y−2 =

-1

3

•(x−1)

y−2 =

-1

3

x+

1

3

3

x+y =

7

3

7

3

I have already completed the question but I am confused on how I got the the answer.

y−y1 = m(x−x1)

y−2 =

-1

3

•(x−1)

y−2 =

-1

3

x+

1

3

y−2 =

-1

3

•(x−1)

y−2 =

-1

3

x+

1

3

1

3

x+y =

7

3

3

x+y =

7

3

3•

1

3

x+3•y = 3•

7

3

7

3

x+3y = 7

How did I get the 7?

Tutors, sign in to answer this question.

Hi Elder;

I assume -1 3 is -1/3, not -13.

Standard form is...

Ax+By=C, neither A nor B equal zero and A is greater than zero.

Slope is -A/B

slope is...-(-1/3)

A=1, B=3

A=1, B=3

x+3y=C

Let's plug-in one point to establish C...

(1)+[(3)(2)]=C

1+6=C

7=C

x+3y=7

"Write the equation of the line which passes through (1, 2) and with slope of - 1 3 in standard form.

I have already completed the question but I am confused on how I got the the answer. How did I get the 7?"

y−y1 = m(x−x1)

y−2 = (-1/3)•(x−1)

y−2 = (-1/3) x + 1/3

(1/3)x + y = 7/3

3•(1/3)x+3•y = 3•7/3

x+3y = 7

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Line is a location of all points with the equal slope.

( X, Y) ( 1, 2 ) Means that all points with the coordinate of ( x. Y ) have the relationship of

Y - 2 = 1/3 , then Alternative solution:

X - 1 Y = 1/3 X + b

2 = 1/3 +b

( Y -2 ) 3 = X -1 b = 2 -1/3 = 5/3

3 Y - 6 = X -1 Y = X/3 + 5/3

3y = X +5

Y = X/3 +5 /3

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