1 improvement of tcp congestion window over ltetanawan44
?
This document discusses improving the performance of TCP congestion control over LTE-Advanced networks. It proposes a new congestion avoidance mechanism that uses the available bandwidth of the connection to better detect the network path capacity and improve congestion avoidance. The mechanism is tested using the NS-2 network simulator to model LTE-Advanced traffic. The document provides background on LTE-Advanced network architecture and existing TCP congestion control mechanisms. It aims to develop an enhanced TCP variant that can efficiently transfer high data rates over the large bandwidth, low latency links of LTE-Advanced networks.
The document provides an overview of routing basics, including: what routers do in finding paths and forwarding packets; the difference between routing and forwarding; how IP route lookup works using longest prefix matching; how routing information databases (RIBs) and forwarding information bases (FIBs) are used; explicit versus default routing; and an introduction to autonomous systems, routing policies, interior gateway protocols (IGPs), exterior gateway protocols (EGPs) like BGP, and how routing and traffic flows work within and between autonomous systems.
Cisco identity services engine (ise) ordering steps & guideIT Tech
?
The document provides information on ordering Cisco Identity Services Engine (ISE) appliances and licenses. It outlines 6 steps for ordering ISE: 1) estimate concurrent endpoints, 2) estimate needed appliances, 3) select appliance type, 4) select support level, 5) select license type, and 6) select services. It describes ISE appliance and virtual appliance options, support options, and function-based and deployment-based licensing options.
The document provides specifications for various Cisco ASA Firepower models, including available RAM and CPU resources for both the ASA and Firepower modules. It lists specifications for the ASA 5505 through 5585-X models as well as the AIP SSC 5, SSM 10, SSM 20, and SSM 40 security modules. RAM available to the ASA and Firepower modules varies between models as does CPU cores and type. The highest-end models include the ASA 5585-X with up to 49GB of RAM and 48 CPU cores accessible to the ASA and Firepower modules.
A VLAN allows a network administrator to logically segment a local area network (LAN) to create multiple broadcast domains on a single physical switch. Devices placed in the same VLAN can communicate via broadcasts within that VLAN but broadcasts are restricted from other VLANs, increasing security. VLANs are commonly used when a network has over 200 devices, heavy broadcast traffic, or different user groups need isolation or access to the same applications across multiple switches. While subnetting a network provides some segmentation, VLANs allow devices in different physical locations and switches to remain on the same logical network without routing.
1 improvement of tcp congestion window over ltetanawan44
?
This document discusses improving the performance of TCP congestion control over LTE-Advanced networks. It proposes a new congestion avoidance mechanism that uses the available bandwidth of the connection to better detect the network path capacity and improve congestion avoidance. The mechanism is tested using the NS-2 network simulator to model LTE-Advanced traffic. The document provides background on LTE-Advanced network architecture and existing TCP congestion control mechanisms. It aims to develop an enhanced TCP variant that can efficiently transfer high data rates over the large bandwidth, low latency links of LTE-Advanced networks.
The document provides an overview of routing basics, including: what routers do in finding paths and forwarding packets; the difference between routing and forwarding; how IP route lookup works using longest prefix matching; how routing information databases (RIBs) and forwarding information bases (FIBs) are used; explicit versus default routing; and an introduction to autonomous systems, routing policies, interior gateway protocols (IGPs), exterior gateway protocols (EGPs) like BGP, and how routing and traffic flows work within and between autonomous systems.
Cisco identity services engine (ise) ordering steps & guideIT Tech
?
The document provides information on ordering Cisco Identity Services Engine (ISE) appliances and licenses. It outlines 6 steps for ordering ISE: 1) estimate concurrent endpoints, 2) estimate needed appliances, 3) select appliance type, 4) select support level, 5) select license type, and 6) select services. It describes ISE appliance and virtual appliance options, support options, and function-based and deployment-based licensing options.
The document provides specifications for various Cisco ASA Firepower models, including available RAM and CPU resources for both the ASA and Firepower modules. It lists specifications for the ASA 5505 through 5585-X models as well as the AIP SSC 5, SSM 10, SSM 20, and SSM 40 security modules. RAM available to the ASA and Firepower modules varies between models as does CPU cores and type. The highest-end models include the ASA 5585-X with up to 49GB of RAM and 48 CPU cores accessible to the ASA and Firepower modules.
