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Suppose we have a linear equation and a point in the plane, then how can one determine on which side of the line the point lies?

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4 Answers 4

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Let your line be given by $ax+by=d$, and call $\vec n=(a,b)$ the normal vector of the line. Let's label the side $\vec n$ points to $+$ and the opposite side $-$. Then for any point $(x,y)$ in the plane, the sign of $$ax+by-d$$ determines which side the point $(x,y)$ is on. Notice that this is $0$ if and only if $(x,y)$ is on the line, so all points not on the line get $+$ or $-$.

Here's a picture illustrating the situation:

A line and its two sides

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  • $\begingroup$ I'm a little confused, are x and y reused here? Shouldn't it be "Then for any point (x1, y1)" ? $\endgroup$
    – EoghanM
    Jun 18, 2020 at 9:37
  • $\begingroup$ When specifying the line as $L = \{ (x,y) \in \mathbb R^2 : ax+by=d \}$, the variables $x$ and $y$ are not bound outside of the definition of $L$. If you prefer to call the point in consideration $(x_1, y_1)$ you are free to do so, but $(x, y)$ is fine as well. $\endgroup$
    – Christoph
    Jun 19, 2020 at 10:17
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Are you meaning that you have a point $(x_0,y_0)$ and a line $y=a+b x$ ? If this is the case, compute $y_*=a+b x_0$ and compare $y_*$ to $y_0$. If $y_0 \gt y_*$, then the point is above the line; if $y_0 \lt y_*$, then the point is below the line; If $y_0 = y_*$, then the point is along the line.

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Suppose the line is

$$ax+by+c=0\iff y=-\frac abx-\frac cb\;,\;\;b\neq 0\;\;(b=0\;\;\text{is a vertical line and a trivial case})$$

This means a point on the plane is on this straight line iff it is of the form $\;\left(x\,,\,-\frac abx-\frac cb\right)\;$ .

Well, now take a general point $\;P=(\alpha\,,\,\beta)\;$ on the plane , and then

$$\begin{align*}\beta<-\frac ac\alpha-\frac cb&\implies P\;\;\text{is on the left half plane of the line}\\ \beta>-\frac ac\alpha-\frac cb&\implies P\;\;\text{is on the right half plane of the line}\\ \beta=-\frac ac\alpha-\frac cb&\implies P\;\;\text{is on the line}\end{align*}$$

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  • $\begingroup$ This doesn't make much sense, you're not picking any "side" of your hyperplane this way. $\endgroup$ Apr 17, 2014 at 10:51
  • $\begingroup$ Am I not, @OlivierBégassat? Then what does "on the left...on the right" mean? When one draws a line on a given plane with the usual convention of axis, there is always "a left" and "a right" of that line (unless it is a horizontal line $\;y=k=$a constant, a particular case easily dealt with). $\endgroup$
    – DonAntonio
    Apr 17, 2014 at 10:53
  • $\begingroup$ My bad, I got confused and for a moment ^^ $\endgroup$ Apr 17, 2014 at 10:59
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Here's a strategy that's independent of how the equation is written.

  • Substitute your point ($P$) into the line equation. If equality still holds, then (of course) $P$ lies on the line, and you're done. Otherwise,

  • Substitute the origin, $O(0,0)$, into the equation. If $P$ and $O$ violate the equality in the same way (eg, both points making the left-hand side of the equation larger than the right-hand side, or both making the right-hand side larger than the left-hand side), then $P$ and $O$ are on the same side of the line; otherwise, the points are on opposite sides of the line.

Note that this strategy works for curves other than lines.

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