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

May 3, 2018

$3 x + y + 5 = 0$ is the required equation of the straight line. graph{(3x+y+5)(x-3y-2)=0 [-8.44, 2.66, -4.17, 1.38]}

#### Explanation:

Any line perpendicular to $a x + b y + c = 0$ is $b x - a y + k = 0$ where k is constant.

Given equation is

$\rightarrow 3 y = x - 2$

$\rightarrow x - 3 y = 2$

Any line perpendicular to $x - 3 y = 2$ will be $3 x + y + k = 0$

As $3 x + y + k = 0$ passes through $\left(- 3 , 4\right)$, we have,

$\rightarrow 3 \cdot \left(- 3\right) + 4 + k = 0$

$\rightarrow - 9 + 4 + k = 0$

$\rightarrow k = 5$

So, the required equation of the straight line is $3 x + y + 5 = 0$