Oblicz pole trójkąta o wierzchołkach A= (-3, -2), B= (3, 5), C= (5 , -3).
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Odpowiedź :

Magda

Odpowiedź:

[tex]A(-3,-2)\ \ ,\ \ B(3,5)\ \ ,\ \ C(5,-3)\\\\P=\frac{1}{2}\cdot|(x_{B}-x_{A})(y_{C}-y_{A})-(y_{B}-y_{A})(x_{C}-x_{A})|\\\\P=\frac{1}{2}\cdot|(3-(-3))(-3-(-2))-(5-(-2))(5-(-3))|\\\\P=\frac{1}{2}\cdot|(3+3)(-3+2)-(5+2)(5+3)|\\\\P=\frac{1}{2}\cdot|6\cdot(-1)-7\cdot8|=\frac{1}{2}\cdot|-6-56|=\frac{1}{2}\cdot|-62|=\frac{1}{\not2_{1}}\cdot\not62^3^1=31\lbrack{j^2}\rbrack[/tex]