A VLAN allows a network administrator to logically segment a local area network (LAN) to create multiple broadcast domains on a single physical switch. Devices placed in the same VLAN can communicate via broadcasts within that VLAN but broadcasts are restricted from other VLANs, increasing security. VLANs are commonly used when a network has over 200 devices, heavy broadcast traffic, or different user groups need isolation or access to the same applications across multiple switches. While subnetting a network provides some segmentation, VLANs allow devices in different physical locations and switches to remain on the same logical network without routing.
This 3-day routing workshop covers key routing concepts including IP routing, routing protocols, IPv6 addressing, and hands-on exercises. Participants will learn about interior routing protocols like OSPF, exterior routing protocols like BGP, routing attributes, path selection, and scaling techniques. Hands-on labs will provide experience with router configuration, OSPF, iBGP, eBGP, route reflectors, and Internet exchange policies. The goal is to give attendees a solid foundation in routing fundamentals and operations.
The document provides information on explaining processes through written texts. It discusses the key features of explanation texts, including presenting information in a logical sequence, using causal language, diagrams, and other visual aids. The document also provides examples of outlines and concept maps that can be used to organize explanations in a clear, step-by-step manner.
Children in Germany attend grade school from ages 6 to 10, focusing on basic subjects. From ages 10 to 15, they attend basic school where they begin to specialize. High school then runs from ages 15 to 18, preparing students for university or vocational training.
1. ?N T?P V? PH??NG TR?NH ???NG TH?NG TRONG M?T PH?NG
So?n: L?u H?i V?nh ¨C GV °Õ´Ç¨¢²Ô Tr??ng THPT NG
I/ L? thuy?t
1/T?a ??: H? t?a ?? Oxy hay (O, ,i j
r r
)
* T?a ?? c?a ?i?m; v¨¦c t?: M(x;y) ( ; ) . .OM x y OM x i y j? = ? = +
uuuur uuuur r r
* ?? d¨¤i c?a m?t v¨¦c t?; ?o?n th?ng:
2 2 2 2
( ) ( ) ( ) ( )A B A B B A B AAB AB x x y y x x y y= = ? + ? = ? + ?
uuur
* Hai v¨¦c t? b?ng nhau:
1 2
1 1 2 2
1 2
( ; ) ( ; )
x x
a x y b x y
y y
=?
= ? ?
=?
r r
* C¨¢c ph¨¦p ³Ù´Ç¨¢²Ô v? v¨¦c t?: Cho 1 1 2 2( ; ), ( ; )a x y b x y
r r
1 2 1 2 1 2 1 2
1 1 1 2 1 2
1 2
1 2
1 2 1 2
1 2 1 2
2 2 2 2
1 2 1 2
/ ( ; ) / ( ; )
/ . ( . ; . ) / . . .
.
/ cung phuong ; 0 :
.
/ . . 0
. . .
/ os( ; )
. .
a b x x y y a b x x y y
k a k x k y a b x x y y
x t x
a b b t
y t y
a b x x y y
a b x x y y
c a b
a b x x y y
+ + = + + + ? = ? ?
+ = + = +
=?
+ ¡Ù ? ? ¡Ê ?
=?
+ ¡Í ? + =
+
+ = =
+ +
r r r r
r r r
r r r r
?
r r
r r
r r
r r
* C¨¢c c?ng th?c li¨ºn quan ??n t?a ?? ?i?m:
+/ M l¨¤ trung ?i?m AB
2
0
2
A B
M
A B
M
x x
x
MA MB
y y
y
+?
=??
? + = ? ?
+? =
??
uuur uuur r
(hay v?i m?i ?i?m O;
1
( )
2
OM OA OB= +
uuuur uuur uuur
)
+/ M chia ?o?n AB theo t? s? k ( A;B ph?n bi?t; k 1¡Ù ) .MA k MB? =
uuur uuur
.
1
.
