Question

what is (7, -3), (4, -8) in standard form?

Answer 1

\[m=\frac{ y^2-y^1 }{x^2-x^1 }\]

Answer 2

\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ 7}}\quad ,&{\color{blue}{ -3}})\quad &({\color{red}{ 4}}\quad ,&{\color{blue}{ -8}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}
\\ \quad \\

y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "0"}\\
\qquad \uparrow\\
\textit{point-slope form}\)

thus it'd end up looking like mx+y=0