Here's the question you clicked on:
mathisblockingmypath
A line passes through (1, –5) and (–3, 7).Write an equation for the line in point-slope form. Rewrite the equation in slope-intercept form.
\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ 1}}\quad ,&{\color{blue}{ 05}})\quad &({\color{red}{ -3}}\quad ,&{\color{blue}{ 7}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values }\\ \qquad \uparrow\\ \textit{point-slope form}\) and to get it in slope-intercept form , \(\bf y = mx+b\) just solve it for "y"
hmm got a 0 there for no good reason =) anyhow \(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ &({\color{red}{ 1}}\quad ,&{\color{blue}{ 5}})\quad &({\color{red}{ -3}}\quad ,&{\color{blue}{ 7}}) \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values }\\ \qquad \uparrow\\ \textit{point-slope form}\)
okay so how would the final answer look