Matematika/Trikampiai: Skirtumas tarp puslapio versijų

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72 eilutė:
:<math>=\frac{1}{2}|(x_1 y_1+x_1 y_2-x_2 y_1-x_2 y_2+x_2 y_2+ x_2 y_3-x_3 y_2- x_3 y_3+x_3 y_3+x_3 y_1-x_1 y_3-x_1 y_1)|=</math>
:<math>=\frac{1}{2}|(x_1 y_2-x_2 y_1+ x_2 y_3-x_3 y_2+x_3 y_1-x_1 y_3)|=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=</math>
:<math>=\frac{1}{2}|(x_2-x_1)(y_2y_3-y_3y_1)-x_2(y_2-y_3)+x_3(y_1-y_2)|=\frac{1}{2}|(x_1-x_2)(y_2-y_3)+x_3(y_1-y_2)|.</math>
 
:'''Pavyzdis'''. Duoti taškai ''A''(1; 1), ''B''(6; 4), ''C''(8; 2). Rasti trikampio ''ABC'' plotą. Randame:
:<math>S_{ABC}=\frac{1}{2}|[(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)]|=\frac{1}{2}|(6-1)(2-1)-(8-1)(4-1)|=\frac{1}{2}|5\cdot 1-7\cdot 3|=\frac{1}{2}|5-21|=\frac{1}{2}|-16|=\frac{16}{2}=8;</math>
:<math>S_{ABC}=S_{ADEC}+S_{BCEF}-S_{ABFD}=\frac{1}{2}|(x_1-x_2)(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac{1}{2}|(1-6)(4-2)+6(2-1)+8(1-4)|=\frac{1}{2}|(-5)\cdot 2+8\cdot (6-3)24|=</math>
:<math>=\frac{1}{2}|-10-24|=</math>