Question

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)?

Answer 1

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

Explanation:

Equation of a line that is perpendicular to `Ax+By=C`

is of the form `Bx-Ay=k` i.e. reversing te coefficients of `x` and `y` and changing sign of one of them.

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

`3x-y=k` and as it passes through `(-3,5)`

we should have `3xx(-3)-5=k`

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

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