Oblicz długość odcinka AB jezeli
a) A(-2,3) i B(3,-1)
b) A(4,-3) i B(-2,5)
c)A(-1,2) i B(5,2)
d) A(0,-2) i b (-3,-3)
|AB|² = (xB-xA)²+(yB-yA)²
a)
|AB|² = (3-(-2))²+(-1-3)²
|AB|² = 5²+(-4)²
|AB|² = 25+16
|AB|² = 41
|AB| = √41
b)
|AB|² = (-2-4)²+(5-(-3))²
|AB|² = (-6)²+8²
|AB|² = 36+64
|AB|² = 100
|AB| = 10
c)
|AB|² = (5-(-1))²+(2-2)²
|AB|² = 6²+0²
|AB|² = 36+0
|AB|² = 36
|AB| = 6
d)
|AB|² = (-3-0)²+(-3-(-2))²
|AB|² = (-3)²+(-1)²
|AB|² = 9+1
|AB|² = 10
|AB| = √10
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|AB|² = (xB-xA)²+(yB-yA)²
a)
|AB|² = (3-(-2))²+(-1-3)²
|AB|² = 5²+(-4)²
|AB|² = 25+16
|AB|² = 41
|AB| = √41
b)
|AB|² = (-2-4)²+(5-(-3))²
|AB|² = (-6)²+8²
|AB|² = 36+64
|AB|² = 100
|AB| = 10
c)
|AB|² = (5-(-1))²+(2-2)²
|AB|² = 6²+0²
|AB|² = 36+0
|AB|² = 36
|AB| = 6
d)
|AB|² = (-3-0)²+(-3-(-2))²
|AB|² = (-3)²+(-1)²
|AB|² = 9+1
|AB|² = 10
|AB| = √10