What is the equation of a line, in general form, that passes through point (1, -2) and has a slope of 1/3
O 3x-y-7=0
x - 3y + 7 = 0
Ox-3y - 7 = 0
O-X + 3y - 7 = 0

Respuesta :

Answer:

x- 3y -7 =0

Step-by-step explanation:

Point slope form is

y-y1 = m(x-x1)  where m is the slope

y - -2 = 1/3 (x-1)

y+2 = 1/3 (x-1)

Multiply each side by 3 to get rid of the fraction

3y +6 = x-1

Subtract 3x from each side

6 = x-3y -1

Subtract 6 from each side

0=x -3y -7

x- 3y -7 =0

The general form is

Ax +by +C = 0  where A , B C are integers and A > 0

Answer:

x-3y-7=0

Step-by-step explanation:

[tex]y=\frac{1}{3} x+c\\-2=\frac{1}{3}\times 1+c\\c=-\frac{7}{3} \\y=\frac{1}{3} x-\frac{7}{3}\\\\3y=x-7, 0=x-3y-7[/tex]