Move every term to the left-hand side to get \(\displaystyle ax+by+c=0 \):
\[y-15=\sqrt2\,x \;\Rightarrow\; -\sqrt2\,x+y-15=0 \]
\[3y-2x=0 \;\Rightarrow\; -2x+3y+0=0 \]
\[5x=3y \;\Rightarrow\; 5x-3y+0=0 \]
\[x=8 \;\Rightarrow\; x+0\cdot y-8=0 \]
\[3y=1 \;\Rightarrow\; 0\cdot x+3y-1=0 \]
\[\begin{array}{|l|l|c|c|c|} \hline \text{Equation} & \text{Standard form} & a & b & c \\ \hline y-15=\sqrt2\,x & -\sqrt2\,x+y-15=0 & -\sqrt2 & 1 & -15 \\ \hline 3y-2x=0 & -2x+3y+0=0 & -2 & 3 & 0 \\ \hline 5x=3y & 5x-3y+0=0 & 5 & -3 & 0 \\ \hline x=8 & x+0\cdot y-8=0 & 1 & 0 & -8 \\ \hline 3y=1 & 0\cdot x+3y-1=0 & 0 & 3 & -1 \\ \hline \end{array} \]
Multiplying a row by \(\displaystyle -1 \) gives an equally valid standard form.
Answer: the table above, with \(\displaystyle (a,b,c)=(-\sqrt2,1,-15),\ (-2,3,0),\ (5,-3,0),\ (1,0,-8),\ (0,3,-1) \)