1
A B
M
A B
M
x k x
x
k
y k y
y
k
??
=?? ?
? ?
?? =
? ??
( hay v?i m?i ?i?m O;
1
( . )
1
OM OA k OB
k
= ?
?
uuuur uuur uuur
)
+/ M l¨¤ tr?ng t?m tam gi¨¢c ABC
3
0
3
A B C
M
A B C
M
x x x
x
MA MB MC
y y y
y
+ +?
=??
? + + = ? ?
+ +? =
??
uuur uuur uuuur r
(hay v?i m?i ?i?m O;
1
( )
3
OM OA OB OC= + +
uuuur uuur uuur uuur
)
* M?t s? t¨ªnh ch?t c?a tam gi¨¢c ABC:
+/ Tam gi¨¢c ABC vu?ng t?i C 2 2 2
. 0 (hay .....)CACB CA CB AB? = ? + = ?
uuur uuur
+/ Tam gi¨¢c ABC c?n t?i B .......BA BC? = ?
uuur uuur
2. +/ Tam gi¨¢c ABC vu?ng c?n t?i A
. 0AB AC
AB AC
? =?
? ?
=??
uuur uuur
uuur uuur
+ Tam gi¨¢c ABC ??u BA BC AC? = =
uuur uuur uuur
2/ Li¨ºn h? t?a ?? v¨¤ b?t ??ng th?c Bunhiacopxki:
V?i hai v¨¦c t? ( ; )u a b
r
v¨¤ ( ; )v x y
r
; ta c¨® 2 2 2 2
. ax+by
os( ; )
. .
u v
c u v
u v a b x y
= =
+ +
r r
r r
r r
2 2 2 2
2 2 2 2
2 2 2 2 2
ax+by
Do os( ; ) 1 1 ax+by .
.
(ax+by) ( ).( )
c u v a b x y
a b x y
a b x y
¡Ü ? ¡Ü ? ¡Ü + +
+ +
? ¡Ü + +
r r
D?u b?ng x?y ra os(u; ) 1c v = ¡À
r r
;u v?
r r
c¨´ng ph??ng ax=by?
3/ ???ng th?ng
a/ ???ng th?ng ?i qua M0(x0;y0) v¨¤ nh?n v¨¦c t? 2 2
( ; ) ( 0)u a b a b+ ¡Ù
r
l¨¤m v¨¦c t? ch? ph??ng c¨®
ph??ng tr¨¬nh tham s? l¨¤ :
0
0
( )
x x at
t R
y y bt
= +?
¡Ê?
= +?
* N?u 0a ¡Ù th¨¬ h? s? g¨®c c?a ???ng th?ng l¨¤ k = b/a.
b/ ???ng th?ng ?i qua M0(x0;y0) v¨¤ nh?n v¨¦c t? ( ; ) ( ; 0)u a b a b ¡Ù
r
l¨¤m v¨¦c t? ch? ph??ng c¨®
ph??ng tr¨¬nh ch¨ªnh t?c l¨¤ : 0 0x x y y
a b
? ?
= .
c/ ???ng th?ng ?i qua M0(x0;y0) v¨¤ c¨® h? s? g¨®c k ; ph??ng tr¨¬nh l¨¤ : y = k(x-x0) + y0
d/ ???ng th?ng ?i qua M0(x0;y0) v¨¤ nh?n v¨¦c t? ( ; ) ( ; 0)n a b a b ¡Ù
r
l¨¤m v¨¦c t? ph¨¢p tuy?n c¨®
ph??ng tr¨¬nh t?ng qu¨¢t l¨¤ : a(x-x0) + b(y-y0) =0
e/ ???ng th?ng ?i qua A(x1;y1) v¨¤ B(x2 ;y2) c¨® ph??ng tr¨¬nh l¨¤ :
? N?u x1 = x2 : Ph??ng tr¨¬nh l¨¤ x = x1
? N?u y1 = y2 : Ph??ng tr¨¬nh l¨¤ y = y1
? N?u
1 2
1 2
x x
y y
¡Ù?
?
¡Ù?
: Ph??ng tr¨¬nh l¨¤
1 1
2 1 2 1
x x y y
x x y y
? ?
