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I need to find a vector that is perpendicular to the line $3x-4y=6$.

I started with calculating slope of the line which I get $3/4$. A perpendicular line/vector would have a negative reciprocal slope if $-4/3$.

I also know that a perpendicular vector has an equation of ac+bd=0 but I wasn't sure if I needed the equation as well.

My thought was to find two points on the line $(2,0)$ and $(10,6)$ which form a vector but I didn't know where to go from there.

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  • $\begingroup$ Vector of what length ? $\endgroup$ – A---B Mar 30 '17 at 21:17
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It's the vector $\vec v(3,-4)$ for a very simple reason: let $M(x,y)$ and $M'(x',y')$ two points on the line. Thus we have $$3x-4y=6=3x'-4y',$$ which implies $\;3(x-x')-4(y-y')=0\iff \vec v\cdot\overrightarrow{M'M}=0.$

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Take two points of the line $(x_0,y_0)$ and $(x_1,y_1)$. then we get:

$$3x_0-4y_0=6\quad (1)\\ 3x_1-4y_1=6\quad (2)$$

Now make $(2)-(1)$

$$3(x_1-x_0)-4(y_1-y_0)=0\Leftrightarrow (3,-4)\cdot (x_1-x_0,y_1-y_0)=0$$

but $(x_1-x_0,y_1-y_0)$ is a vector that give us the direction of the line, then $$(3,-4)$$ is perpendicular to the line.

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