Question #bccac

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Question #bccac

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Answer 1 · 2017-12-03T01:50:35

By converting the equation in point-slope form to slope-intercept form, we obtain $y = \frac{8}{9} x + \frac{8}{3}$ $y=\frac{8}{9}x+\frac{8}{3}$

Explanation:

Given the slope of a line and a point on it, we can write its equation in point-slope form, denoted by:

$y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$

Where $m$ $m$ is the slope, and $(x_{1}, y_{1})$ $(x_1,y_1)$ is a point on the line.

Let’s plug those values in:

$y - (- 8) = \frac{8}{9} (x - 6)$ $y-(-8)=\frac{8}{9}(x-6)$

$\Rightarrow y + 8 = \frac{8}{9} (x - 6)$ $\implies y+8=\frac{8}{9}(x-6)$

We can now move around terms to convert this to slope-intercept form:

$y = 8 = \frac{8}{9} (x - 6)$ $y=8=\frac{8}{9}(x-6)$

$\Rightarrow y + 8 = \frac{8}{9} x - \frac{16}{3}$ $\implies y+8=\frac{8}{9}x-\frac{16}{3}$

$\Rightarrow y = \frac{8}{9} x + \frac{8}{3}$ $\implies y=\frac{8}{9}x+\frac{8}{3}$

Answer 2 · 2017-12-03T08:14:07

$y = \frac{8}{9} x - \frac{40}{3}$ $y=8/9x-40/3$

Explanation:

$the equation of a line in slope-intercept form$ $"the equation of a line in "color(blue)"slope-intercept form"$ is.

$color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))$

$"where m is the slope and b the y-intercept"$

$"here "m=8/9$

$rArry=8/9x+blarrcolor(blue)"is the partial equation"$

$"to find b substitute "(6,-8)" into the partial equation"$

$-8=(8/9xx6)+brArrb=-8-16/3=-40/3$

$rArry=8/9x-40/3larrcolor(red)"in slope-interceot form"$

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