A system of linear equations in two variables may have no solution. In such a case, the equations represent __________ lines.

Find the linear approximation of the function $f(x)=\sqrt{4-x}$ at $a=0$
Use L(x) to approximate the numbers $\sqrt{3.9}$ and $\sqrt{3.99}$ Round to four decimal places

I am stucked on the following challenge: "If the line determined by two distinct points ${\displaystyle (x_1,y_1)}$ and ${\displaystyle (x_2,y_2)}$ is not vertical, and therefore has slope ${\displaystyle \frac{(y_2-y_1)}{(x_2-x_1)}}$, show that the point-slope form of its equation is the same regardless of which point is used as the given point." Okay, we can separate ${\displaystyle (x_0,y_0)}$ from the form to get:
${\displaystyle y(x_2-x_1)-x(y_2-y_1)=y_0(x_2-x_1)-x_0(y_2-y_1)}$
But how exclude this point ${\displaystyle (x_0,y_0)}$ and leave only ${\displaystyle x,y,x_1,y_1,x_2,y_2}$ in the equation? UPDATE: There is a solution for this challenge:
${\displaystyle (y_1-y_2)x+(x_2-x_1)y=x_2{y}_1-x_1{y}_2}$
From the answer I found that
${\displaystyle y_2(x-x_1)-y_1(x-x_2)=y(x_2-x_1)}$
but why this is true?

If we were given two points on a linear equation ${\displaystyle (x_1,y_1),(x_2,y_2)}$, it is quite easy to find the slope and use substitution to find the slope intercept form ${\displaystyle y=mx+b}$, to graph it.
Is it possible to solve for b strictly in terms of ${\displaystyle x_1,y_1,x_2,y_2}$?

Use an inverse matrix to solve system of linear equations.
(a)
2x-y=-3
2x+y=7
(b)
2x-y=-1
2x+y=-3

Complex form of Fourier's Theorem
Any periodic function can be decomposed into a linear combination of complex exponentials.
Proof
Consider a complex exponential with period $T_0$:
$e^{i\frac{2\pi }{T_0}t}$
Hence, we have:
${\int }_0^{T_0}{e}^{i\frac{2\pi }{T_0}t}dt=0\dots \dots (1)$
Let us claim that for a periodic function $x(t)$ we can write:
$x(t)\approx \sum _{k=-N}^N{C}_k{e}^{i\frac{2\pi k}{T_0}t},$
where $2N+1$ is the number of frequency components used. As $N\to \infty$, we have:
$x(t)=\sum _{k=-\infty }^{\infty }{C}_k{e}^{i\frac{2\pi k}{T_0}t}\dots \dots (2)$
Consider:
$v_k(t):=e^{i\frac{2\pi k}{T_0}t}\dots \dots (3)$
Then we have:
$v_k(t+T_0)=v_k(t)$
Hence, $v_k(t)$ is a periodic function. Now,
${\int }_0^{T_0}{v}_k(t)v_l^{\ast }(t)dt$
$={\int }_0^{T_0}{e}^{i\frac{2\pi (k-l)}{T_0}t}dt$
$=0(k\ne l)$
$T_0(k=l)\dots \dots (4)$
Thus, $v_k(t)$ is orthogonal. Moreover, if we assume equation (2) to be valid , then we can multiply both its sides by $v_l^{\ast }(t)$ and integrate over $T_0$ we get:
$C_k=\frac{1}{T_0}{\int }_0^{T_0}x(t)v_k^{\ast }(t)dt\dots \dots (5)$
Putting this value in equation (2) we get back $x(t)$. Hence, any periodic function can be decomposed into a linear combination of complex exponentials.
Real form of Fourier's Theorem
An arbitrary periodic function $F(t)$ with period $T$ can be decomposed into a linear combination of the functions $f_n(t)$ and $g_n(t)$ where,
$f_n(t)=\sin \frac{2\pi nt}{T}$
$g_n(t)=\cos \frac{2\pi nt}{T}$
Mathematically,
$F(t)=b_0+b_1{g}_1(t)+b_2{g}_2(t)+\dots \dots +a_1{f}_1(t)+a_2{f}_2(t)+\dots \dots ,$
where $n$ is a non-negative integer and all of $a_i$,$b_i$ are real.
Problem
Is there a similar proof for the real form of Fourier's Theorem as for the complex form? I couldn't get one on the internet or in any book. Does this mean that it can be derived from the complex form? If yes then how? Any help would be appreciated.

${\displaystyle A=(-5,1,3),B=(-2,6,-3),C=(-4,3,4)}$
I also have a line with points:
${\displaystyle D=(-2,7,6),E=(-1,8,-2)}$
The vector equation form of the line:
$((-2),(7),(6))+r((1),(1),(8)):|r\in RR$
The vector equation form of the plane:
$((-5),(1),(3))+s((3),(5),(-6))+t((1),(2),(1)):|s,t\in RR$
I am supposed to compute the intersection between these two.
What I know about intersection is that if I have a linear equation with no solution, then there is no intersection. If i have a Linear equation with one solution, then there is one intersection. If i have a linear equation with infinitely many solutions, then the line is contained within the plane.
I the answer is that the line is contained within the plane. I know the answer, but how is the solution squired is beyond me.

Solve the following system of equations.
$x_{n+1}=2y_n-z_n$
$y_{n+1}=y_n$
$z_{n+1}=x_n-2y_n+2z_n$
What is the solution in general for $x_0$, $y_0$, $z_0$ arbitrary?
It is intended to be solved using Jordan Normal Form of the Linear Algebra knowledge. I have no idea how to start, can anyone give a hint?