What is the slope-intercept form of the equation of the line that contains the points (3, 8) and (6, 5) ?

Question

What is the slope-intercept form of the equation of the line that contains the points (3, 8) and (6, 5) ?

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    0
    2021-10-11T03:36:16+00:00

    First for the question you need to find the slope the formula for slope is:
     \frac{y 2 - y1}{x2 - x1}  =  \: slope
    so now you substitute the points you have into this forumla like this:
     \frac{5 - 8}{6 - 3}  =  \frac{ - 3}{ - 3}  = 1
    Now we know the slope is 1. We now have to find what b will be in the equation y= mx + b.
    To find this out, you must choose one of the points given( you can choose either one you will still get a right answer) For this I will choose point (3,8). In the point 3 is the x value and 8 is the y value. With these numbers and the slope we found we will put them into the equation y = mx + b form it will now look like this:

    8 = 1(3) + b now solve

    8 = 3 + b
    -3 -3

    5 = b. Now you have all the numbers to put write the complete equation. The answer is this:

    y = x + 5

    0
    2021-10-11T03:36:41+00:00

    \bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{8})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-8}{6-3}\implies \cfrac{-3}{3}\implies -1 \\\\\\ \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-8=-1(x-3) \\\\\\ y-8=-x+3\implies y=-x+11

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