=
? ?
f/ Ch¨² ? :
? N?u 2 2
( ; ) ( 0)u a b a b+ ¡Ù
r
l¨¤ 1 v¨¦c t? ch? ph??ng c?a (d) th¨¬ k. ( ; ) 0u ka kb k= ? ¡Ù
r
c?ng l¨¤ m?t
v¨¦c t? ch? ph??ng c?a (d).
? N?u 2 2
( ; ) ( 0)n a b a b+ ¡Ù
r
l¨¤ 1 v¨¦c t? ch? ph??ng c?a (d) th¨¬ k. ( ; ) 0n ka kb k= ? ¡Ù
r
c?ng l¨¤ m?t
v¨¦c t? ch? ph??ng c?a (d).
? N?u 2 2
( ; ) ( 0)u a b a b+ ¡Ù
r
l¨¤ 1 v¨¦c t? ch? ph??ng c?a (d) th¨¬ ( ; )n b a= ?
r
l¨¤ m?t v¨¦c t? ph¨¢p
tuy?n c?a (d).
4/ G¨®c gi?a hai ???ng th?ng
Gi? s? ¦Á l¨¤ g¨®c gi?a hai ???ng th?ng c¨® v¨¦c t? ph¨¢p tuy?n theo th? t? l¨¤ 1 2;n n
ur uur
1 2
1 2
1 2
n .
os = cos(n ; )
n . n
n
c n¦Á? =
uur uur
uur uur
uur uur (c¨® th? t¨ªnh theo v¨¦c t? ch? ph??ng)
3. 5/ Kho?ng c¨¢ch t? ?i?m M ??n (d) : ax + by +c = 0
M
2 2
ax
( ;( )) Mby c
d M d
a b
+ +
=
+
6/ ???ng ph?n gi¨¢c c?a g¨®c h?p b?i hai ???ng th?ng
Cho (d1) : a1x + b1y +c1 = 0 v¨¤ (d2) : a2x + b2y +c2 = 0
Ph??ng tr¨¬nh ???ng ph?n gi¨¢c c?a g¨®c h?p b?i (d1) ; (d2) l¨¤ :
1 1 1 2 2 2
2 2 2 2
1 1 2 2
a x b y c a x b y c
a b a b
+ + + +
=
+ +
II/ B¨¤i t?p
B¨¤i 1 : A-2010
Cho tam gi¨¢c ABC c?n t?i A(6;6), ???ng th?ng ?i qua trung ?i?m canh AB v¨¤ AC c¨® ph??ng tr¨¬nh
x + y ¨C 4 = 0. T¨¬m t?a ?? B v¨¤ C bi?t E(1;-3) n?m tr¨ºn ???ng cao ?i qua C c?a tam gi¨¢c ?? cho.
B¨¤i 2: B-2010
Cho tam gi¨¢c ABC vu?ng t?i A; ??nh C(-4;1); ph?n gi¨¢c trong g¨®c A c¨® ph??ng tr¨¬nh
x + y -5 = 0. Vi?t ph??ng tr¨¬nh BC, bi?t di?n t¨ªch tam gi¨¢c ABC b?ng 24 v¨¤ ??nh A c¨® ho¨¤nh ??
d??ng.
B¨¤i 3: D-2010
? C?u VIa: Cho tam gi¨¢c ABC; A(3;-7); tr?c t?m H(3;-1), t?m ???ng tr¨°n ngo?i ti?p I(-2;0).
X¨¢c ??nh t?a ?? C bi?t ho¨¤nh ?? C d??ng.
? C?u VIb: Cho A(0;2); ? l¨¤ ???ng th?ng ?i qua O. G?i H l¨¤ h¨¬nh chi?u vu?ng g¨®c c?a A l¨ºn
? . Vi?t ph??ng tr¨¬nh ? bi?t kho?ng c¨¢ch t? H ??n tr?c ho¨¤nh b?ng AH.
