First Method.

You are given that the line goes through $(0,\var{b})$ and $(-1,\var{b-a})$ and the equation of the line is of the form $y=ax+b$

Hence:

1) At $x=0$ we have $y=\var{b}$, and this gives $\var{b}=a \times 0 +b =b$ on putting $x=0$ into $y=ax+b$.

So $b=\var{b}$.

2) At $x=-1$ we have $y=\var{b-a}$, and this gives $\var{b-a}=a \times (-1) +b =\simplify[all,!collectNumbers]{-a+{b}}$ on putting $x=-1$ into $y=ax+b$.

On rearranging we obtain $a=\simplify[all,!collectNumbers]{{b}-{b-a}}=\var{a}$. 

So $a=\var{a}$.

So the equation of the line is $\simplify{y={a}*x+{b}}$.

Second Method.

The equation $y=ax+b$ tells us that the graph crosses the $y$-axis (when $x=0$) at $y=b$.

So looking at the graph we immediately see that $b=\var{b}$.

$a$ is the gradient of the line and is given by the change from $(-1,\var{b-a})$ to $(0,\var{b})$:

$$a=\frac{\text{Change in y}}{\text{Change in x}}=\frac{\simplify[all,!collectNumbers]{({b-a}-{b})}}{-1-0}=\var{a}$$