kazah
megoldása
2,
a, Ábra
b,
`d_(AB)` = `root()((x_A-x_B)^2+(y_A-y_B)^2)` = `root()((1-(-2))^2+(1-2)^2)` = `root()(10)`
`d_(BC)` = `root()((x_B-x_C)^2+(y_B-y_C)^2)` = `root()((1-(-1))^2+(1-(-1))^2)` = `root()(8)`
`d_(AC)` = `root()((x_A-x_C)^2+(y_A-y_C)^2)` = `root()((-2-(-1))^2+(2-(-1))^2)` = `root()(10)`
K = `d_(AB)+d_(AC)+d_(BC)` = `2*root()(10)+root()(8)` `approx` 9,153
c,
I. A BC oldal egyenlete:
`m_(BC)` = `(y_C-y_B)/(x_C-x_B)` = `(-1-1)/(-1-1)` = 1
`b_(BC)=y_B-m_(BC)*x_B` = `1-1*1` = 0
y = x
Az oldalra merőleges egyenes meredeksége:
`m_(m_A)` = `-1/m_(BC)` = -1
`y_A=m_(m_A)*x_A+b_(m_A)`
`b_(m_A)` = `y_A-m_(m_A)*x_A` = `2-(-1)*(-2)` = 0
y = -x
A (0;0) és a (2;-2) pont távolsága:
d = `root()((2-0)^2+(-2-0)^2)` = `2*root()(2)` `approx` 2,82.
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