> with(geometry):
point(B,-4,7),point(S,4,8):
l3:=line(BC,[B,S]); l3:= BC
> Equation(BC,[x,y]);
ABC tomonlarining uzunliklari:
> with(geometry):
> triangle(ABC, [point(A,10,3), point(B,-4,7), point(C,4,8)]):
distance(A,B);distance(A,C);distance(B,C);
> sides(ABC);
2)AB va AC tomonlar orasidagi burchak
formulaga asosan topiladi:
, : ,
> with(geometry):
_EnvHorizontalName := 'x': _EnvVerticalName := 'y':
line(AB, 82-4*x-14*y=0), line(AC, 68-5*x-6*y = 0);
> hi:=FindAngle(AB, AC);
> hi:=evalf(hi);
3) AD balandlikni topish uchun A(10;3) nuqtadan o‘tuvchi to‘g‘ir chiziqlar dasatasi tenglaasini yozamiz, undan BC tomonga perpendikulyar bo‘lgan to‘g‘ri chiziqni ajratamiz va tenglamasini yozamiz.
y – y1 = k ( x – x1 )
da x1 = 10, y1 = 3 bo‘lsa,
y – 3 = k ( x – 10 )
bu to‘g‘ri chiziqlar dastasidan BC( ) ga perpendikulyarini ajratish uchun k ni quyidagi perpendikulyarlik shartidan foydalanib topamiz:
, dan k = -8
Bu holda AD balandilik tenglamasi:
y – 3 = 8 ( x – 10 ), 8x + y – 83 = 0 (AD)
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