Verified answer

coeficiente angular m = -2/3

--> equação da reta

y = -2x/3 + b

ponto A(4,-2)

-2 = -2.4/3 + b

b = -2 + 8/3 = 2/3

y = -2x/3 + 2/3

m=-2/3

ponto(4,-2)

y2-y1/x2-x1

ponto(4,-2)

m=y-(-2)/x-4

m=y+2/x-4=-2/3

2(x-4)= -3(y+2)

2x-8=-3y-6

2x-8+6+3y=0

2x+3y-2=0

Y-Y0=m(X-X0)

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

Y+2=-2/3X+8/3

2/3X+Y=8/3-2

2/3X+Y=2/3

2/3X+Y-2/3=0

Y-y=m.(X-X)

substituindo temos

Y+2= -2/3.X-4/3

3Y+6=-2X-4

3Y+10+2X=0

(y - -2)/(x-4)= -2/3

(y+2)/(x-4)= -2/3

3(y+2)=-2*(x-4)

3y+6=2x-8

2x-3y-8-6=0

2x-3y-14=0

m = (y-y0)/(x-x0)

-2/3 = (y-(-2))/(x-4)

-2/3 = (y+2)/(x-4)

-2x/3+8/3=y+2

y+2+2x/3-8/3=0

y+2x/3-2/3=0