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What is the most appropriate way to write equation of a straight line through a given point $(\alpha,\beta)$?

If you write it using the standard form
$y- \beta = m (x - \alpha )$
Where m is the gradient of the line.
There is a problem of how to obtain equation of the line when it is vertical.
But if you write it as
$b(y- \beta)= a(x - \alpha )$ , here you have chance of obtaining equation for any situation.
I have come across this issue when finding equation of tangents to a given circle. What do you think about this?

Added for further clarification,
Think about the equation of the tangent drawn at the point $(\alpha,\beta)$ to the circle $x^2 + y^2 +2gx +2fy + c =0$ , where $(\alpha,\beta)$ is a point on the given circle.
Here you have the standard form,
$x\alpha + y\beta + g(x+\alpha)+f(y+\beta) +c = 0$ and by this you can get vertical tangents too. Do you know why ?
I suggest the reason is here to derive this standard form you are using something similar to $b(y - \beta) = a( x - \alpha)$.
Therefore I think when we need to find equation of a straight line through a given point much better way is to use the form,
$b(y - \beta) = a( x - \alpha)$.
Otherwise you need to check whether x = α is another possibility or not.

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    $\begingroup$ What course is this for? In College Algebra, I like to use slope-intercept form for simplicity. In Calc 1, I prefer point-slope as a preview of Taylor series. In Calc 3, I use "standard" form for its relationship to vector dot products. $\endgroup$
    – TomKern
    Commented May 1, 2022 at 16:44
  • $\begingroup$ @ Tomkern this issue you may have in grade 12 coordinate geometry. $\endgroup$ Commented May 1, 2022 at 17:27
  • $\begingroup$ Computer Vision solves this problem by using homogeneous coordinates: a line will be defined as a.x + b.y + c = 0. The slope is a/b. Has many advantages, for example a point will also be defined by a 3-element vector, and the intersection of two lines is just pt=l1 X l2, and the the line defined by two points will be written as l = p1 x p2. (with "X" the cross product) $\endgroup$
    – kebs
    Commented May 2, 2022 at 12:12
  • $\begingroup$ Most appropriate for what? What do you want to achieve? $\endgroup$
    – Pablo H
    Commented May 6, 2022 at 22:25
  • $\begingroup$ @ Pablo - as I have already mentioned if you want to cover all the cases of straight lines through a given point, specially when you are asked to find tangents drawn to a curve through a given point . $\endgroup$ Commented May 7, 2022 at 4:13

3 Answers 3

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Assuming we are talking about an algebra class because of the secondary-education tag, the reason it is okay to use "point-slope form," i.e.

$y - y_1 = m(x - x_1)$

as the default way to write the equation of a line is that the two most important uses of lines do not care about vertical lines at all:

  • Using the tangent line as the linear approximation to a function in calculus
  • Using a line of best fit correlating two variables in statistics

You could certainly do other things if you wanted. But is it worth it to use a nonstandard format to catch a relatively unimportant corner case? I don't think so.

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    $\begingroup$ @ Chris - My issue is by , y- β = m ( x - α ) y , vertical tangents can not be obtained. Then you have to find it separately, but if you use the form b(y- β ) = a ( x - α ) all cases can be covered . As an example what is the best way you can suggest to use when finding equation of tangents drawn to a given circle through a given point? $\endgroup$ Commented May 2, 2022 at 0:43
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    $\begingroup$ My point is that the vertical lines are the least interesting, because they don't help you approximate any functions in calculus and they don't help you predict anything in statistics, so it's okay to treat them like a weird case. $\endgroup$ Commented May 2, 2022 at 4:11
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    $\begingroup$ @JanakaRodrigo I would make Chris's point differently, or as he'll probably disagree, it's simply a different point: In calculus (not geometry), the function is the primary object of interest. In the ordinary, boring case, $m=f'(a)$ exists (and is finite), the tangent line is not vertical, and one may calculate without thinking; but when the tangent is vertical, things are more interesting, $f'(a)$ does not exist, and you have to think what to do because the basic calculus rules do not apply. But the ordinary case is common and covered by the pt-slope form. One can treat $x=a$ as an exception. $\endgroup$
    – user1815
    Commented May 3, 2022 at 5:09
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@TomKern mentioned the "standard" form above. I believe what he has in mind is ax+by=c, which allows for both horizontal and vertical lines.

