The general equation of a straight line is $y= mx + c$ but when a line is parallel to y-axis this equation doesn't work because the value of the slope is $1 \over 0$(taking 1 unit for rise and 0 for run), so I tried to solve it with respect to x so that the value of m gets inversed that is $0 \over 1$ which can be used to solve it.
$$\implies x=\frac{y-c}{m}$$
for a line parallel to y-axis the slope is $m={1\over0}$, substituting m in the above equation we get
\begin{align}x=(y-c)\div m \\ \implies x = (y-c) \div {1\over0}\\ \implies x=(y-c)\times{0\over1} \\ \implies x=0\end{align}
But $x = 0$ means that the line is overlapping y-axis and the correct answer should be $x = c$ where c is x-intercept. Where did I went wrong? and why can't we use this equation for this condition?
For a straight line parallel to y-axis, using the formula $x = \frac{y-c}{m}$ gives $x = 0$
coordinate systems
Related Solutions
Yes you work is correct but it will be good to state in the beginning that the lines intersect only if $a \ne b$. For $a = b$, both equations in fact represent the same line.
You can also simplify your work a bit. Say, the intersection point is $(x_1, y_1)$. Then you have,
$\frac{x_1}{a}+\frac{y_1}{b}=1...(i)$
$\frac{x_1}{b}+\frac{y_1}{a}=1...(ii)$
From $(i)$ and $(ii)$, $\frac{x_1}{a}+\frac{y_1}{b} = \frac{x_1}{b}+\frac{y_1}{a}$
$x_1 (\frac{1}{a} - \frac{1}{b}) = y_1 (\frac{1}{a} - \frac{1}{b})$
If $a \ne b$, you have $~ y_1 = x_1$
So the equation of line passing through the origin and intersection of the given lines is,
$(y-0) = \frac{y_1 - 0}{x_1 - 0} ~ (x-0) ~ $ or $ ~ y = x$
I think your problem is regarding this equation:
$$\tan(45^{\circ})=\pm\frac{-\frac{3}{4}-m_2}{1-\frac{3}{4}m_2} \tag{1} \label{1}$$ (You are using equation tags (i) and (1) both creating confusion among the readers)
Your misconception: You think the two values of $\theta$ one can get by solving $\tan{\theta}= |\frac{m_1-m_2}{1+m_1m_2}|$ represent acute angle and the obtuse angle respectively between the lines with slopes $m_1$ and $m_2$.
Fact: In the above formula $\theta$ is always the acute angle between the lines with slopes $m_1$ and $m_2$. $\theta$ is simply equal to $|\tan^{-1}{m_1}-\tan^{-1}{m_2}|$ from which one derives the above formula.(In your question, of course $\theta=45°$ is acute and for that acute angle only, you are getting two values of $m_2$)
Further information: The equation of lines passing through point $(x_1,y_1)$ making an angle $\alpha$ with the line $y=mx+c$ are given by:
$y-y_1=\tan(\theta-\alpha)(x-x_1)$ and $y-y_1= \tan(\theta-\alpha)(x-x_1)$, where $\tan{\theta}=m$
Here the two lines $CD$ and $CV$ passing through $C(x_1,y_1)$ make an angle $\alpha$ with line $AB$ having slope $\tan{\theta}$. You'll always get exactly two such lines and their slopes will always be $\tan{(\theta+\alpha)}$ and $\tan{(\theta+180°-\alpha)}=\tan{(\theta-\alpha)}$. These are the two slopes you get while solving for $m_2$ in $(\ref{1})$. Notice that the angles $\theta + \alpha$ and $\theta - \alpha$ are not acute obtuse pair.
Note: Acute and obtuse have nothing to do here unless you were given an inclination angle which is obtuse, for which you simply have to find the corresponding acute angle between the lines and use that for $\theta$ in $\tan{\theta}= |\frac{m_1-m_2}{1+m_1m_2}|$ or alternatively, you can use this formula: $\tan{\theta_{obtuse}}= - |\frac{m_1-m_2}{1+m_1m_2}|$ where $\theta_{obtuse}$ is the obtuse angle between the lines with slopes $m_1$ and $m_2$.
Answering some questions after reading your edit:
Here, $\pm$ has been included to include both the acute and the obtuse angles that are formed when two lines with slopes $m_1$ and $m_2$ intersect each other.
No. You are completely wrong there. $\pm$ is there because the angle $\theta$ can be measured from two different directions namely clockwise and anticlockwise.This fact is responsible for two different lines satisfying one condition of inclination angle. See the figure attached in my answer.
Why am I getting 2 values of $m_2$ when there is only one value of $m_2$ in the general form of the equation?
You might be thinking $(\ref{1})$ is one equation, but that equation is actually a simpler way of writing two equations connected with "or". This is pretty obvious. Why can't you expect two different solutions of two different equations?
Best Answer
$m=\frac10$ is a nonsensical expression. As such, you should come as no surprise that anything you write after that expression will also not make any sense.
When the line is vertical, it's quite inaccurate to speak of it's slope as being equal to $\frac10$. It helps to be accurate when talking about mathematical terms, and to keep in mind what the definitions of what you are talking about are.
The definition of the slope of a line is that it is the value $k$ for which the equation $$y=kx + n$$ describes the line. In more detail, it's the value for which the set $$\{(x, y)| y=kx + n\}$$ is equal to the line.
If the line is vertical, it is relatively easy to show that no such value $k$ exists, and therefore, by the definition of the slope, the line does not have a slope. The slope is not equal to $\frac10$.