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2 years ago

Find the point slope equation for the line that passes through the points (7,-21) (-4,23)

1 answer:

5 0

$\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-21})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{23}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{23-(-21)}{-4-7}\implies \cfrac{23+21}{-11}\\\\\\ \cfrac{44}{-11}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-21)=-4(x-7)\implies y+21=-4(x-7)$

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Write an equation that represents the line. (-3,-6) and (2,-2) use exact numbers

Ilia_Sergeevich [38]

Given:

The line passes through (-3,-6) and (2,-2).

To find:

The equation of line.

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If a line passes through two points $(x_1,y_1)\text{ and }(x_2,y_2)$ , then the equation of line is

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

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$y+6=\dfrac{-2+6}{2+3}(x+3)$

$y+6=\dfrac{4}{5}(x+3)$

$y+6=\dfrac{4}{5}(x)+\dfrac{4}{5}(3)$

Subtract 6 from both sides.

$y=\dfrac{4}{5}(x)+\dfrac{12}{5}-6$

$y=\dfrac{4}{5}(x)+\dfrac{12-30}{5}$

$y=\dfrac{4}{5}(x)+\dfrac{-18}{5}$

$y=\dfrac{4}{5}(x)-\dfrac{18}{5}$

Therefore, the equation of line is $y=\dfrac{4}{5}(x)-\dfrac{18}{5}$ .

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2 years ago

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Lapatulllka [165]

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