Write the equation of the line in point-slope form that passes through (5, -1) and has a slope of 2/3

   "  y  =  (2/3)x  −  (13/3) " .

Note:  "point-slope" form;  or "slope-intercept form" is:
 
"y = mx + b" ; 

in which "y" is on the left-hand side of the equation, with NO coefficient (except for the "implied coefficient of "1"); 

m = the slope; and is the coefficient of "x" ; 

b = the y-intercept; or the value of "x" at the point which "y = 0" .

We are given the following point on the line:  (5, -1).   

We are given the slope, " m = 2/3 " .

 Note the formula:

y − y₁ = m(x − x₁) ;   Given "x₁ = 5" ;  "y₁ = -1" ;  "m = 2/3" ;

Plug in these known values into the formula:

        y − (-1) = (2/3)*(x − 5) ;  

to get:

          y + 1 = (2/3)x − (2/3)*(5) ;

Note:  5* (2/3) = 5/1  * 2/3 = (5*2)/(1*3) = 10/3

          y + 1  = (2/3)x  −  10/3  ;

Subtract "1" from EACH side of the equation:

         y + 1 − 1 = (2/3)x  −  (10/3) − (1) ;    {Note:  "3/3 = 1"}.

Rewrite as:

              y + 1 − 1 = (2/3)x  −  (10/3) − (3/3) ;
 
to get:

              y  =  (2/3)x  −  (13/3) .