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Find another point on the line that passes through the point (4,5) and has a slope of 2. Explain how you obtained your answer.

User EWizardII
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1 Answer

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well, we know its slope and we also know a point on it, so hmmm without further ado, we can get its equation, once we have the equation, we can get any point we like on it pretty much, so


(\stackrel{x_1}{4}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ 2 \\\\\\ \begin{array}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}{5}=\stackrel{m}{2}(x-\stackrel{x_1}{4}) \\\\\\ y-5=2x-8\implies y=2x-3

well then hmmm to get some other point hmmm let's pick a random "x" hmmm say dunno x = -13, so


\underline{x=-13}\hspace{5em}y=2(\stackrel{x}{-13})-3\implies y=-26-3\implies \underline{y=-29} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \underset{another~point}{(-13~~,~~-29)}~\hfill

User DanSogaard
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