СУ задача

[antoniy]

Screenshot_8.jpg (83.25 KiB) Прегледано 313 пъти

[peyo]

Screenshot_8.jpg

Ok, решаваме директно. Намираме координатите на 3-те върха на триъгълника и му намираме лицето:

Код: Избери целия код

from sympy import *
var("x,y")
l = x-4*y+3
m = 3*x-y-2
n = 5*x+2*y-29

p1,p2,p3 = (Point(x,y).subs(sol) for sol in (solve([l,m],(x,y)),solve([m,n],(x,y)),solve([n,l],(x,y))))

print(p1,p2,p3)

print(Triangle(p1,p2,p3).area)

Point2D(1, 1) Point2D(3, 7) Point2D(5, 2)
-11

[ammornil]

$l:\quad x-4y+3=0 \Rightarrow y=\dfrac{1}{4}x+\dfrac{3}{4} \\[6pt] m: \quad 3x-y-2=0 \Rightarrow y=3x-2 \\[6pt] n: \quad 5x+2y-29=0 \Rightarrow y=-\dfrac{5}{2}x+\dfrac{29}{2} \\[6pt] l\cap{}m=A(x_{A},y{A}): \quad \dfrac{1}{4}x_{A}+\dfrac{3}{4}=3x_{A}-2 \Leftrightarrow 11x_{A}=11 \Leftrightarrow x_{A}=1 \\[6pt] \quad y_{A}=3x_{A}-2= 3\cdot{}1 -2= 1 \Rightarrow A(1,1) \\[6pt] l\cap{}n=C(x_{C},y_{C}): \quad \dfrac{1}{4}x_{C}+\dfrac{3}{4}= -\dfrac{5}{2}x_{C}+\dfrac{29}{2} \Leftrightarrow 11x_{C}= 55 \Leftrightarrow x_{C}=5 \\[6pt] \quad y_{C}= \dfrac{1}{4}x_{C}+ \dfrac{3}{4}= \dfrac{1}{4}\cdot{}5 +\dfrac{3}{4}= 2 \Rightarrow C(5,2) \\[6pt] m\cap{}n=B(x_{B},y): \quad 3x_{B}-2= -\dfrac{5}{2}x_{B} +\dfrac{29}{2} \Leftrightarrow 11x_{B}=33 \Leftrightarrow x_{B}=3 \\[6pt] \quad y_{B}=3x_{B} -2= 3\cdot{}3 -2= 7 \Rightarrow B(3,7)\\[12pt] A(1,1), \quad B(3,7), \quad C(5,2) \Rightarrow S_{ABC}= \dfrac{1}{2}\cdot{}\begin{vmatrix} x_{A}(y_{B}-y_{C}) +x_{B}(y_{C}-y_{A}) +x_{C}(y_{A}-y_{B}) \end{vmatrix} \\[6pt] S_{ABC}= \dfrac{1}{2}\cdot{}\begin{vmatrix} 1(7-2) +3(2-1) +5(1-7)\end{vmatrix} =\dfrac{1}{2}\cdot{}\begin{vmatrix} 5 +3 -30 \end{vmatrix}= \dfrac{1}{2}\cdot{}22= 11$