How do you write an equation of a line in slope intercept form if the line has a slope of $- \frac{2}{3}$ $-2/3$ and contains the point $(2, - 2)$ $(2,-2)$ ?

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Answer 1 · 2018-04-16T04:45:23

$2 x + 3 y + 2 = 0$ $2x + 3y + 2 = 0$

$S l o p e = m = - \frac{2}{3}, (x_{1}, y_{1}) = (2, - 2)$ $Slope = m = -2/3, (x_1, y_1) = (2, -2)$

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$Point - Slope Form (y - y_{1}) = m \cdot (x - x_{1})$ $"Point - Slope Form " (y - y_1) = m * (x - x_1)$

$y + 2 = - (\frac{2}{3}) \cdot (x - 2)$ $y + 2 = -(2/3) * ( x - 2)$

$3 y + 6 = - 2 x + 4$ $3y + 6 = -2x + 4$

$2 x + 3 y + 2 = 0$ $2x + 3y + 2 = 0$

How do you write an equation of a line in slope intercept form if the line has a slope of -2/3−23-2/3 and contains the point (2,-2)(2,−2)(2,-2)?