16k=−40−14h−2k=203 rd eq .
−14h−2k=20−2
¿7h+k=−10
k=−10−7 h→tentative value of k
GROUP SEATWORKS
Number 1
Parallelogram (6 , 0) (2 , 5) (-3 , -1) and (1, -6) find the point of intersection
(2 ,3 ) (1 ,−6 )
x=12(2+1)
¿ 13(3)
¿112∨1.5
y=12(5−6)
¿ 12(−1)
¿ −12
∨−.5
1st point (1.5 , -0.5)
(6 , 0) (-3 , -1)
x=12(6−3)
¿ 12(3)
¿112∨1.5
y=12(0−1)
¿ 12(−1)
¿ −12
∨−0.5
2nd point (1.5 , -0.5)
Therefore the points of intersection is (1.5 , -0.5)
#2Given: (-7 , 4) and (6 , -4) divided into 3 equal parts.
x=x1+r (x¿¿1−x1)¿
¿−7+ 13(6+7)
¿−7+ 13(13)
¿ −7+133
¿ −83
y= y1+r ( y¿¿2− y1)¿
¿4+ 13(−4−4)
¿4+ 13(−8)
¿12+(−8)
3
¿ 43
x=12( x2+x1)
¿ 12 (−8
3+6)
¿ 12 (−8+18
3 )
y=12( y¿¿2+ y1)¿
¿ 12 ( 4
3+ (−4 ))
¿ 12 ( 4−12
3 )
¿ 12 ( 10
3 )¿ 5
3
¿ 12 (−8
3 )¿ −4
3
Therefore the two points are
(−83
,43 )∧( 5
3,−43 )consecutively .
ASSIGNMENTS
#1
Find CF, AC, FB, & FC
CF=x2−x1
¿2+1
¿3
AC=x2−x1
¿−1+3
¿−2
FB=x2−x1
¿−2−2
¿−4
FC=x2−x1
¿−1−2
¿−3
EXERCISES
#1
1. If AB = 3, then find AC, BC & CA
AC=x2−x1
¿6−1
¿5
BC=x2−x1
¿6−4
¿2
CA=x2−x1
¿1−6
¿−5
2. Find BC, AC and CA
BC=x2−x1
¿4+1
AC=x2−x1
¿4−0
CA=x2−x1
¿0−4
¿5 ¿4 ¿−4
#2
1. Find the distance between 2 points (6 , 3) (4 , 3)
d=√(x¿¿2−x1)2+( y¿¿2− y1)
2 ¿¿
¿√ (4−6 )2+ (3−3 )2
¿√4
¿2
2. Find the distance between 2 points (-3 , 1) (9 , 6)
d=√(x¿¿2−x1)2+( y¿¿2− y1)
2 ¿¿
¿√ (9−3 )2+ (6−1 )2
¿√ (12 )2+ (5 )2
¿√169
¿13
#3
1.m=y2− y1
x2−x1
2.d=√(x¿¿2−x1)2+( y¿¿2− y1)
2 ¿¿
3.m=12( y¿¿2+ y1)m=1
2¿¿¿
4. x=x1+r (x¿¿2−x1)¿
y= y1+r ( y¿¿2− y1)¿
5. (4 ,1 )∧ (x ,3 )
−2 x+8=6
−2 x=6−8
−2x=−2−2
x=1m=−23
Republic of the PhilippinesSORSOGON STATE COLLEGE
Sorsogon City CampusSorsogon City
COURSE SYLLABUSNALYTIC GEOMETRY AND CALCULUS
Area: College of EducationYear Level: 3rd Year / BEEDSemester: 2nd Semester / S.Y. 2009 – 2010
I. COURSE DESCRIPTION
The course with related topics on lines, circles and conic sections plus description of their properties are discussed. In calculus, elaboration or concepts of functions and its properties in terms of definite and integral components are likewise expounded.
Pre-requisite: Math 1 for Physical ScienceCredits:3 unitsTime duration: 54 hours
II. OBJECTIVESA. General
To develop a better understanding of the concept of analytic geometry for education students. It provides the students the basic concepts, which are needed in calculus and in many other areas of mathematics. It helps the students analyze and solve problems involving lines, circles and conic sections. Moreover, it familiarize students of the different coordinate systems and provides them skill in applying such systems in the solution of real world problems.
B. Specific1. Determine the different form of equations of lines, circles and conics.2. Reduce equation of a line into its normal form.3. Determine the direct distance from a line to a point.4. Acquire and develop critical – thinking skill through problem solving.5. Write equations of the family lines, circles and conics.6. Obtain the derivatives and antiderivatives of functions.
III. COURSE CONTENTUnit I. Fundamental Concepts 10 hours
1.1 Directed Line Segment1.2 Cartesian Coordinate System1.3 Distance between two points1.4 Division of a Line Segment1.5 Slope of a Line
1.5.1 Slope of Perpendicular Lines1.5.2 Slope of Parallel Lines1.5.3 Angle Between Two Lines
Unit II. The Lines2.1 Equations of Line 10 hours
2.1.1 Two Point Form2.1.2 Intercept Form2.1.3 Slope Intercept Form2.1.4 Other Form of Equation of Lines2.1.5 Family of Lines
2.2 Directed Distance from a Line to a Point
Unit III. The Circle 13 hours3.1 The Circle
3.1.1 Standard Form of the Equation of a Circle3.1.2 Equation of a Circle with a center at the Origin
3.1.3 General Equation of a Circle3.1.4 Circles Determine by Geometric Conditions
Unit IV. The Conic Section 15 hours4.1 The Parabola
4.1.1 Parabola with Vertex at the Origin4.1.2 Parabola with Vertex at (h , k)
4.2 Ellipse4.2.1 Ellipse with Center at the Origin4.2.2 Ellipse with Center at (h , k)
4.3 Hyperbola4.3.1 Hyperbola with Center at the Origin4.3.2 Hyperbola with Center at (h , k)
Unit V. Differential Calculus 6 hours5.1 Functions5.2 The Derivatives
5.2.1 The Derivatives of a Function5.2.2 Rules of Derivation5.2.3 Derivatives of Higher Order5.2.4 Derivatives of Products and Quotients
IV. METHODOLOGY / TEACHING STRATEGIESLecture / DiscussionsDiscoveryGroup Activity / Peer TeachingBoard Work / Seat Work / Problem Solving Approach
V. REQUIREMENTSMidterm and Final ExaminationQuizzes / AssignmentProject (worksheet compilation) / Seminar WorkshopRecitation
VI. GRADING SYSTEMSQuizzes / Long Tests 20%Recitation 20%Midterm and Final Exam 30%Project 20%Attendance / Behavior 10% Total 100%
VII. REFERENCES1. Modern Analytic Geometry by Feliciano and Uy2. Analytic Geomety by Gordon Fuller3. Analytic Geometry by Aguaviva and Yap4. Introduction to Analytic Geometry and Calculus by Deuna and Lamayo5. Analytic Geometry by Quirino and Mijares
Prepared by:
MS. ELLA H. GREFALDOInstructor I
Approved by:
DR. RITZELDA A. DERI
DIR. DERGIO DEYTO
Campus Director