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)
Soluções para a tarefa
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
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