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How to derive the Descartes equation of a line in a coordinate system? $ax+by+c=0$

I searched in proofwiki but I didn't find, and in KhanAcademy and youTube but I didn't find anything related to the derivation of the Descartes equation of a line in a coordinate system.

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    $\begingroup$ The slope of a line is a constant. $\endgroup$ – David Mitra Jan 22 '14 at 19:21
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This is probably an overkill, but because I did not find it in the internet I post this answer here.

Let me make a definition. Let $d$ denote the Euclidean distance function. Such that $d$, the distance between two points, is defined by the usual "distance formula".

Definition: a line is a theoretical continuous figure, and collection of an infinite amount of points, such that for any three points $A$, $B$, and $C$ lying on the figure we have $d(A,C)=d(A,B)+d(B,C)$ if $B$ lies between $A$ and $C$. That is to say that the $x$ coordinate of point $C$ is between the $x$ coordinates of the other points and the same with the $y$ coordinates.

Does this correspond to your intuition on what a line is (think about it).

Derivation of equation of the line in two dimensions:

Let $A,B,C$ be points in $\mathbb{R}^2$ satisfying the criteria in the definition above. Let $\Delta x_1$ be the horizontal distance from $A$ to $B$ i.e. $x_b-x_a$. Let $\Delta y_1$ be the vertical distance from $A$ to $B$. Let $\Delta x_2$ and $\Delta y_2$ be the horizontal and vertical distances respectively from $B$ to $C$. By definition above we wish to find when,

$$\sqrt{\Delta x_1^2+\Delta y_1^2}+\sqrt{\Delta x_2^2+\Delta y_2^2}=\sqrt{(\Delta x_1+\Delta x_2)^2+(\Delta y_1+\Delta y_2)^2}$$

Can you prove why? Squaring gives,

$$\Delta x_1^2+\Delta x_2^2+\Delta y_1^2+\Delta y_2^2+2\sqrt{(\Delta x_1^2+\Delta y_1^2)(\Delta x_2^2+\Delta y_2^2)}=\Delta x_1^2+\Delta x_2^2+\Delta y_1^2+\Delta y_2^2+2(\Delta x_1 \Delta x_2+ \Delta y_1 \Delta y_2)$$

Cancellation, and squaring again gives,

$$4(\Delta x_1^2+\Delta y_1^2)(\Delta x_2^2+\Delta y_2^2)=4(\Delta x_1 \Delta x_2+ \Delta y_1 \Delta y_2)^2$$

So that,

$$\Delta x_1^2 \Delta x_2^2+\Delta x_1^2 \Delta y_2^2+\Delta y_1^2 \Delta x_2^2+\Delta y_1^2 \Delta y_2^2=(\Delta x_1 \Delta x_2)^2+(\Delta y_1 \Delta y_2)^2+2 \Delta x_1 \Delta x_2 \Delta y_1 \Delta y_2$$

Which gives,

$$(\Delta x_1 \Delta y_2-\Delta x_2 \Delta y_1)^2=0$$

Finally we get!

$$\Delta x_1 \Delta y_2=\Delta x_2 \Delta y_1$$

Which means,

$$\frac{\Delta y_1}{\Delta x_1}=\frac{\Delta y_2}{\Delta x_2}$$

This leads us to a very important fact, for lines the quantity $\frac{\Delta y}{\Delta x}$ between two points is constant. Let us call this quantity the slope and denote it as $m$. Then we have,

$$\frac{\Delta y_1}{\Delta x_1}=m$$

Let $B$ have coordinates $(x,y)$ and $A$ have coordinates $(a,b)$.

$$\frac{y-b}{x-a}=m$$

$$y=m(x-a)+b$$

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If you do as David suggests, then $dy/dx = m$ (a constant). Then,

$$y(x) = \int m dx = mx + C,$$

where $C$ is another constant. Hence, $ax + by + c = 0,$ with $a=m, b=-1, c=C.$

Something like that?

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  • $\begingroup$ What's that thing that looks like a s in the middle ? $\endgroup$ – rope Jan 22 '14 at 19:45
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    $\begingroup$ The integral sign $\int$? The question was tagged "calculus" so I thought that would be familiar? $\endgroup$ – John Jan 22 '14 at 19:51
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    $\begingroup$ No I didn't tag the question calculus, it's someone other who did it $\endgroup$ – rope Jan 22 '14 at 20:03
  • $\begingroup$ No this answer doesn't answer my question i have no idea what the integral sign means $\endgroup$ – rope Jan 23 '14 at 12:58

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