What is the equation of the line with slope -2/3 and passes through the point (3, 2)?

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Answer 1 · 2016-06-09T23:45:15

The equation of the line in slope-intercept form becomes

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

The slope intercept form of a line is $y = m x + b$ $y=mx+b$

For this problem we are given the slope as $- \frac{2}{3}$ $-2/3$ and a point on the line of $(3, 2)$ $(3,2)$

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

We plug in the values and then solve for the $b$ $b$ term which is the
y-intercept.

$2 = - \frac{2}{3} (3) + b$ $2 =-2/3(3) +b$

$2 = - 2 + b$ $2 =-2 +b$

Now isolate the $b$ $b$ term.

$2 +2 =cancel(-2) cancel(+2)+b$

$b= 4$

The equation of the line in slope-intercept form becomes

$y = -2/3 x + 4$