punkt S jest środkiem odcinka AB. Oblicz a i b. A ( 3a, 2 ) B ( a, 2b ) S ( b,a )
a)
A = ( -3; a), B = (2 ; 5) oraz S = (0 ; 5)
mamy
Błędne dane , bo [-3 +2]/2 = -1/2 = - 0,5 ,a jest podane 0
b)
A = ( 0; 4), B = (a ; b) oraz S = ( 2 ; -1)
Mamy
[0 +a]/2 = 2 oraz [4 + b]/2 = -1
a = 2*2 = 4 oraz 4 + b = -1*2 = -2
a = 4 oraz b = -2 - 4 = -6
Odp. a = 4 oraz b = -6
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c)
A = ( a; 2 ) , B = (2a ; b) oraz S = (b ; a)
[a + 2a]/2 = b oraz [2 + b]/2 = a
3a =2b oraz 2 + b = 2a
b = (3/2) a oraz 2a = (3/2) a +2
b = (3/2) a oraz 2a -1,5 a = 2
b = (3/2) a oraz 0,5 a = 2 / * 2
a = 4 oraz b = 1,5 *4 = 6
Odp. a = 4 oraz b = 6
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a)
A = ( -3; a), B = (2 ; 5) oraz S = (0 ; 5)
mamy
Błędne dane , bo [-3 +2]/2 = -1/2 = - 0,5 ,a jest podane 0
b)
A = ( 0; 4), B = (a ; b) oraz S = ( 2 ; -1)
Mamy
[0 +a]/2 = 2 oraz [4 + b]/2 = -1
a = 2*2 = 4 oraz 4 + b = -1*2 = -2
a = 4 oraz b = -2 - 4 = -6
Odp. a = 4 oraz b = -6
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c)
A = ( a; 2 ) , B = (2a ; b) oraz S = (b ; a)
Mamy
[a + 2a]/2 = b oraz [2 + b]/2 = a
3a =2b oraz 2 + b = 2a
b = (3/2) a oraz 2a = (3/2) a +2
b = (3/2) a oraz 2a -1,5 a = 2
b = (3/2) a oraz 0,5 a = 2 / * 2
a = 4 oraz b = 1,5 *4 = 6
Odp. a = 4 oraz b = 6
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