Write the equation of a line in slope-intercept form that passes through the points: (1,5) and (3,13)

The answer is:  " y = 4x + 1 " .

Explanation:

The "slope-intercept form" equation of a line is:

         " y = mx + b " ; 

in which:

     m = the slope (and is the co-efficient of "x" ) ; 

     b = the "y-intercept" (of the line). 

     "y" exists as a "stand-alone" variable on the "left-hand side" of the equation.


Given two points on that line:

 (1, 5) ; and (3, 13). 

Let's refer to these points as:

(x₁ , y₁) and (x₂ , y₂) ; 

So;  x₁ = 1 ; y₁ = 5 ;  x₂ = 3;  y₂ = 13

To solve for the slope, "m", of the line:

m = (y₂ − y₁) / (x₂ − x₁)  =  (13 − 5) / (3 − 1) = (8/2) = 4 . 

So, the slope, "m" , equals "4" .

Now, to write the equation:

Note:  " y − y₁ = m(x - x₁) " ; 

We know that  " y₁ = 5 " ;  and that:  " x₁ = 1 " ; and that "m = 4" ; 

So;   " y − 5 = 4(x − 1) " ; 

Note the "distributive property" of multiplication :

   a(b + c)  =   ab + ac ; 

   a(b − c)  =   ab − ac .

So, taking the "right-hand side" of the equation:

    " 4(x − 1)  =  (4 * x) − (4 * 1)  =  4x − 4 " ;  

and rewrite the equation:

       " y − 5  =  4x − 4 " ; 

Add "5" to each side of the equation; to isolate "y" on one side of the equation:

       " y − 5 + 5  =  4x − 4 + 5 " ; 

to get:

       " y = 4x + 1 " ;

which is written in "slope-intercept form" :

      " y = mx + b " ;  in which " m = 4 " ; and " b = 1 " .

2/1 or just 2. sorry if i get this wrong