What is the general equation of lines going through 'a' particular point?

Question

I want to know the general equation of lines going through a single point where there will be arbitary constants which will change and cause the line to rotate in a circle and consequently the center will be the given point

Answer 1

For any point $\bf x_0$ in $n$-dimensional space, suppose that $\bf v$ is the unit-length vector. Every line through $\bf x_0$ can be presented as $\bf \bf x_0+\lambda\bf v$ with $\lambda \in\mathbb R$ (or the respective field/group of the vector space). $\bf v$ gives you the direction of the line.

For instance in $\mathbb R^2$, every point $(x,y)$ on a line through $(x_0,y_0)$ can be presented as: $$ (x,y)=(x_0,y_0)+\lambda(\cos\theta,\sin\theta) $$ where $\theta$ gives you the direction.

Answer 2

Its known as family of lines its $$(ax+by+c)+\lambda(dx+ey+f)=0$$ where $\lambda$=arbitrary constant.

EDIT

if we know the angle we can use rotation matrix to get new equation which is

$$\left(\begin{matrix} \cos\theta& \sin\theta\\-\sin\theta & \cos\theta\end{matrix}\right)$$

Answer 3

The general equation for a line is either $y=ax+b$ or $ax+by=c$, depending a bit on the setting. If you take your favourite one of those, swap $x$ and $y$ for the corresponding coordinates of your point, you get a relation between $a$ and $b$ (and possibly $c$) that must be fulfilled if the line is to go through your point. Vary the coefficients $a$ and $b$ (and $c$)$ within that constraint, and you get all the different lines that go through your point.