B¨¤i 4: D-2009
? C?u VIa: Cho tam gi¨¢c ABC; M(2;0) l¨¤ trung ?i?m c?a AB. ???ng trung tuy?n v¨¤ ???ng
cao ?i qua A l?n l??t c¨® ph??ng tr¨¬nh l¨¤: 7x-2y-3=0 v¨¤ 6x-y-4=0. Vi?t ph??ng tr¨¬nh AC.
? C?u VIb: Cho ???ng tr¨°n (C): (x-1)2
+y2
=1. G?i I l¨¤ t?m c?a (C). X¨¢c ??nh t?a ?? ?i?m M
thu?c (C) sao cho ¡¤ 0
30IMO = .
B¨¤i 5 : B-2009
? C?u VIb: Cho tam gi¨¢c ABC c?n t?i A(-1;4). C¨¢c ??nh B;C thu?c ? : x-y-4=0. X¨¢c ??nh t?a
?? B ; C bi?t di?n t¨ªch tam gi¨¢c ABC b?ng 18.
B¨¤i 6: A-2009
? C?u VIa: Cho h¨¬nh ch? nh?t ABCD c¨® I(6;2) l¨¤ giao ?i?m hai ???ng ch¨¦o AC v¨¤ BD. ?i?m
M(1;5) thu?c AB v¨¤ trung ?i?m E c?a c?nh CD thu?c ? : x+y-5=0. Vi?t ph??ng tr¨¬nh AB.
? C?u VIb: Cho ???ng tr¨°n (C): x2
+y2
+4x+4y+6 =0 v¨¤ ? :x+my-2m+3=0. G?i I l¨¤ t?m c?a
(C). T¨¬m m ?? ? c?t (C) t?i hai ?i?m ph?n bi?t A; B sao cho di?n t¨ªch tam gi¨¢c IAB max.
B¨¤i 7: D-2008
Trong Oxy; cho (P) c¨® ph??ng tr¨¬nh 2
16y x= ; A(1;4). Hai ?i?m ph?n bi?t B; C kh?ng tr¨´ng v?i A
di ??ng tr¨ºn (P) sao cho 0
90BAC¡Ï = . CMR ???ng th?ng BC lu?n ?i qua m?t ?i?m c? ??nh.
B¨¤i 8: B-2008
Trong Oxy; cho tam gi¨¢c ABC; h¨¬nh chi?u vu?ng g¨®c c?a C l¨ºn AB l¨¤ H(-1;-1); ???ng ph?n gi¨¢c
trong c?a g¨®c A: x-y+2=0; ???ng cao k? t? B: 4x+3y-1=0; T¨¬m t?a ?? C?
B¨¤i 9: B-2007
Cho A(2;2); (d1): x+y-2=0; (d2): x+y-8=0. T¨¬m t?a ?? B thu?c (d1) ; C thu?c (d2) sao cho tam gi¨¢c
ABC vu?ng c?n t?i A.
4. B¨¤i 10: B-2006
Cho ???ng tr¨°n (C): x2
+y2
-2x-6y+6 =0 v¨¤ M(-3 ;1). G?i T1 ; T2 l¨¤ c¨¢c ti?p ?i?m c?a c¨¢c ti?p tuy?n
k? t? M ??n (C). Vi?t ph??ng tr¨¬nh T1T2.
B¨¤i 11: A-2006
Cho (d1) : x+y+3=0 ; (d2) : x-y-4=0 ; (d3) : x-2y=0.
T¨¬m t?a ?? M thu?c (d3) sao cho kho?ng c¨¢ch t? M ??n d1 b?ng hai l?n kho?ng c¨¢ch t? M ??n d2.
B¨¤i 12: A-2005
Cho (d1) : x-y=0 ; (d2) : 2x+y-1=0. T¨¬m t?a ?? c¨¢c ??nh c?a h¨¬nh vu?ng ABCD bi?t A thu?c d1 ; C
thu?c d2 v¨¤ B ; D thu?c tr?c ho¨¤nh.
B¨¤i 13: D-2004
Trong Oxy; cho tam gi¨¢c ABC; A(-1;0); B(4;0); C(0;m). T¨¬m t?a ?? tr?ng t?m G theo m; m 0¡Ù .
T¨¬m m ?? tam gi¨¢c GAB vu?ng t?i G.
