HOLA! les agradeceria mucho si me ayudan con esta tarea de geometria analitica por favor! : encontrar la ecuacion de la recta (forma dos puntos) 1.-pasa por (2,3) y (4,6) 2.-pasa por (3,1) y (5,1) 3.-pasa por (4,-3) y (5,-1) 4.-pasa por (2,0) y (2,6)

2

Respuestas

2012-09-26T07:35:06+02:00

1
(x1, y1) = (2,3)
(x2,y2) = (4,6)
pendiente = (y2-y1)/(x2-x1)= (6-3)/(4-2)=3/2
ecuacion=> y-y1=m(x-x1)
y-3=(3/2)(x-2) -> y = (3/2)x-(3/2)*2+3 -> y= (3/2)x -3 +3
ecuacion = y = (3/2)x

2
(x1, y1) = (3,1)
(x2,y2) = (5,1)
pendiente = (1-1)/(5-3) = 0/2 =0
ecuacion => y-y1 = m(x-x1) -> y-1=0(x-3)-> y-1 =0
ecuacion-> y=1

3
(x1, y1) = (4,-3)
(x2,y2) = (5,-1)
pendiente=(-1 - -3)/(5-4) = 2/(-1)= -2
ecuacion=> y- -3 = (-2)(x-4) -> y+3 = (-2)(x-4) -> y=-2x +8 -3
ecuacion -> y= -2x +5

4
(x1, y1) = (2,0)
(x2,y2) = (2,6)
pendiente = (6-0)/(2-2) = 6/0
ecuacion=> y-0 = (6/0)(x-2) -> 0(y-0)= 6(x-2) -> x-2=0
ecuacion-> x=2

2012-09-26T07:58:05+02:00

1:

m= (y2-y1)/(x2-x1)

m= (6-3)/(4-2)

m= 3/2

 

2:

m= (1-1)/(5-3)

m= 0/2

m = 0

 

3: 

m= (-1+3)/(5-4)

m= 2/1

M= 2

 

 

4:

m= (6-0)/(2-2)

m = 6/0

m= 0