Ejercicios resueltos de geometria analítica.pdf
Short Description
Ejercicios resueltos de geometría analítica con sus respectivas gráficas....
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( 3, 1)
m =
−
−
1 2
( 3, 1) ,(0, 3) ,(3, 4) ,(4, 1) ( 2, 1) ,(2, 2) ,(5, 2)
BC
− −
− − − (2, −2) ,(−8, 4) ,(5, 3) (12, 1) ,(−3, −3) ,(2, −1) (0, 1) ,(3, 5) ,(7, 2) ,(4, −2) A (3, 8) ,B (2, −1) ,C (6 (6, −1)
−
D
AD (1, 1) ,(3, 5) ,(11, 6) ,(9, 2)
p =
a+b+c 2
area a ´ rea =
p ( p
−
(0, 0) ,(1, 2) ,(3, 4) a) ( p b) ( p c)
−
−
−
(Consulta)
A ( 1, 1) ,B (3, 5) ,C (5 (5,
−
−3)
A = 0, C = 0, D=0 =0, C y 2 + Dx + Ey + F = 0
V ´ ertice ertice
p
( 3, 1)
−
tan θ = m =
Cateto opuesto opuesto Cateto adyacente adyacente
=
− 12
m =
−
1 2
( 3, 1) ,(0, 3) ,(3, 4) ,(4, 1)
− −
P 1 ( 3,
− −1)
P 2 (0, 3) P 3 (3, 4) P 4 (4,
−1)
−
− − −
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
= 5 = a
X 1 )2 + (Y 2
(X 2
(0
( 3))2
2
− Y ) + (3 − (−1)) 1
2
(0 + 3)2 + (3 + 1)2 (3)2 + (4)2
√ 9 + 16 √ 25
− −
dP 2 P 3
=
(X 2
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
= 3, 2 = b
(3
X 1 )2 + (Y 2
0)2 + (4
− 3)
2
− Y ) 1
2
(3)2 + (1)2
√ 9 + 1 √ 10
− − −
X 1 )2 + (Y 2
dP 3 P 4
=
(X 2
dP 3 P 4
=
dP 3 P 4
=
dP 3 P 4
=
dP 3 P 4
=
dP 3 P 4
= 5, 1 = c
(4
2
− Y ) 1
3)2 + ( 1
2
− − 4)
(1)2 + ( 5)2
√ 1 + 25 √ 26
− − −
dP 4 P 1
=
(X 2
dP 1 P 4
=
dP 1 P 4
=
dP 1 P 4
=
dP 1 P 4
=
dP 1 P 4
=
dP 1 P 4
= 7 = d
X 1 )2 + (Y 2
2
− Y ) (4 ( 3)) + (−1 − (−1)) (4 + 3) + (−1 + 1) 2
2
(7)2 + (0)2
√ 49 + 0 √ 49
1
2
2
P er´ımetro ımetro
= a+b+c+d
P er´ımetro ımetro = 5 + 3 + 3, 2 + 5 , 1 + 7 P er´ımetro ımetro = 20, 3
( 2, 1) ,(2, 2) ,(5, 2)
− −
P 1 ( 2,
− −1)
P 2 (2, 2) P 3 (5,
−2)
−
− − −
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
= 5 = a
(2
25
− − −
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
= 5 = b
=
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
X 1 )2 + (Y 2
(X 2
2
− Y ) 1
2)2 + ( 2
(5
2
− − 2)
(3)2 + ( 4)2
√ 9 + 16 √ 25
− − − −
X 1 )2 + (Y 2
(X 2
2
− Y ) (5 ( 2)) + (−2 − (−1)) (5 + 2) + (−2 + 1) 1
2
2
2
2
(7)2 + ( 1)2
√ 49 + 1 √ 50
dP 1 P 3
=
(25) (2) (2) √ (25)
dP 1 P 3
= 5 2 = c
a = b
(2, 2) ,( 8, 4) ,(5, 3)
−
P 1 (2,
−2)
2
√ 16 + 9 √
dP 2 P 3
dP 1 P 3
1
(4)2 + (3)2
=
=
( 2))2
2
− Y ) + (2 − (−1))
(2 + 2)2 + (2 + 1)2
dP 2 P 3
dP 1 P 3
X 1 )2 + (Y 2
(X 2
−
P 2 ( 8, 4)
−
P 3 (5, 3)
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p1 p3
=
m p1 p3
=
m p1 p3
=
m p1 p3
=
y2 x2
−y −x 4 − (−2) −8 − 2 4+2 −10 − 106 − 35 = m 1
1
1
y2 x2
−y −x 3 − (−2) 5−2 1
1
3+2 3 5 = m 3 3
m1
m3
(12, 1) ,( 3, 3) ,(2, 1)
− −
P 1 (12, 1) P 2 ( 3,
− −3) P (2, −1) 3
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p1 p2
=
y2 x2
−y −x −2 − 1 −3 − 12 −3 −15 1
1
3 15 1 = m 1 5
−
m p2 p3
=
m p2 p3
=
m p2 p3
=
m p2 p3
=
y2 x2
−y −x −1 − (−2) 2 − (−3) −1 + 2 1
1
2+3 1 = m 2 5
m1 = m 2
(0, 1) ,(3, 5) ,(7, 2) ,(4, 2)
−
