how to find the equation of a line when you are given a point on that line and its normal vector in $\Bbb R^2$

Question

I see the answer for this in $\Bbb R^3$, but it's just confusing, can someone show me how to or point me to a place that shows me how to solve a problem like this (in $\Bbb R^2$). I am given a point $(3,5)$ on a line and a normal vector $\langle 2,4\rangle$ and asked to give the equation of the line in normal form, general form and parametric form.

Answer 1

A way to easily remember the formula to be used is to understand the geometric intuition behind this formula: the points $A=(3,5)$ and $M=(x,y)$ on the line have the same projection on the normal vector $\vec n=\langle 2,4\rangle$, i.e. using the dot product: $$A\cdot\vec n=M\cdot\vec n\quad\text{ or }\quad 3\cdot2+5\cdot4=3x+5y. $$

As to the parametric form, you have to find a directing vector for the line: since you're given $\vec n=\langle 2,4\rangle$, you can take $\;\vec u=\langle -4,2\rangle$ to obtain $$M=A+t\vec u=(3-4t,5+2t).$$

Answer 2

I'm not sure what 'normal form' and 'general form' are, but I'll try to help nonetheless. Perhaps it's useful to include these (standard) forms in your question.

A normal vector of a line with cartesian equation $$\color{blue}{a}x+\color{blue}{b}y+c=0 \tag{$*$}$$ is the vector $(\color{blue}{a},\color{blue}{b})$ so if you are given this normal vector, substitute the given point into the equation $(*)$ to find $c$.

Alternatively, use the following form for a line with a given normal vector as above and through the point $(\color{green}{x_0},\color{green}{y_0})$ directly: $$\color{blue}{a}(x-\color{green}{x_0})+\color{blue}{b}(y-\color{green}{y_0})=0$$ And then perhaps you have already seen a procedure to convert one form (type of equation) into another?