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)?

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Answer 1 · 2017-02-01T04:57:54

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

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]}