Morena15
Respondido

Dê a equação geral da reta em cada caso:

 

a)A(1,1) e B(-1,-5)

 

b)A(-2,-2) e B(2,4)

 

c)A(-1,5) e B(6,5)

 

d)A(12,5) e B(12,-4)

Resposta :

homemformiga

a) A(1,1) B(-1,-5)

Matriz:

 1, 1, 1, 1, 1
-1,-5, 1,-1,-5
 x, y, 1, x, y

 

[(1.(-5).1)+(1.1.x)+(1.(-1).y)]-[(1.(-1).1)+(1.1.y)+(1.-5.x)]
[-5+x-y]-[-1+y-5x]=0 =>
-5+x-y+1-y+5x=0 =>
-5+1+x+5x-y-y=0 =>

-4+6x-2y=0

 

b)A(-2,-2) B (2,4)

Matriz:

-2,-2, 1,-2,-2
 2, 4, 1, 2, 4
 x, y, 1, x, y


[(-2.4.1)+(-2.1.x)+(1.2.y)]-[(-2.2.1)+(-2.1.y)+(1.4.x)]
[-8-2x+2y]-[-4-2y+4x] =>
-8-2x+2y+4+2y-4x=0 => -8+4-2x-4x+2y+2y=0 =>

-4-6x+4y=0

 

c) A(-1,5) B(6,5)

Matriz:

-1, 5, 1,-1, 5
 6, 5, 1, 6, 5
 x, y, 1, x, y

 

[(-1.5.1)+(5.1.x)+(1.6.y)]-[(5.6.1)+(-1.1.y)+(1.5.x)]=0
[-5+5x+6y]-[30-y+5x]=0 => -5+5x+6y-30+y-5x=0
-5-30+5x-5x+6y+y=0
-35+0+7y=0

 

d)A(12,5)B(12,-4)

Matriz:

12, 5, 1, 12, 5
12,-4, 1, 12,-4
x,  y, 1,  x, y

 

[(12.(-4).1)+(5.1.x)+(1.12.y)]-[(5.12.1)+(12.1.y)+(1.(-4).x)]=0
[-48+5x+12y]-[60+12y-4x]=0 => -48+5x+12y-60-12y+4x=0
-48-60+5x+4x+12y-12y=0  =>

 

-108+9x=0

gabyin

a) coeficente algular (m)= y-yo/ x - xo

m= -5-1/-1-1

m= 3

m= 1/3

Y-1= 3.(X-1)

Y-1-3x+3=0

Y- 3x +2= 0

 

b)m= 4+2/2+2

m= 6/4

m=3/2

Y-2= (3/2).(X-4)

Y -(3/2)X + 4=0

 

c) m= 5-5/6+1

m= 0

Y-5= 0.(X-6)

Y-5-0x+6=0

y+1=0

 

d) m= -4-5/12-12

m= não existe

Ou seja, a reta não possue coeficiente angular, e assim ela é vertical. 

X= 12

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