Question

1. M(2;-2)si P(0,4).Calculati coordonatele mijlocului [MP]
2.A(-1,0) si B(3,2).Ecuatial dreptei AB
3.E(-1,3) si F(3,2). Calculati distanta de la E LA F.
4.ū=3ī-2ῑ ; ӯ=ī+4ῑ. Aflati coordonatele ē=2ū-ӯ
5.Aflati panta dreptei AB, A(0,-2) si B(-1,1).
6.A(1,-2),B(2,0) si C(1,3). Ecuatia dreptei care trece prin A si este ⊥ BC
7.m=? daca ū1=5ī+mῑ si ū2=-ī+(m-1)ῑ sunt coliniare.

Answer 1

1. Mx+Px/2=2+0/2 =1 (coordonata x) ; My +Py/2=-2+4/2= 1; Mij:(1;1);
2.AB ec= (X-XA)/XB-XA=(Y-YA)/YB-YA=(x+1)/4=y/2;4y=2x+2;2y=x+1;x-2y+1=0
3.radical((Xf-Xe)^2 -(Yf-Ye)^2)= radical(17);
4.2u= 6i -4j; e= 6i-4j -i -4j= 5i -8j;
5.panta = (yB-yA)/xB-xA=3/-1=-3;
6.y-yA=m(x-xA)(formula ecuatiei printr-un punct cu panta);mA*mBC=-1;
mBC=3+0/-1=-3; -3mA=-1; mA=1/3;
y+2=1/3(x-1); y+2=x/3 -1/3; y=(-6-1+x)/3=(-7+x)/3; y=(x-7)/3; 3y=x-7;x-3y-7=0
7.daca vec sunt coliniari rezulta : 5/-1=m/m-1;formula(a1/a2=b1/b2);
-5=m/m-1;  m=-5m +5;  6m=5;  m=5/6;