P 1 (0, 1) P 2 (3, 5) P 3 (7, 2) P 4 (4,
−2)
m p1 p2
=
m p1 p2
=
m p1 p2
=
m p2 p3
=
m p2 p3
=
m p2 p3
=
m p2 p3
=
m p3 p4
=
m p3 p4
=
m p3 p4
=
m p3 p4
=
m p1 p4
=
m p1 p4
=
m p1 p4
=
m p1 p4
=
m1 y m2
y2 x2
−y −x 5−1 3−0
1 1
4 = m 1 3 y2 x2
−y −x 2−5 7−3 −3
1 1
4 3 = m 2 4
−
y2 x2
−y −x −2 − 2 4−7 −4 −3
1 1
4 = m 3 3 y2 x2
−y −x −2 − 1 4−1 −3
1 1
4 3 = m 4 4
−
m3 y m4
A (3, 8) ,B (2, BC
−1) ,C (6 (6, −1)
D AD
D
BC
P m (X m , Y m) = P m
P mBC = P mBC = P mBC = P mBC = P mBC = P mBC = D (4,
−1)
x1+ x2 y 1 + y2
2
;
2
− − −− − − x1+ x2 y 1 + y2
; 2 2 2 + 6 1 + ( 1) ; 2 2 8 1 1 ; 2 2 2 4; 2 2 4; 2 (4; 1) = D
−
AD
dAD
=
dAD
=
dAD
=
dAD
=
dAD
=
− − −
X 1 )2 + (Y 2
(X 2
(4
3)2 + ( 1
2
− Y ) 1
2
− − 8)
(1)2 + ( 9)2
√ 1 + 81 √ 82
(1, 1) ,(3, 5) ,(11, 6) ,(9, 2)
P 1 (1, 1) P 2 (3, 5) P 3 (11, 6) P 4 (9, 2)
m1 = m3
−y −x 5−1 3−1
1
=
m p1 p2
=
m p1 p2 m p1 p2
4 2 = 2 = m 1
m p2 p3
=
1
=
y2 x2
−y −x 6−5 11 − 3
1 1
m p2 p3
=
m p2 p3
=
1 = m 2 8
m p3 p4
=
y2 x2
−y −x 2−6 9 − 11 −4 −2
1 1
m p3 p4
=
m p3 p4
=
m p3 p4 m p3 p4
4 2 = 2 = m 3
m p1 p4
=
m p1 p4
=
m p1 p4
=
=
y2 x2
−y −x 2−1 9−1
1 1
1 = m 4 8
m2 = m4
(0, 0) ,(1, 2) ,(3, 4) ´rea = p ( p a) ( p area a
y2 x2
m p1 p2
− −
p =
p =
− b) ( p − c)
a+b+c
2
´rea = area a
p ( p − a) ( p − b) ( p − c)
a+b+c 2
P 1 (0, 0) P 2 (1, 2) P 3 (3,
−4)
− −
dP 1 P 2
=
(X 2
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
=
dP 1 P 2
= 2, 2 = a
(1
X 1 )2 + (Y 2
0)2 + (2
− 0)
2
− Y ) 1
2
(1)2 + (2)2
√ 1 + 4 √ 5
− − −
X 1 )2 + (Y 2
dP 2 P 3
=
(X 2
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
=
dP 2 P 3
= 2 10
dP 2 P 3
= 6, 3 = b
(3
1)2 + ( 4
1
2
− − 2)
(2)2 + ( 6)2
√ 4 + 36 √ √ 40
2
− Y )
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
dP 1 P 3
=
− − − −
X 1 )2 + (Y 2
(X 2
2
− Y ) (5 ( 2)) + (−2 − (−1)) (5 + 2) + (−2 + 1) 1
2
2
2
2
(7)2 + ( 1)2
√ 49 + 1 √ 50
dP 1 P 3
=
(25) (2) (2) √ (25)
dP 1 P 3
= 5 2 = c
´rea = area a p
p ( p
p =
− a) ( p − b) ( p − c)
a+b+c
2 2, 2 + 6 , 3 + 5 p = 2 13, 5 p = 2 p = 6, 8 p
area a ´rea =
´rea = area a ´rea = area a area a ´rea =
p ( p
− a) ( p − b) ( p − c) 6, 8 (6, 8 − 2, 2)(6, 8 − 6, 3)(6, 8 − 5) 6, 8 (4, 3)(0, 2)(1, 5) 8, 8
area a ´rea = 2, 9
(Consulta )
α
β
α < β < 90
β = = 90
α = β
α > β
A ( 1, 1) ,B (3, 5) ,C (5 (5,
−
−3)
x2 + y 2 + Dx + Ey + F = 0 A ( 1, 1)
−
( 1) + (1) + D ( 1) + E (1) + F = 0
−
−
C y 2 + Dx + Ey E y + F = 0
A = 0, C = 0, D =0 =0, p
V ´ e´rtice ertice
C y2 + Dx + Ey + F = 0 C
(C y 2 + Dx + Ey + F = 0) C y2 + Dx C
= D
Dx Ey F + + = 0 C C C
Ey C
F C
= E ,
= C
y 2 + Dx + Ey + F = 0
2
y + Ey +
2
E
2
y + y + y +
E E E
−F − − Dx +
=
−F − − Dx + E 4
2
=
−
F E 2 D x+ + D 4D
−
4 F E 2 D x+ 4D
2
2
2
2
2
2
E
=
2
2
=
(y
− −
2
− k)
= 4 p (x
= Ecuacio´n de la par abola ´bola a
− h)
v ertice e´rtice V = (h, k ) v ertice e´rtice = p
− − E
2
;
4 p = p =
4F E 2 4D
−D
− D4
− −
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