How do you write an equation of a line passing through (3, -4), perpendicular to  3y=x-2?

Oct 4, 2016

$\text{equation of line passing through (3,-4) is } y = - 3 x + 5$

Explanation:

$\textcolor{b l u e}{3 y = x - 2}$

$\textcolor{b l u e}{y = \frac{1}{3} x - \frac{2}{3}}$

$\text{the slope of blue line is : "m_"blue} = \frac{1}{3}$

${m}_{\text{red":"the slope of red line}}$

$\text{so " alpha=90 " , " m_"blue" * m_"red} = - 1$

$\frac{1}{3} \cdot {m}_{\text{red}} = - 1$

${m}_{\text{red}} = - 3$

$y - {y}_{1} = {m}_{\text{red}} \cdot \left(x - {x}_{1}\right)$

$y + 4 = - 3 \left(x - 3\right)$

$y + 4 = - 3 x + 9$

$y = - 3 x + 9 - 4$

$y = - 3 x + 5$