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.
Best Answer
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.