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2010-04-05T20:41:28+02:00

A = (0; 4) , B = (6; 0)
pr. AB:
y = ax +b
4 =a*0 +b ---> b = 4
0 =a*6 + 4 ---> -6a = 4 --> a = -4/6 = -2/3
zatem y = (-2/3) x + 4
Szukam punktu C
y = (-2/3) x + 4 oraz y = x
(-2/3) x + 4 = x
x + (2/3)x = 4
(5/3)x = 4
x = 4*(3/5) = 12/5
y = x = 12/5
C = (12/5 ; 12/5)
-->
AC = [12/5 - 0; 12/5 - 20/5] = [12/5 ; -8/5]
I AC I² = (12/5)² + (-8/5)² = 144/25 + 64/25 = 210/25
I AC I = √210/5
-->
CB = [30/5 - 12/5; 0 - 12/5] = [18/5; -12/5]
I CB I² = (18/5)² + (-12/5)² = 324/25 + 144/25 = 468/25
I CB I = √468/5
I AC I : I CB I = [√210 /5] : [√468 /5] = √210 : √468 = √(105/234}


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