[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{11})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{17}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{17-11}{4-2}\implies \cfrac{6}{2}\implies 3 \\\\\\ \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-11=3(x-2)\implies y-11=3x-6[/tex]
[tex]\bf y=3x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
Answer:
[tex]y = 3x + 5[/tex]
Step-by-step explanation:
First, find the rate of change [slope]:
-y₁ + y₂\-x₁ + x₂ = m
[tex]\frac{-11 + 17}{-2 + 4} = \frac{6}{2} = 3[/tex]
Now, plug the coordinates into the Slope-Intercept Formula instead of the Point-Slope Formula because you get it done much swiftly. It does not matter which ordered pair you choose:
17 = 3[4] + b
12
5 = b
[tex]y = 3x + 5[/tex]
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11 = 3[2] + b
6
5 = b
[tex]y = 3x + 5[/tex]
You see? I told you it did not matter which ordered pair you choose because you will always get the exact same result.
I am joyous to assist you anytime.
