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2010-01-31T00:51:24+01:00
A =(5; 10) , B = (-7 ;2) , S = (1 ; 4/3)
I AB I = 2* I CD I
S - punkt przecięcia sie przekątnych trapezu ABCD.
D = (x1; y1)
wektor BS = 2* wektor SD
wektor BS = [1+7;4/3 - 2] = [8; -2/3]
wektor 2*SD = 2*[x1 -1;y1 -4/3] =[2x1 -2; 2 y1 -8/3]
Mamy
2x1 - 2 = 8 oraz 2y1 - 8/3 = -2/3
2x1 = 10 oraz 2y1 = 6/3 = 2
x1 = 5 oraz y1 = 1
D = ( 5 ; 1)
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C = (x2 ; y2)
wektor AS = 2 *wektor SC
wektor AS = [1-5; 4/3 - 10] = [-4; -26/3]
2*wektor SC = 2*[x2 - 1; y2 - 4/3] = [2x2 -2 ; 2y2 - 8/3]
Mamy
2x2 - 2 = -4 oraz 2y2 - 8/3 = -26/3
2x2 = -2 oraz 2y2 = -18/3
x2 = -1 oraz y2 = -9/3 = -3
C = (-1 ; -3)
Odp. C = ( -1 ; -3) , D = (5 ; 1)
Spr.
wektor AB = [-7-5 ; 2 -10] = [-12; -8]
AB =√(-12)²+(-8)² = √144+64 = √208 = 4√13
wektor CD = [5 +1; 1 + 3] = [6 ; 4]
CD = √6² + 4² = √ 36+16 = √ 52 = √4*√13 = 2√13
Zatem AB = 2* CD
26 3 26