Question

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

Answer 1

x-3y-7=0

Explanation:

`y=(1)/(3) x+c\\-2=(1)/(3)* 1+c\\c=-(7)/(3) \\y=(1)/(3) x-(7)/(3)\\\\3y=x-7, 0=x-3y-7`

Answer 2

x- 3y -7 =0

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