direction vector of line

Question

Could someone explain in simple words (maybe with a drawing) why the direction vector of

$$ \alpha x+\beta y+\gamma=0 $$

is equal to

$$ (\beta,-\alpha) $$

Answer 1

Let's break this down step by step and use a simple visual representation to understand why the direction vector of the line given by the equation $αx + βy + γ = 0$ is $(β, -α)$.

1. The equation $αx + βy + γ = 0$ represents a straight line in a $2D$ coordinate system (a plane). It's in what's called the general form of the equation of a line.

2. To find the direction vector of this line, we can convert this equation into a slope-intercept form, which is more intuitive. The slope-intercept form of a line is $y = mx + b$, where $m$ is the slope, and $b$ is the $y$-intercept.

3. To convert $αx + βy + γ = 0$ into the slope-intercept form, we solve for $y$: $$βy = -αx - γ \tag 1$$ $$y = {-α \over β}x - {γ \over β} \tag 2$$

4. Now, we can see that the slope $m$ of this line is $-α/β$.

5. The direction vector of a line can be represented as a vector $(dx, dy)$, where $dx$ is the change in the $x$-coordinate and $dy$ is the change in the $y$-coordinate as you move along the line. In other words, it tells you how much you move in the $x$-direction and the $y$-direction when you take one step along the line.

6. In our case, the direction vector $(dx, dy)$ is $(1, -α/β)$.

So, in simple words, the direction vector $(1, -α/β)$ tells us that as you move along the line $αx + βy + γ = 0$, for every $1$ unit you move to the right (in the positive $x$-direction), you will move $-α/β$ units downward (in the negative $y$-direction).

So, for each $1$ unit you move to the right, you move $-α/β$ units downward, which is why the direction vector is $(1, -α/β)$.