How do you write the equation of the line with a slope of 2/3 and passing through the point (6,-2)?

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Answer 1 · 2016-08-30T11:46:00

We begin by writing $y = \frac{2}{3} x + b$ $y=2/3x+b$

Now plug in the values for $x$ $x$ and $y$ $y$ from the point $(6, - 2)$ $(6,-2)$

$- 2 = \frac{2}{3} \cdot 6 + b \to - 2 = 4 + b \to b = - 6$ $-2=2/3*6+b->-2=4+b->b=-6$

So the equation in slope-intercept form is:
$y = \frac{2}{3} x - 6$ $y=2/3x-6$
graph{2/3x-6 [-5.25, 14.75, -7.44, 2.56]}

Answer 2 · 2016-08-30T12:26:13

$y = \frac{2}{3} x - 6$ $y = 2/3x-6$

If we are given the slope and one point, substitute these values into the following formula which is based on the definition for slope.

$y - y_{1} = m (x - x_{1}) \times \times \leftarrow m = \frac{2}{3} and (6, - 2)$ $y-y_1 = m(x-x_1) color(white)(xxxx) larr m = 2/3 and (6,-2)$

$y - (- 2) = \frac{2}{3} (x - 6) \times \times \leftarrow$ $y -(-2) = 2/3(x -6)color(white)(xxxx)larr$ multiply out and simplify.

$y + 2 = \frac{2}{3} x - 4$ $y+2 = 2/3x-4$

$y = \frac{2}{3} x - 4 - 2$ $y = 2/3x-4-2$

$y = \frac{2}{3} x - 6$ $y = 2/3x-6$