Question

Punkt S(a-2b, a + b) jest środkiem odcinka o końcach A(−1, 3) i B(3, 5) jeżeli:

A. a = 3 i b = 1

B. a = - 3 i b= 1

C. a = - 3 i b = −1

D. a = 3 i b = −1

Answer 1

Odpowiedź:

A = ( - 1 , 3 ) , B = (3 , 5 ) , S = ( a - 2b ; a + b)

xa = - 1 , xb = 3 , xs = a - 2b, ya = 3 , yb = 5 , ys = a + b

xs = (xa + xb)/2 = (- 1 + 3)/2 = 2/2 = 1

ys = (ya + yb)/2 = (3 + 5)/2 = 8/2 = 4

Układ równań

a - 2b = 1

a + b = 4 | * 2

a - 2b = 1

2a + 2b = 8

dodajemy równania

a +2a - 2b + 2b = 1 + 8

3a = 9

a = 9/3 = 3

a + b = 4

3 + b = 4

b = 4 - 3 = 1

Odp: A