How do you write an equation in slope-intercept form for an equation of a line perpendicular to $y = - 3 x + 1$ $y = -3x +1$ and that intersects at point $(1, \frac{2}{3})$ $(1, 2/3)$ ?

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How do you write an equation in slope-intercept form for an equation of a line perpendicular to $y = - 3 x + 1$ $y = -3x +1$ and that intersects at point $(1, \frac{2}{3})$ $(1, 2/3)$ ?

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Answer 1 · 2018-03-29T07:30:28

$3 y = x + 1$ $3y=x+1$

Explanation:

The slope of this line is $- 3$ $-3$ . Slope of a line perpendicular to it it $\frac{1}{3}$ $1/3$ .
Because $m_{1} \cdot m_{2} = - 1$ $m_1*m_2=-1$ where $m_{1}$ $m_1$ and $m_{2}$ $m_2$ are the slope of perpendicular lines

Now, the equation of the line is $y = \frac{x}{3} + c$ $y=x/3+c$

To find $c$ $c$ , put a point which in this case is $(1, \frac{2}{3})$ $(1,2/3)$

$\frac{2}{3} = \frac{1}{3} + c$ $2/3=1/3+c$ $\Rightarrow$ $=>$ We get that $c = \frac{1}{3}$ $c=1/3$

So required equation is $y = \frac{x}{3} + \frac{1}{3}$ $y=x/3+1/3$ or $3 y = x + 1$ $3y=x+1$

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How do you write an equation in slope-intercept form for an equation of a line perpendicular to $y = - 3 x + 1$ $y = -3x +1$ and that intersects at point $(1, \frac{2}{3})$ $(1, 2/3)$ ?

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Explanation:

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How do you write an equation in slope-intercept form for an equation of a line perpendicular to $y = - 3 x + 1$ $y = -3x +1$ and that intersects at point $(1, \frac{2}{3})$ $(1, 2/3)$ ?