B¨¤i 14: B-2004
Cho A(1;1); B(4;-3). T¨¬m C thu?c ???ng th?ng x-2y-1=0 sao cho kho?ng c¨¢ch t? C ??n AB b?ng 6.
B¨¤i 15: A-2004
Cho A(0;2); B(- 3 ;-1). T¨¬m t?a ?? tr?c t?m v¨¤ t?m ???ng tr¨°n ngo?i ti?p tam gi¨¢c OAB.
B¨¤i 16: B-2003
Cho tam gi¨¢c ABC c¨® AB=AC; ¡¤ 0
90BAC = . ?i?m M(1;-1) l¨¤ trung ?i?m BC v¨¤ G(
2
3
;0) l¨¤ tr?ng t?m
tam gi¨¢c ABC. T¨¬m t?a ?? A;B;C.
B¨¤i 17: B-2002
Trong Oxy; cho h¨¬nh ch? nh?t ABCD c¨® t?m I(1/2;0); ???ng th?ng AB c¨® pt: x-2y+2 = 0 v¨¤ AB
=2AD. T¨¬m t?a ?? A; B; C; D bi?t ho¨¤nh ?? A ?m.
B¨¤i 18: 2002
Trong Oxy; cho tam gi¨¢c ABC vu?ng t?i A. ???ng th?ng BC c¨® ph??ng tr¨¬nh: 3 3 0x y? ? = ; A
v¨¤ B thu?c Ox. B¨¢n k¨ªnh ???ng tr¨°n n?i ti?p tam gi¨¢c ABC b?ng 2. T¨¬m t?a ?? tr?ng t?m G c?a
tam gi¨¢c.
B¨¤i 19: Trong Oxy; cho tam gi¨¢c ABC; A(-5;6); B(-4;-1); C(4;3).
a/ T¨¬m to? ?? D ?? ABCD l¨¤ h¨¬nh b¨¬nh h¨¤nh.
b/ T¨¬m to? ?? h¨¬nh chi?u vu?ng g¨®c c?a A l¨ºn BC.
B¨¤i 20: Trong Oxy; cho tam gi¨¢c ABC; A(0;2); B(-2;-2); C(4;-2). H l¨¤ ch?n ???ng cao h? t? B; M
v¨¤ N l?n l??t l¨¤ trung ?i?m AB v¨¤ BC. L?p ph??ng tr¨¬nh ???ng tr¨°n qua H; M; N.
B¨¤i 21: Trong Oxy; cho ???ng tr¨°n (C): 2 2
( 2) ( 3) 2x y? + ? = ; ???ng th?ng (d): x-y-2=0.
T¨¬m M thu?c (C) ?? kho?ng c¨¢ch t? M ??n (d):
a/ max? b/ min?
B¨¤i 22: Trong Oxy; cho tam gi¨¢c ABC; C(-2;3). ???ng cao k? t? A c¨® ph??ng tr¨¬nh: 3x-2y-25=0;
???ng ph?n gi¨¢c trong g¨®c B c¨® ph??ng tr¨¬nh: x-y=0. L?p ph??ng tr¨¬nh AC?
B¨¤i 23: Trong Oxy; cho h¨¬nh vu?ng ABCD; CD c¨® ph??ng tr¨¬nh: 4x-3y+4=0; M(2;3) th?c BC;
N(1;1) thu?c AB. Vi?t ph??ng tr¨¬nh c¨¢c c?nh c¨°n l?i.
B¨¤i 24: Trong Oxy; cho parabol (P): 2
4y x= . L?p ph??ng tr¨¬nh c¨¢c c?nh c?a tam gi¨¢c ABC bi?t A
tr¨´ng ??nh O; hai ?i?m B; C thu?c (P); tr?c t?m tr¨´ng v?i ti¨ºu ?i?m c?a (P).
B¨¤i 25: Trong Oxy; cho tam gi¨¢c ABC; A(5;2); ???ng trung tr?c c?a BC c¨® ph??ng tr¨¬nh x+y-6=0;
???ng trung tuy?n CM c¨® ph??ng tr¨¬nh 2x-y+3=0. T¨¬m t?a ?? A; B; C.
