Oblicz długości odcinków AB, BC, AC, jeśli:
a) A = (2, 8), B = (-2, 5), C = (6, -1),
b) A = (-3, 4), B = (5, -1), C = (5, 9),
c) A = (-2, 1), B = (0, 6), C = (6, 2).


Odpowiedź :

a]

[tex]A=(2,8), \ B=(-2,5), \ C=(6,-1)\\\\|AB|=\sqrt{(-2-2)^2+(5-8)^2}=\sqrt{(-4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\\\\|BC|=\sqrt{(6-(-2))^2+(-1-5)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=10\\\\|AC|=\sqrt{(6-2)^2+(-1-8)^2}=\sqrt{4^2+(-9)^2}=\sqrt{16+81}=\sqrt{97}[/tex]

b]

[tex]A=(-3,4), \ B=(5,-1), \ C=(5,9)\\\\|AB|=\sqrt{(5-(-3))^2+(-1-4)^2}=\sqrt{8^2+(-5)^2}=\sqrt{64+25}=\sqrt{89}\\\\|BC|=\sqrt{(5-5)^2+(9-(-1))^2}=\sqrt{0^2+10^2}=\sqrt{100}=10\\\\|AC|=\sqrt{(5-(-3))^2+(9-4)^2}=\sqrt{8^2+5^2}=\sqrt{64+25}=\sqrt{89}[/tex]

c]

[tex]A=(-2,1), \ B=(0,6), \ C=(6,2)\\\\|AB|=\sqrt{(0-(-2))^2+(6-1)^2}=\sqrt{2^2+5^2}=\sqrt{4+25}=\sqrt{29}\\\\|BC|=\sqrt{(6-0)^2+(2-6)^2}=\sqrt{6^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\\\\|AC|=\sqrt{(6-(-2))^2+(2-1)^2}=\sqrt{8^2+1^2}=\sqrt{64+1}=\sqrt{65}[/tex]