207k views
3 votes
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

User Gbdcool
by
8.4k points

2 Answers

5 votes

Answer:

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

User Jennise
by
9.0k points
4 votes

Answer:

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

User Cfreak
by
7.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories