Pls help! I’m stuck!

Mathematics

1 answer:

nadezda [96]3 years ago

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Answer:

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PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)

ExtremeBDS [4]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 3 &,& -4~) 
% (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}$

$\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23$

side note: multiplying by the LCD of both sides is just to get rid of the denominators

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