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How to compute distance between a point and a line

January 27, 2016 Leave a comment

This semester (Even Semester, 2015-2016) I am teaching a new course, that is called Analytic Geometry or Geometry Analytic, both are the same, I think. Don’t get me wrong, the course is not new, it is just I would be teaching this course the first time. When they told me that I was going to be the lecturer of this course, they did not give me a standard, they did not give me a textbook they normally use, or a list of topics that I must cover, perhaps because this course is not a compulsory course. So, I have a freedom, I can choose topics that I want to teach in Analytic Geometry. The first thing I did, was to browse the internet any textbook about Analytic Geometry or any lecture note. In the end, I pick a textbook (Indonesian book) and I choose some topics of my interest to be the material of this course.

Okay, enough for the background, I am sure you don’t want to hear anymore on that. Let’s get back to main point of this note. Long story short, a straight line became one of the topics in my course, and I was interested on how the formula to compute distance between a point and a line is derived.In this case, I only consider two dimensional case and by distance, I mean the shortest distance between such a point and a certain line, which also means the perpendicular distance. If you don’t remember what is the formula, let me recall you.

Consider a straight line l, given by the following equation:

l : ax + by + c=0.

Suppose we have a point P(x_p,y_p), then the distance between the point P and the straight line l is

d = \frac{|ax_p+by_p+c|}{\sqrt{a^2+b^2}}.fig1-crop

There are other proofs on how to derive the above formula, but I really really like the proof I am going to show you below.  I will divide this note into two section. The first section will talk about a straight line and how to get an equation of a straight line. The latter section will derive the formula.

1. A straight line equation

In high school, I think you must know what conditions we need to have in order to obtain a straight line equation. If we have two points in XY-coordinate we can compute the line equation using the following:

\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}.

If we have a gradient and a point, we can also compute the line equation using the following:

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

Of course, in a problem, we won’t get this information so easily. We have work a bit more to get either two points or one point and a gradient and then we are able to obtain the equation of the straight line.

Before I teach this course, I can only conclude that if you want to know the equation of a straight line you need to know either:
(i) two points, or
(ii) a point and gradient of the line.
But, actually there is a third condition, and if we know this condition, we can also compute the equation of a straight line. In the third condition, a line is assumed to be the tangent line of a certain circle. Thus, the characteristics we need to know to form a line are the radius of the circle and the angle between the radius and the positif x-axis. See the figure below.fig2-crop

In the figure, we can see that a line can be determined uniquely if the radius of the circle n and the angle \alpha are known. In the first section of this note, we shall derive how to define a line equation given these two conditions.
Suppose we have a straight line, n are \alpha are given. See the figure below. Consider the point P(x,y). We are going to find the relationship of x and yfig3-crop

The radius n, which is |ON|,  can be computed by adding |OM| and |MN|. We shall consider |OM| first. Consider triangle OMQ which is a right triangle at M. We have the relationship:

|OM|=|OQ|\cos \alpha=x \cos \alpha.

Consider another triangle, PM'Q, which is also a right triangle at M'. We have the following relationship

|PM'|=|PQ|\sin \alpha=y \sin \alpha as it can be computed that the angle \angle PQM is also \alpha.

Therefore, we have n=x \cos \alpha + y \sin \alpha, which is the equation of straight line given that n and \alpha is known.

Before we discuss how to derive the formula to compute the distance between a point and a line, I would like to discuss how to find n and \alpha if we know the general equation of a straight line ax+by+c=0. Consider a line equationax+by+c=0, then we move $c$ to the right hand side and multiple both sides with a non zero constant kkax+kby=-kc. We shall choose k such that ka=\cos \alpha and kb=\sin \alpha. Therefore, we have

k = \pm \frac{1}{\sqrt{a^2+b^2}} and

our equation of line becomes:

\pm(\frac{a}{\sqrt{a^2+b^2}}x+\frac{b}{\sqrt{a^2+b^2}}y)=\pm(-\frac{c}{\sqrt{a^2+b^2}}).

Here, we have that the right hand side of the above equation is n=\pm(-\frac{c}{\sqrt{a^2+b^2}}), and we choose the positive value to get n. The angle \alpha can also be computed once we determine n.

2. Distance between a point and a line

To compute a perpendicular distance between a point and a line, we shall use the above result. Consider a straight line l:x \cos \alpha+y \sin \alpha = n_1 and a point P in the following figure.fig4-crop

fig5-crop

We want to compute d. In order to do that, we need to make another line that is parallel to the line l and passing through point P. This line, because it is parallel to l and, has the following equation:

x \cos \alpha+y \sin \alpha = n_2

Thus the distance between the point P and the line l can be easily computed by considering the absolute difference between n_1 and n_2 as follow:

d = |n_2-n_1|=|x_p \cos \alpha + y_p \sin \alpha - n_1|

As we know from the first section that we can substitute the latter expression into:

d = |x_p \frac{a}{\sqrt{a^2+b^2}}+ y_p\frac{b}{\sqrt{a^2+b^2}}-(-\frac{c}{\sqrt{a^2+b^2}} )|

or

d=\frac{|ax_p+by_p+c|}{\sqrt{a^2+b^2}}

which is the same as the formula we mentioned in the beginning of our note.

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