Equation of the line is $Ax+By+C=0$

This can be written as

$y= \bigg( -\large\frac{A}{B} \bigg) x + \bigg( -\large\frac{C}{B} \bigg)$

This is of the form $y=mx+C$

Hence slope of the form $m = -\large\frac{A}{B}$

We know that parallel lines have same slope. Hence the equation of the line passing through the point $(x_1,y_1)$ and slope $ -\large\frac{A}{B}$ is

$ y-y_1= \bigg( -\large\frac{A}{B} \bigg) (x-x_1)$

$ \Rightarrow B(y-y_1)=-A(x-x_1)$

$ \Rightarrow A(x-x_1)+B(y-y_1)=0$

Hence the line through the point $(x_1,y_1)$ and parallel to the line $Ax+By+C=0$ is $A(x-x_1)+B(y-y_1)=0$.

Hence proved.