Bai Tap Phep Tinh Tien
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Transcript of Bai Tap Phep Tinh Tien
BI TP PHP TNH TIN
BI TP PHP TNH TIN
Bi 1: Trong mt phng ta Oxy cho im A(-3;3 ) , B(1 ;3 ) v ng trn (C ) c tm I( 3 ;1 ) , bn knh R = 1 . Mt ng thng (d) c phng trnh : x + y 1 = 0 . Tm trn ( d) mt im M v trn ng trn (C) im M sao cho :
Bi 2: Trong mt phng ta Oxy cho im A(-3; 3 ) v B(-1 ; 6) .
a) Tm to im M l nh ca im M ( 4 ; -5 ) qua php tnh tin
b) Xc nh phng trnh tng qut ca ng thng d1 l nh ca ng thng d c phng trnh : qua php tnh tin
c) Xc nh phng trnh ng trn ( C1 ) l nh ca ng trn ( C) c phng trnh : qua php tnh tin
Bi 3: Cho hai ng trn (C1) :
(C2 ) :
v mt ng thng (d) : x + 2y 10 = 0
Tm nh ca ng thng (d) trong php tnh tin bin ng trn (C1) thnh (C2 ) ?
Bi 4: Cho hai ng thng :
(d) : x + y + 6 = 0
() : x + y 4 = 0
v mt ng thng m : x y = 0
Bit rng tn ti mt php tnh tin vi cng phng vi m bin ng thng (d) thnh () . Tm nh ca ng trn (C) : trong php tnh tin ni trn ?
Bi 5: Cho tam gic ABC . V hnh ch nht BCDE bn ngoi tam gic . Gi d1 , d2 ln lt l ng thng qua D v E vung gc vi Ab , AC . Gi K l giao im ca d1 v d2.
a) Php tnh tin bin d1 , d2 thnh hai ng thng no , bin K thnh im no?
b) Suy ra AK vung gc vi BC ? Bi 6: Cho on thng AB c nh v mt ng trn c nh (O) . C l mt im di ng trn ( O). V hnh bnh hnh ABCD .
a) Tm tp hp nhng im D . V tp hp ny ?
b) V tam gic u CDE . Tm tp hp nhng im E . V tp hp ny ?
Bi 7:
a) Tm nh ca parapol (P) : y = x2 2x 3 qua php tnh tin theo vecto = (2; -5)
b) Bit rng tn ti mt php tnh tin bin ng trn : thnh ng trn :
Tm php tnh tin ni trn ? Bi 8: Cho php tnh tin c biu thc ta :
EMBED Equation.DSMT4 .a) Tm nh ca ng trn ( C) : qua ?
b) Tm m bit ng thng : khng thay i (bt bin) qua
PAGE 1Phm L Thanh .. ..Ton 11(NC)
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