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I have a line, parallel to the $y$-axis, that passes through a point, P:

$$P(1/2,-3/5)$$

What is the equation of the line?

What I tried:

$$(y−y_0)=m(x−x_0)$$

$$(y+3/5)=m(x−1/2)$$

Though, there exists no $m$, therefore I can't figure out what the equation of the line is.

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  • $\begingroup$ slope of the line=(diff in y coordinates)/(diff in x coordinates), but diff in x coordinates=0, division by zero is not defined, which means slope of the vertical line is not defined, all the points on the vertical line are at the same distance from y axis, so $x=1/2$ is your desired equation $\endgroup$
    – Vikram
    Mar 1, 2015 at 14:58

3 Answers 3

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Hint: Let $(x_0, y_0) = (\frac{1}{2}, - \frac{3}{5})$ then $x = \frac{1}{2}$.

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  • $\begingroup$ Thank you, that's what I did. There exists no m, though. $\endgroup$
    – Cesare
    Mar 1, 2015 at 14:52
  • $\begingroup$ Yes, there isn't $m$. $\endgroup$ Mar 1, 2015 at 14:58
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A line parallel to the $y$-axis has a fixed $x$-value, therefore its equation is of the form $$x=c,\quad c\in\mathbb{R}$$ where you can easily see that as you vary the value of $y$, you are drawing a line perpendicular to the $x$-axis. Since the point $(x_0,y_0)=(1/2,-3/5)$ belongs to that line, then all the points with the same $x$-value will do too. Finally you have: $$x=\dfrac{1}{2}$$


Another way to see the problem is to take a vector along the $y$-axis - for instance, the unit vector $\vec{j}=(0,1)$ - and an arbitrary vector $\vec{u}$ that defines your line, and require them to be parallel: i.e. they must be related by a constant factor.

Since you already know the coordinates of a point $(x_0,y_0)$ on your line, you can define $$\vec{u}=(x-x_0,y-y_0)=(x-1/2,y+3/5)$$ Now for the parallelism condition: $$\vec{u}=k\vec{j},\quad k\in\mathbb{R}$$ From which you get the following system of equations: \begin{cases}x-1/2=0\\y+3/5=k,\quad k\in\mathbb{R}\end{cases} The first line gives $x=1/2$, while the second tells you that $y$ can take any possible value as you vary $k$. The vector $\vec{u}$ that defines your line turns out be $\vec{u}=(0,k)$, which simply means that once you've fixed a value for $x$, then any value of $y$ is possible. And again, since $x_0=1/2$ belongs to the line, the equation of the line must be $x=1/2$.

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  • $\begingroup$ @amWhy Usual abuse of notation in physics, sorry... $\endgroup$
    – Demosthene
    Mar 1, 2015 at 15:01
  • $\begingroup$ Thank you for your answer! I understand now. $\endgroup$
    – Cesare
    Mar 1, 2015 at 15:12
  • $\begingroup$ Thanks for your edit! Very much appreciated! $\endgroup$
    – Cesare
    Mar 1, 2015 at 15:20
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HINT: The equation of the line has the form $x=c$ with a constant $c$ (there exists no coefficient $m$!). Through which Point $x$ does this line pass?

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  • $\begingroup$ Thanks for your answer! The line passes through 1/2. $\endgroup$
    – Cesare
    Mar 1, 2015 at 14:58

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