Answer:
[y−5=4−1{x−(−3)}
\begin{gathered}\longrightarrow y -5=\dfrac{-1}{4}(x+3)\\\end{gathered}⟶y−5=4−1(x+3)
\begin{gathered}\longrightarrow 4(y -5 ) =-1(x +3) \\\end{gathered}⟶4(y−5)=−1(x+3)
\begin{gathered}\longrightarrow 4y -20 = - x -3 \\\end{gathered}⟶4y−20=−x−3
\begin{gathered}\longrightarrow x + 4y -20+3=0\\\end{gathered}⟶x+4y−20+3=0
\begin{gathered}\longrightarrow \underset{Standard \ Form }{\underbrace{\underline{\underline{ x + 4y -17=0}}}} \\\end{gathered}⟶Standard Formx+4y−17=0
[tex][y−5=4−1{x−(−3)}
\begin{gathered}\longrightarrow y -5=\dfrac{-1}{4}(x+3)\\\end{gathered}⟶y−5=4−1(x+3)
\begin{gathered}\longrightarrow 4(y -5 ) =-1(x +3) \\\end{gathered}⟶4(y−5)=−1(x+3)
\begin{gathered}\longrightarrow 4y -20 = - x -3 \\\end{gathered}⟶4y−20=−x−3
\begin{gathered}\longrightarrow x + 4y -20+3=0\\\end{gathered}⟶x+4y−20+3=0
\begin{gathered}\longrightarrow \underset{Standard \ Form }{\underbrace{\underline{\underline{ x + 4y -17=0}}}} \\\end{gathered}⟶Standard Formx+4y−17=0
[/tex]