B¨¤i 26: Trong Oxy; cho tam gi¨¢c ABC; A thu?c (d): x-4y-2=0; BC song song v?i (d); ???ng cao
BH c¨® ph??ng tr¨¬nh: x+y+3=0; M(1;1) l¨¤ trung ?i?m c?a AC. T¨¬m t?a ?? A; B; C.
5. B¨¤i 27: Trong Oxy; L?p ph??ng tr¨¬nh qua A(1;1) c¨¢ch ??u B(-2;3) v¨¤ C(0;4).
B¨¤i 28: Trong Oxy; cho tam gi¨¢c ABC; B(1;0); hai ???ng cao c¨® ph??ng tr¨¬nh l?n l??t l¨¤: x-
2y+1=0; 3x+y-1=0. T¨ªnh di?n t¨ªch tam gi¨¢c ABC.
B¨¤i 29: Trong Oxy; cho tam gi¨¢c ABC c?n t?i A; G(
4 1
;
3 3
) l¨¤ tr?ng t?m; ???ng th?ng ch?a c?nh BC
c¨® ph??ng tr¨¬nh l¨¤: x-2y-4=0; ???ng BG c¨® pt: 7x-4y-8=0. T¨¬m t?a ?? A; B; C.
B¨¤i 30: L?p ph??ng tr¨¬nh ???ng th?ng qua I(-2;0); c?t (d1): 2x-y+5=0 v¨¤ c?t (d2): x+y-3=0 l?n l??t
t?i A v¨¤ B sao cho: 2IA IB=
uur uur
B¨¤i 31: Cho ???ng th?ng (d1): x+y+5=0 v¨¤ (d2): x+2y-7=0. A(2;3); T¨¬m B thu?c (d1); C thu?c (d2)
sao cho tam gi¨¢c ABC c¨® tr?ng t?m G(2;0).
B¨¤i 32: Ll?p ph??ng tr¨¬nh ???ng th?ng qua M(
5
;2
2
); c?t (d1): x-2y=0 v¨¤ c?t (d2): 2x-y=0 l?n l??t
t?i A v¨¤ B sao cho: M l¨¤ trung ?i?m AB.
B¨¤i 33: Trong Oxy; cho ???ng th?ng (d): 2x-y-5=0 v¨¤ A(1;2); B(4;1). T¨¬m M thu?c (d) sao cho
MA MB? max.
B¨¤i 34: Trong Oxy; cho h¨¬nh vu?ng ABCD; CD c¨® ph??ng tr¨¬nh: 4x-3y+4=0; M(2;3) thu?c BC;
N(1;1) thu?c AB. L?p ph??ng tr¨¬nh AD.
B¨¤i 35: Trong Oxy; l?p ph??ng tr¨¬nh (d1); (d2) l?n l??t ?i qua A(4;0); B(0;5) v¨¤ nh?n (d): 2x-2y-
1=0 l¨¤ ph?n gi¨¢c.
B¨¤i 36: Cho tam gi¨¢c ABC c?n t?i A; ???ng th?ng AB: 2x-y+5=0; ???ng th?ng AC:3x+6y-1=0;
M(2;-1) thu?c BC. L?p ph??ng tr¨¬nh c?nh BC.
B¨¤i 37: Trong Oxy; cho tam gi¨¢c ABC; A(-1;2); B(2;0); C(-3;1).
a/ T¨¬m t?m ???ng tr¨°n ngo?i ti?p tam gi¨¢c ABC.
b/ T¨¬m M thu?c BC sao cho
1
3
AMB ABCS S? ?= .
B¨¤i 38: Cho h¨¬nh b¨¬nh h¨¤nh ABCD c¨® di?n t¨ªch b?ng 4. Bi?t A(1; 0); B(2;0); giao ?i?m hai ???ng
ch¨¦o I thu?c ???ng th?ng y =x. T¨¬m t?a ?? C; D?
B¨¤i 39: Cho (d1): x-2y=0; (d2): 2x-y=0; M(
5
;2
2
). L?p ph??ng tr¨¬nh ???ng th?ng ?i qua M c?t (d1);
(d2) t?i A v¨¤ B sao cho:
a/ M l¨¤ trung ?i?m AB b/ MB=2MA.
