# How do you find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5)?

Feb 1, 2017

Desired equation is $3 x - y + 14 = 0$

#### Explanation:

Equation of a line that is perpendicular to $A x + B y = C$

is of the form $B x - A y = k$ i.e. reversing te coefficients of $x$ and $y$ and changing sign of one of them.

Hence, the line perpendicular to $x + 3 y = 6$ has equation of the form

$3 x - y = k$ and as it passes through $\left(- 3 , 5\right)$

we should have $3 \times \left(- 3\right) - 5 = k$

and therefore $k = - 9 - 5 = - 14$

and desired equation is $3 x - y = - 14$ or $3 x - y + 14 = 0$
graph{(3x-y+14)(x+3y-6)=0 [-11.38, 8.62, -3.12, 6.88]}