Another form, rarely used, but more like the equations of conics, is:

$\frac{x}{a}+\frac{y}{b}=1$

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  • $\begingroup$ @ Sue -you have to include the point (α,β ) in your equation, otherwise we can easily use the general form ax +by+c =0 . $\endgroup$ Commented May 2, 2022 at 0:51
  • $\begingroup$ I did. I labeled it differently. The standard way we label a point in a conic is with (h,k). (Maybe it's different in other countries? $\endgroup$
    – Sue VanHattum
    Commented May 2, 2022 at 3:20
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    $\begingroup$ @ Jean Marie- both of the equations mentioned are not representing straight lines going through the given point. Here I'm talking about that particular case. $\endgroup$ Commented May 2, 2022 at 13:26
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    $\begingroup$ Thanks, @JeanMarieBecker. I'll fix that. $\endgroup$
    – Sue VanHattum
    Commented May 2, 2022 at 15:25
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    $\begingroup$ @JanakaRodrigo Let $b\rightarrow\infty$ and you get an equation of the vertical line $x=a$, and let $a\rightarrow\infty$ to get an equation of the horizontal line $y=b$. Sue, the "Another form" (intercept-intercept form?) is dimensionless/homogeneous, which can be nice in some situations (not enough to make it less rarely used, though). — Similarly, the "symmetric equation" $(x-x_0)/a=(y-y_0)/b$ is both dimensionless/homogeneous and can represent horizontal/vertical lines in the limit as $a$ or $b$ go to infinity. $a$ and $b$ may be thought of as the $x$ and $y$ velocities, respectively. $\endgroup$
    – user1815
    Commented May 3, 2022 at 5:25
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TL;DR

I try not to get overly hung up on notation, and prefer to teach some kind of principle. In this setting, the principle is one of geometric transformation using basic techniques. Many equations for lines result, including some of the standard guys, like

$$ ax + by + c = 0 \qquad\text{and}\qquad y = mx + b. $$

A student should be comfortable with any equation for a line, and should understand the limitations (e.g. $y=mx+b$ is great for seeing the $y$-intercept and slope, but can't describe vertical lines; $ax + by + c = 0$ is very general, but the underlying geometry is largely obfuscated; etc.).


In my precalculus classes, I generally start the discussion of functions with more general relations, i.e. subsets of the plane. In set builder notation, I might define a set as follows:

$$ S = \{ (x,y) : \text{[some formula]} \}. $$

Translating or scaling such a set is relatively straight-forward. For example, to move the set to the right $h$ units, we have to increase all of the $x$-coordinates by $h$. Thus a right-translation of $S$ by $h$ is the set

$$ \tau_h S = \{ (x+h,y) : \text{[some formula]} \}. $$

(I tend not to use this notation with precalculus students—I generally use something more ad hoc like $S_{\to h}$—but the notation $\tau_h$ for a translation is pretty standard in higher math, so I'll assume that the audience here is okay with it.)

For example, the unit circle is

$$ \{ (x,y) : x^2 + y^2 = 1 \}, $$

and a right translation of that set is given by

$$ \{ (x+h, y) : x^2 + y^2 = 1 \}. $$

However, I don't really like having the translation on the left-hand side of the set builder notation, so I make a change of variables: let $\overline{x} = x+h$. Then I have the set

$$ \{ (\overline{x}, y) : x^2 + y^2 = 1, \overline{x} = x+h \}. $$

This is gross, so let's try to simplify this:

$$ x^2 + y^2 = 1 \land \overline{x} = x+h \iff x^2 + y^2 = 1 \land x = \overline{x}-h \iff (\overline{x}-h)^2 + y^2 = 1. $$