B¨¤i 40: Cho h¨¬nh thoi c¨® m?t ???ng ch¨¦o: x+2y-7=0; m?t c?nh: x+3y-3=0; m?t ??nh (0;1). L?p
ph??ng tr¨¬nh c¨¢c c?nh.
B¨¤i 41: Cho tam gi¨¢c ABC; A(-6;3); B(-4;3); C(9;2).
a/ L?p ph??ng tr¨¬nh ???ng ph?n gi¨¢c trong g¨®c A.
b/ T¨¬m M tr¨ºn AB; ?i?m N tr¨ºn AC sao cho MN//BC v¨¤ AM=CN.
B¨¤i 42: Cho tam gi¨¢c ABC; A(-6;3); B(-4;3); C(9;2). T¨¬m P thu?c ???ng ph?n gi¨¢c trong g¨®c A sao
cho ABPC l¨¤ h¨¬nh thang.
B¨¤i 43: Cho tam gi¨¢c ABC; A(-2;3); tr?c t?m H tr¨´ng v?i trung ?i?m c?a ???ng cao AK. ???ng
cao BM c¨® h? s? g¨®c b?ng 2. T¨¬m t?a ?? B; C.
B¨¤i 44: Cho tam gi¨¢c ABC c¨® tr?ng t?m G(-2;-1); AB: 4x+y+15=0; AC: 2x+5y+3=0. T¨¬m tr¨ºn
???ng cao AH c?a tam gi¨¢c ?i?m M sao cho tam gi¨¢c BMC vu?ng t?i M.
B¨¤i 45: Cho A(1;0); B(3;-1); (d):x-2y-1=0. T¨¬m C thu?c (d) sao cho 6ABCS? = .
6. B¨¤i 46: Cho tam gi¨¢c ABC; c?nh AB: y=2x; c?nh AC: y=
1 1
4 4
x? + ; tr?ng t?m G(
8 7
;
3 3
). T¨ªnh di?n
t¨ªch tam gi¨¢c ABC.
B¨¤i 47: T¨¬m to? ?? ?i?m M tr¨ºn (d): x-2y-2=0 sao cho 2 2
2MA MB+ nh? nh?t; A(0;1); B(3;4).
B¨¤i 48: Cho A(2;-1); B(1;-2); tr?ng t?m G thu?c (d):x+y-2=0. T¨¬m C bi?t di?n t¨ªch tam gi¨¢c ABC
b?ng 3/2.
B¨¤i 49: T¨¬m M n?m ph¨ªa tr¨ºn Ox sao cho g¨®c MAB=300
; g¨®c AMB = 900
; A(-2;0); B(2;0).
B¨¤i 50: Cho tam gi¨¢c ABC vu?ng t?i A; A(-1;4); B(1;-4); M(
1
2;
2
) thu?c BC. T¨¬m t?a ?? C?
7. B¨¤i 46: Cho tam gi¨¢c ABC; c?nh AB: y=2x; c?nh AC: y=
1 1
4 4
x? + ; tr?ng t?m G(
8 7
;
3 3
). T¨ªnh di?n
t¨ªch tam gi¨¢c ABC.
B¨¤i 47: T¨¬m to? ?? ?i?m M tr¨ºn (d): x-2y-2=0 sao cho 2 2
2MA MB+ nh? nh?t; A(0;1); B(3;4).
B¨¤i 48: Cho A(2;-1); B(1;-2); tr?ng t?m G thu?c (d):x+y-2=0. T¨¬m C bi?t di?n t¨ªch tam gi¨¢c ABC
b?ng 3/2.
B¨¤i 49: T¨¬m M n?m ph¨ªa tr¨ºn Ox sao cho g¨®c MAB=300
; g¨®c AMB = 900
; A(-2;0); B(2;0).
B¨¤i 50: Cho tam gi¨¢c ABC vu?ng t?i A; A(-1;4); B(1;-4); M(
1
2;
2
) thu?c BC. T¨¬m t?a ?? C?