Equation of line and its points 

In the xy coordinate system if (a,b) and (a+3,b+k) are two points on the line defined by equation (the equation is kind of faded in text , but it seems to be like x=3y-7) then k =
A)9 , B)3 , C)1 , D)1  (Ans is 1)


Any suggestions on how that answer was calculated ?

The slope of the line through $(a,b)$ and $(a+3, b+k)$ is $\frac{b+k-b}{a+3-a}$, which is $\frac{k}{3}$.
The slope of the line $x=3y-7$ is $\frac{1}{3}$. This is because the equation can be rewritten as $3y=x+7$, and then in standard slope-intercept form as $y=\frac{1}{3}x+\frac{7}{3}$.
These slopes are equal $\frac{k}{3}$ and $\frac{1}{3}$ are equal.
Another way: Because $(a,b)$ is on the line, we have $a=3b-7$. Because $(a+3,b+k)$ is on the line, we have $a+3=3(b+k)-7$, that is, $a+3=3b+3k-7$. 
Since $a=3b-7$, we conclude that $3=3k$.
