# How do you write an equation of a line passing through (6, 4), perpendicular to y=3x - 2?

Jul 15, 2016

$x + 3 y - 18 = 0$

#### Explanation:

Slope intercept form of equation of a line is $y = m x + c$, where $m$ is its slope and $c$ is its intercept on $y$-axis. As such slope of $y = 3 x - 2$ is $3$.

As product of slopes of two lines perpendicular to each other is $- 1$, hence slope of line perpendicular to $y = 3 x - 2$ is $- \frac{1}{3}$.

Equation of a line passing through $\left({x}_{1} , {y}_{1}\right)$ and having a slope of $m$ is $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$. Hence equation of line passing through $\left(6 , 4\right)$ and slope of $- \frac{1}{3}$ is

$\left(y - 4\right) = - \frac{1}{3} \left(x - 6\right)$ or

$3 y - 12 = - x + 6$ or

$x + 3 y - 12 - 6 = 0$ i.e.

$x + 3 y - 18 = 0$