In the question, it has been asked to find and draw the equivalence classes of the relation $∼$ on $(\Bbb R\times\Bbb R)\setminus\{(0, 0)\}$ which is defined as
$$(x_1, x_2) \sim (y_1, y_2)~\text{if}~(y_1, y_2) = \alpha(x_1, x_2)\text{, for some }\alpha \neq 0$$
On solving, I have obtained the equivalence class as
$$\begin{align}[(a,b)]&=\{(x,y)\in\Bbb R:(x,y)\sim(a,b)\}\\& =\{(x,y)\in\Bbb R:(x,y)=\alpha(a,b)\}\end{align}$$
On solving this, I got $x=\alpha a$ and $y=\alpha b$ and $x-y=\alpha(a-b)$ which will correspond to the set of straight lines having slope $1$ and $y$-intercept $-\alpha(a-b)$ excluding the straight line passing.
But in the solution of this question, the equivalence class is given as the set of all straight lines passing through the origin(which is excluded).
I have just begun discrete math, and I am not getting any idea of the solution.
It would be really helpful if someone could help me with this.
There is no problem with your algebra, but with which values are varying and which are constants.
Notice that $\alpha$ is a function of $x,y$, not some independent constant. We might do well to write it as $\alpha_{x,y}$ to emphasise this fact. This is the nuanced difference between your set, $$[(a,b)]=\{(x,y)\in\Bbb R:(x,y)=\alpha(a,b)\}$$ and the more formal $$[(a,b)]=\{(x,y)\in\Bbb R:\exists \alpha\in\mathbb{R}\setminus\{0\}:(x,y)=\alpha(a,b)\}$$
You have rearranged $x=\alpha_{x,y} a,y=\alpha_{x,y} b$ to $y=x-\alpha_{x,y} (a-b)$. Now we might see more clearly why this is not the equation of a straight line: $\alpha_{x,y}$ will vary as $x$ varies.
For instance, for some particular values $x,y$, it might be that $\alpha_{x,y}=1$ and $\alpha_{x+1,y}=3$. Here, the change that increasing $x$ by $1$ will have on $y$ is $+1-2(a-b)$, so it is not correct to say that the gradient is $1$.
Instead, we need to eliminate $\alpha_{x,y}$ from our equation. We can do that by approaching the algebra a bit differently:
$$\begin{align} &x=\alpha_{x,y} a,&y=\alpha_{x,y} b\\ &\alpha_{x,y}=x/a,&y=\alpha_{x,y} b\\ y&=\frac{b}{a}x \end{align}$$
Now we genuinely do have the equation of a straight line, with gradient $b/a$ and $y$-intercept $0$.
Varying $a,b$ can produce any possible real number: for instance, to achieve a straight line through the origin with gradient $r$, take $a=1,b=r$.
I'll leave you with this: there is one exceptional case, a straight line through the origin that does not have the form $y=rx$ for some $r\in\mathbb{R}$. What is the equation of that line, and what values of $a,b$ produce it?