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ANSWER KEY: LESSON 1: USING THE DISTANCE FORMULA IN PROVING GEOMETRIC PROPERTIES ACTIVITY 1 1 .TRUE 2. TRUE 3. TRUE 4. (7,9) 5. (-8,9) 6. TRUE
- TRUE 8. 5.8 9. TRUE 10. 12.6 11. 8.6 12. (-8,9)
- (-5,4) 14. TRUE 15. TRUE ACTIVITY 2 THE WORD IS ( ACTION) PART II.
- E 2. A 3. D 4. B 5. C Activity 3 A. 1. D ¿) 2. C ( 0 , 0 ) , B (− 2 a , b ) c. B (− a ,d ) ,C (− b ,c ) , D (− a , 0 )
- D (− a , 0 ) ; A (− a , b ) ; B ( a , b ) B. 1. M (− x , 0 ) ; R ( x , 0 ) ; B (0,4) 2. A ( 0 , d ) ; B ( 0 , 0 ) , C ( d , 0 )
- M (− 3 a , 0 ) A (− 2 a , 5 b ) ; T ( 2 a , 5 b ) , E ( 3 a , 0 ) 4. B ( 1 , 6 ) ; E ( 6 , 6 ) ; N ( 5 , 0 ) ;Y (0,0)
- (^) H (− x , 0 ) ; O ¿ 6. A ( 0 , x ) , B (− x , 0 ) ; C ( 0 , − x ) ; D ( x , 0 ) Activity 4
- The segment joining the midpoints of two sides of a triangle is half as long as the third side parallel to it. Let B ( 0 , 0 ) ; A ( 0 , 2 a ) , C ( 2 b , 0 ) E is the midpoint of AB and D is the midpoint of AC By midpoint Formula: E =
0 + 2 a
=( 0 , a )
D =(
0 + 2 B
2 a + 0
=( b , a ) Getting the distance of ED and BC to determine the length:
DE =√( b − 0 )
2 +( a − a ) 2 = b
BC =√( 2 b − 0 )
2 +( 0 − 0 ) 2
=√ 4 b
2 = 2 b DE = b and BC = 2 b ; DE^ is^
BC .
- Diagonals of a rectangle are equal Let A (− 2 a , b ) ; B ( 2 a , b ) ; C ( 2 a , 0 ) ; D (− 2 a , 0 ) Using the Distance Formula to find the length of each diagonal:
AC =√( 2 a + 2 a )
2
=√ 4 a
2
BD =√( 2 a + 2 a )
2 +( b − 0 ) 2
=√ 4 a
2
Since AC = BD =√ 4 a^2 + b^2 then the diagonals are equal.
- The median of a trapezoid is equal to half the sum of the two bases. Let A (− a , b ). B ( a ,b ) ; C ( c , 0 ) ; D (− c , 0 ) ; E
− c + a , b
; F ( c − a , b 2
Get the sum of the distances of AB and CD. Compare to the sum of the distance of EF
AB =√(− a − a )
2 +( b − b ) 2
=√ 4 a
2 = 2 a
CD =√( c + c )
2 +( 0 ) 2
=√ 4 c
2 = 2 c AB + CD = 2 a + 2 c EF =
( c − a )−(− c + a ) 2
b 2
b
2
=√( 2 c − 2 a )
2 = 2 c − 2 a
- Diagonals of a rhombus bisect each other Let B ( 0 , 0 ) , A ( c , b ) , D ( a + c ,b ) ; C ( a , 0 ) Get the midpoint of the two diagonals. If they are the same then the midpoint is the point of intersection Let E be the midpoint of AC
E =(
c + a 2
0 + b
c + a 2
b 2
Let E be the midpoint of BD
E =(
0 + c + a 2
0 + b
c + a 2
b 2
Since they have the same midpoint, then Point E is the point of intersection LESSON 2 : EQUATION OF A CIRCLE ACTIVITY 1
- (^) x^2 + y^2 = 25
- ( x － 4 ) 2 +( y － 2 ) 2 = 64
- (^) ( x + 3 )^2 +( y － 6 )^2 = 16
- (^) ( x + 5 )^2 +( y + 1 )^2 = 36
- (^) ( x −
2 +( y + 5 ) 2 =
- (^) ( x − 3 )^2 +( y +
2
- x 2 +( y + 8 ) 2 =
- (^) ( x − 3 )^2 + y^2 = 4
- ( x^ －^3 ) 2 +( y + 8 ) 2 =
- (^) ( x －
2 +( y −
2
- (^) ( x − 3 )^2 +( y − 2 )^2 = 16 ( 3 , 2 ) 4
- (^ x^ +^3 ) 2
12. ( x − 6 )^2 +( y − 1 )^2 = 24 ( 6 , 1 ) 2 √ 6
13. x^2 +( y − 7 )^2 = 50 ( 0 , 7 ) 5 √ 2
- (^) ( x −
2 +¿ (^
15. ( x − 2 )^2 + y^2 = 75 ( 2 , 0 ) 5 √ 3
ACTIVITY 4
- (^) ( x − 1 ) 2 + ¿;^ C^ (^1 ,^^4 ) ;^ R =^8
- ¿; C (
) ; R =
3. x^2 + ¿; C ( 0 , − 4 ) ;R =√ 68
4. ( x + 6 ¿ ¿^2 +¿; C (− 6 , − 3 ) ; R =√ 68
5. ¿; C (− 5 , − 2 ) ; R = 4 √ 2
6. ( x + 3 )^2 + ¿; C (− 3 , − 2 ) ;R =√ 3
7. ( x + 4 )^2 +( y + 2 )^2 = 30 ; C (− 4 , − 2 ) ; R =√ 30
8. ( x + 6 )^2 + ¿; C (− 6 , − 3 ) ; R =√ 33
9. ( x + 1 )^2 +¿; C (− 1 , − 1 ) ; R =√ 7
10. ( x + 4 )^2 +( y + 6 )^2 = 44 ; C (− 4 , − 6 ) ; R = 2 √ 11
LESSON 3:GRAPHING CIRCLE ACTIVITY 1 Part I.
- out 2. Out 3. In 4. On 5. In 6. Out 7. In 8. On 9. In 10. Out 11. Out
- in 13. Out 14. Out 15. Out ACTIVITY 2 Part I
- C ( 0 , 0 ) ;r = 10 circle.
- C(0, -6); r = 0 point circle.
- C(6, 0); r = 5 √^3 circle.
- C(4, 2); r = 9 circle
- C(-5, 5); r = 8 circle.
- C(-2, 3); r = 0 point circle.
7. C(0, -4); r = √− 9 no circle.
- C(-3, -3/2); r =
√^39
It has a graph.
9. C(-1, 3); r =√− 2 no circle
10. C(0, 3); r =√ 13 circle.
Part II
