a)
[tex]\it AB^2=(x_B-x_A)^2+(y_B-y_A)^2=(4-2)^2+(-6-5)^2=2^2+(-11)^2=\\ \\ =4+121=125 \Rightarrow AB=\sqrt{125}=\sqrt{25\cdot5}=5\sqrt5\\ \\ \\ AC^2=(x_C-x_A)^2+(y_C-y_A)^2=(-2-2)^2+(-3-5)^2=(-4)^2+(-8)^2=\\ \\ =16+64=80 \Rightarrow AC=\sqrt{80}=\sqrt{16\cdot5}=4\sqrt5\\ \\ \\ BC^2=(x_C-x_B)^2+(y_C-y_B)^2=(-2-4)^2+(-3+6)^2=(-6)^2+(3)^2=\\ \\ =36+9=45 \Rightarrow BC=\sqrt{45}=\sqrt{9\cdot5}=3\sqrt5[/tex]
b)
[tex]\it AC^2+BC^2=80+45=125=AB^2[/tex]
Din reciproca teoremei lui Pitagora ⇒ ΔACB -dreptunghic în C.
d( C, AB) = h (înălțimea corespunzătoare ipotenuzei)
[tex]\it h=\dfrac{c_1\cdot c_2}{ip}=\dfrac{\ \ 3\sqrt5\cdot4\sqrt5^{(\sqrt5}}{5\sqrt5}=\dfrac{12\sqrt5}{5}[/tex]