Putting this back into the set builder notation gives

$$ \{ (\overline{x},y) : \overline{x}^2 + y^2 \}. $$

Replacing $\overline{x}$ with $x$ (this is just a renaming, if a little confusing) gives the result that

$$ \{ (x+h,y) : x^2 + y^2 = 1 \} = \{ (x,y) : (x-h)^2 + y^2 = 1 \}. $$

Similar kinds of arguments apply for vertical translations (which give one a $y-k$ on the right), and for scalings (which give factors of $\frac{1}{a}$ for a horizontal scaling, and $\frac{1}{b}$ for a vertical scaling). Once you have established this idea (that transformations of a set are most easily understood on the left, by doing the "obvious" thing; but can be moved to the right by doing the "opposite" thing), it is possible to reason about lines:

  1. The primitive line is the set $\{ (x,y) : x = y\}$. Since we "know" how to apply transformations to sets, I'll drop the set builder notation, abuse language a little bit, and refer to "the primitive line $x=y$).

  2. Given a point $P_1 = (x_1, y_1)$ in the plane, the primitive line can be scaled to pass through $P_1$. The idea is to scale the coordinates so that $(1,1)$ gets sent to $P_1$ (this is not the only way to reason about lines, but it is the one which makes the computation simplest). This requires a horizontal scaling by a factor of $x_1$, and a vertical scaling by a factor of $y_1$. Thus the line $$ \frac{x}{x_1} = \frac{y}{y_1} $$ is a line through the origin and the point $P_1$.

  3. Given a second point $P_2 = (x_2, y_2)$, reduce the problem to something which has already been solved. Translate both $P_1$ and $P_2$ so that $P_2$ lands on the origin—that is, apply a horizontal translation of $-x_2$ and a vertical translation of $-y_2$ to that two point set. $P_1$ ends up at the point $$P_1' = (x_1 - x_2, y_1 - y_2), $$ and from step (2), the line through the origin and $P_1'$ is given by $$ \frac{x}{x_1 - x_2} = \frac{y}{y_1 - y_2}. $$ Note that this only makes sense if $x_1 - x_2 \ne 0$ and also $y_1 - y_2 \ne 0$.

  4. To deal with the case that one or the other is zero (they can't both be zero, because then $P_1$ and $P_2$ not distinct), we can "clear the denominators" to get the equation $$ (y_1 - y_2) x = (x_1 - x_2) y. $$

  5. Finally, the origin needs to be translated back to $P_2$, which requires a horizontal translation of $x_2$ units, and a vertical translation of $y_2$ units, which gives the equation $$ (y_1 - y_2) (x - x_2) = (x_1 - x_2) (y - y_2). \tag{$\ast$}$$

This is, essentially, a variation of the "point-point" equation for a line, which is usually given as something like $$ y - y_2 = \underbrace{\frac{y_1 - y_2}{x_1 - x_2}}_{\text{$=m$, the slope}}(x-x_2). $$ On the other hand, if ($\ast$) is expanded out, we get $$ (y_1 - y_2) x - (x_2 - x_1) y - (y_1 - y_2)x_2 + (x_1 - x_2)y_2 = 0 \implies ax + by + c = 0,$$ where \begin{align} a &= y_1 - y_2, \\ b &= -(x_2 - x_1), \text{ and} \\ c &= - (y_1 - y_2)x_2 + (x_1 - x_2)y_2. \end{align} Note that this formulation describes any line, and that it generalizes nicely to the idea of specifying a co-dimension one subspace of $\mathbb{R}^n$ in terms of a normal vector. In this setting the slope (if it exists) is $b/a$.

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  • $\begingroup$ @ Xander - your 5th case is what used in obtaining equation of the tangent drawn to a circle at a given point on the circle which I mentioned in the final part of my question. That is why you can include all cases there. $\endgroup$ Commented May 3, 2022 at 2:27

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