Решите пж номер 39.13(а)

Question

Решите пж
номер 39.13(а)

---

Answer 1

A(2,3),B(-1,4)
tg a = k = (y2-y1)/(x2-x1)=(4-3)/(-1-2)=1/(-3)=-1/3
y-y1=k(x-x1)
y-3=-1/3(x-2) /.3
3y-9=-(x-2)
3y-9=-x+2
x+3y-11=0

Comments

3y=-x+11, y=-x/3+11/3