[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-6})~\hspace{10em} \stackrel{slope}{m}\implies 5 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{5}(x-\stackrel{x_1}{1}) \\\\\\ y+6=5x-5\implies y=5x-11[/tex]
Answer: [tex]y=5x-11[/tex]
Step-by-step explanation:
The equation of the line in Slope-Intercept form is:
[tex]y=mx+b[/tex]
Where "m" is the slope and "b" is the y-intercept
We know that the slope of this line is:
[tex]m=5[/tex]
And its that passes through the point (1,-6), then we can substitute values into [tex]y=mx+b[/tex] and solve for "b":
[tex]-6=5(1)+b\\\\-6-5=b\\\\b=-11[/tex]
Then, the equation of this line in Slope-Intercept form is:
[tex]y=5x-11[/tex]
