What is this equation in point-slope form of the line passing through (-2,1) and (4,13)

The answer is:  " y = 2x + 5 ".
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{Refer to the graph of the equation; as an "attached image".}.
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Explanation: 
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Note: The equation of a line; in "point slope form"; also known as:  "slope-intercept form";  is: " y = mx + b "

which is:
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" y = mx + b " ;

in which:
"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:
"m" is the slope of the line; which is what we want to solve for;
"b" is the "y-intercept"; or more precisely, the value of "x"
(that is; the "x-coordinate") of the point at which "y = 0";
that is, the value of "x" ; or the "x-coordinate" of the point at which
the graph of the equation crosses the "x-axis".
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Given the coordinate of any TWO (2) points on the line:

Calculate the slope, "m"  =  (y₂ − y₁) / (x₂ − x₁) ;

Given the two points: "(-2,1)" and "(4,13)" ;
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(x₁, y₁) ↔ (-2, 1) ;  x₁, = -2 ;   y₁ = 1 ;

(x₂, y₂) ↔ (4, 13);  x₂ = 4 ; y₂ = 13 ;
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 →  m =  (y₂ − y₁) / (x₂ − x₁)  = (13 − 1) /  [(4 − (-2)]  ; 
                                             = 12 / (4+2) ;
                                             = 12 / 6 ;
                                             = 2 ;
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So, we know that:  "m = 2" .
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 In "point-slope form" :
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  → " y = mx + b " ;

So, we can write:  y = 2x + b ; 

We need to solve for "b".
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  →   When "y = 0" , what does "x" equal?  The answer to this question, is the value for "b" (the "y-intercept" of the line of graph of this equation).
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One method is to plot these points on a graph;  and, using our known slope, "m = 2" ;  visually inspect the graph, plot other points; noting that the slope, "2", is "rise over the run"; i.e. the "change in y" over "the change in x";
   (i.e. "Δy /Δxgraph the equation) ; and determine the "x-coordinate" of the y-intercept).

The other method is to solve for "b" ; as follows:
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  →  y = 2x + b ;

Subtract "2x" from EACH SIDE of the equation; to isolate "b" on one side of the equation; and to solve for "b" ;
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  →  y − 2x = 2x + b − 2x ;
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to get:  y − 2x = b ;
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  ↔  b = y − 2x ;

  ↔ b = -2x + y ;

To solve for "b" ; take EITHER of the 2 points given;  and plug in the values for "x" and "y" into the equation, to solve for "b"
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  Let us try:  " (-2, 1) " :   x = -2,  y = 1  ; 
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  → b = -2x + y = -2(-2) + 1 = 4 + 1 = 5 ; →  b = 5 .
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Alternately, and to confirm that "b = 5" ;  (Check our work):
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  Let us try the second point:  " (4, 13) " :
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 → b = -2x + y =  -2(4) + 13 = -8 + 13 = 5 ;  → b = 5 . Yes!
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 So; we have:   →  y = 2x + b ;   and we know:  " b = 5" ; so we can write the equation:
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        " y = 2x + 5 " .
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The answer is:  " y = 2x + 5 ".
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{Refer to the graph of the equation; as an "attached image"}:
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