What is the equation of the line in standard form?

So I find the slope first which is
m=(y2-y1/x2-x1)
m=(2-1/3--3)
m= (1/6)

Then, I put the numbers in slope intercept form first,
y-1=(1/6)(x-(-3))

Now change it to standard form:
( y-1=(1/6)(x-(-3)) ) times 6
6y-6=1(x+3) minus x on both sides and plus 6 on both sides and get the
6y-x=9

And you can plug the numbers in to check

Answer x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​

step-by-step explanation: