OBLICZ DLUGOSC ODCINKA AB, JEZELI; a ; A=(2;4) B = (7;9) b ; A=(-1;2) B= (2;-1) c; A= (-3;-4) B=(-7 1)
IABI= √[(xb-xa)²+(yb-ya)²]
a) A=(2;4) B=(7;9)
IABI = √[(7-2)²+(9-4)²]=√[5²+5²] = √[25+25] = √50 = 5√2
b) A=(-1;2) B(2;-1)
IABI =√[(2- (-2))² + (-1-(-1))²] = √[16+0] = √16=4
c) IABI = √[(-7+3)² + (1+4)²] = √[16+25] = √41
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IABI= √[(xb-xa)²+(yb-ya)²]
a) A=(2;4) B=(7;9)
IABI = √[(7-2)²+(9-4)²]=√[5²+5²] = √[25+25] = √50 = 5√2
b) A=(-1;2) B(2;-1)
IABI =√[(2- (-2))² + (-1-(-1))²] = √[16+0] = √16=4
c) IABI = √[(-7+3)² + (1+4)²] = √[16+25] = √41