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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.
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
 
How did I get the 7?
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3 Answers

Hi Elder;
(1, 2) and with slope of - 1 3
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
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 = 2 + 1/3 = 6/3 + 1/3

3•(1/3)x+3•y = 3•7/3

x+3y = 7
 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

Comments

slope is negative 1/3