IDNLearn.com makes it easy to find the right answers to your questions. Discover in-depth answers from knowledgeable professionals, providing you with the information you need.
Sagot :
to get the equation of any straight line, we simply need two points off of it, let's use the ones in the table.
[tex]\begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} -1&-10\\ 3&14\\ \cline{1-2} \end{array}\hspace{5em} (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{14}-\stackrel{y1}{(-10)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-1)}}} \implies \cfrac{14 +10}{3 +1}\implies \cfrac{24}{4}\implies 6[/tex]
[tex]\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}{(-10)}=\stackrel{m}{6}(x-\stackrel{x_1}{(-1)}) \\\\\\ y+10=6(x+1)\implies y+10=6x+6\implies y=6x-4[/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.
