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I found set builder form of writing equation of straight line like $\left\{(8t+8,2t+5)∣t∈R\right\}$ on this answer. I found this way of the equation of line unique. I'm seeing this first time and interested in knowing how to convert any linear equation of the form $ax + by + c =0 $ in the set builder notations like $\left\{(8t+8,2t+5)∣t∈R\right\}$. What's the basic idea of representing equations in this form?

I would be grateful if anyone could tell me how to convert the linear equation of a straight line in set builder notations. I searched well on the internet but could not find any good article on this topic. Neither do our schools teach this way of writing equations.

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    Google "parametric equation line". – David Mitra Jan 05 '22 at 15:29
  • Each line can be represented in this form in many different ways, since you can choose any two points on the line to correspond to $t=0$ and $t=1$. One way is to just set $t=x$; this works unless the line is vertical (i.e. $b=0$), in which case $t=y$ works instead. – Karl Jan 05 '22 at 15:34
  • As @DavidMitra comments, the important concept here is parametric representation, not the fact that you then describe the line in set builder notation. – Ethan Bolker Jan 05 '22 at 15:40

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What you are asking for is a parametric equation of a line, expressed in set-builder notation. To start, the line $3x + 2y - 6 = 0$ can be written in set-builder notation as $\{(x,y)\,|\,\forall x\forall y \in \mathbb{R} \land y = -\frac{3}{2}x + 3\}$.

In the example, you can modify $(x,y)$ to be functions of $t$ such that $(x,y) = (2t, -3t + 3)$. There are a lot of ways to express $(x,y)$ into $(f(t), g(t))$, especially when $y = f(x)$ represents a line.

In general, it is of the form $\left\{\big[f(t), g(t)\big]\,\Big|\,t\in\mathbb{R}\right\}$ where $(x,y) = (f(t), g(t))$ and expressing the parametric equation into the form $y = f(x)$ should yield a linear equation in terms of $x$.

soupless
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  • @Utkarsh You can also do that, as long as it satisfies the conditions. For example, $x - y + 2 = 0$ when $x \in [-2, 0]$ and $y = [0, 2]$ can be expressed as $(x,y) = (\sin t - 1, \sin t + 1)$ for $t \in [-\pi/2, \pi/2]$. – soupless Jan 05 '22 at 16:04