One of the axioms of geometry states that through any two points it is possible to draw a single straight line. This axiom testifies that there is a unique numerical expression that uniquely describes the specified one-dimensional geometric object. Consider in the article the question of how to write the equation of a straight line passing through two points.
What is a point and a line?
Before considering the question of constructing in space and on the plane a straight line of an equation passing through a pair of different points, one should define the specified geometric objects.
A point is uniquely determined by a set of coordinates in a given system of coordinate axes. In addition to them, there are no more characteristics for the point. She is a zero-dimensional object.
When talking about a straight line, each person imagines a line depicted on a white sheet of paper. At the same time, it is possible to give an exact geometric definitionthis object. A straight line is such a collection of points for which the connection of each of them with all the others will give a set of parallel vectors.
This definition is used when setting the vector equation of a straight line, which will be discussed below.
Since any line can be marked with a segment of arbitrary length, it is said to be a one-dimensional geometric object.
Number vector function
An equation through two points of a passing straight line can be written in different forms. In three-dimensional and two-dimensional spaces, the main and intuitively understandable numerical expression is a vector.
Assume that there is some directed segment u¯(a; b; c). In 3D space, the vector u¯ can start at an arbitrary point, so its coordinates define an infinite set of parallel vectors. However, if we choose a specific point P(x0; y0; z0) and put it as the beginning of the vector u¯, then, multiplying this vector by an arbitrary real number λ, one can obtain all points of one straight line in space. That is, the vector equation will be written as:
(x; y; z)=(x0; y0; z0) + λ(a; b; c)
Obviously, for the case on the plane, the numerical function takes the form:
(x; y)=(x0; y0) + λ(a; b)
The advantage of this type of equation compared to the others (in segments, canonical,general form) lies in the fact that it explicitly contains the coordinates of the direction vector. The latter is often used to determine whether lines are parallel or perpendicular.
General in segments and canonical function for a straight line in two-dimensional space
When solving problems, sometimes you need to write the equation of a straight line passing through two points in a certain, specific form. Therefore, other ways of specifying this geometric object in two-dimensional space should be given (for simplicity, we consider the case on the plane).
Let's start with a general equation. It has the form:
Ax + By + C=0
As a rule, on the plane the equation of a straight line is written in this form, only y is explicitly defined through x.
Now transform the expression above as follows:
Ax + By=-C=>
x/(-C/A) + y/(-C/B)=1
This expression is called an equation in segments, since the denominator for each variable shows how long the line segment cuts off on the corresponding coordinate axis relative to the starting point (0; 0).
It remains to give an example of the canonical equation. To do this, we write the vector equality explicitly:
x=x0+ λa;
y=y0+ λb
Let's express the parameter λ from here and equate the resulting equalities:
λ=(x - x0)/a;
λ=(y - y0)/b;
(x -x0)/a=(y - y0)/b
The last equality is called the equation in canonical or symmetric form.
Each of them can be converted to vector and vice versa.
The equation of a straight line passing through two points: a compilation technique
Back to the question of the article. Suppose there are two points in space:
M(x1; y1; z1) and N(x 2; y2; z2)
The only straight line passes through them, the equation of which is very easy to compose in vector form. To do this, we calculate the coordinates of the directed segment MN¯, we have:
MN¯=N - M=(x2-x1; y2- y1; z2-z1)
It is not difficult to guess that this vector will be the guide for the straight line, the equation of which must be obtained. Knowing that it also passes through M and N, you can use the coordinates of any of them for a vector expression. Then the desired equation takes the form:
(x; y; z)=M + λMN¯=>
(x; y; z)=(x1; y1; z1) + λ(x2-x1; y2-y1; z2-z1)
For the case in two-dimensional space, we obtain a similar equality without the participation of the variable z.
Once the vector equality for a line is written, it can be translated into any other form that the problem requires.
Task:write a general equation
It is known that a straight line passes through the points with coordinates (-1; 4) and (3; 2). It is necessary to formulate the equation of a straight line passing through them, in a general form, expressing y in terms of x.
To solve the problem, we first write the equation in vector form. The vector (guide) coordinates are:
(3; 2) - (-1; 4)=(4; -2)
Then the vector form of the equation of the straight line is the following:
(x; y)=(-1; 4) + λ(4; -2)
It remains to write it in general form in the form y(x). We rewrite this equality explicitly, express the parameter λ and exclude it from the equation:
x=-1 + 4λ=>λ=(x+1)/4;
y=4 - 2λ=> λ=(4-y)/2;
(x+1)/4=(4-y)/2
From the resulting canonical equation, we express y and come to the answer to the question of the problem:
y=-0.5x + 3.5
The validity of this equality can be checked by substituting the coordinates of the points specified in the problem statement.
Problem: a straight line passing through the center of the segment
Now let's solve one interesting problem. Suppose that two points M(2; 1) and N(5; 0) are given. It is known that a straight line passes through the midpoint of the segment that connects the points and is perpendicular to it. Write the equation of a straight line passing through the middle of the segment in vector form.
The desired numerical expression can be formed by calculating the coordinate of this center and determining the direction vector, whichsegment makes an angle 90o.
The midpoint of the segment is:
S=(M + N)/2=(3, 5; 0, 5)
Now let's calculate the coordinates of the vector MN¯:
MN¯=N - M=(3; -1)
Since the direction vector for the desired line is perpendicular to MN¯, their scalar product is equal to zero. This allows you to calculate the unknown coordinates (a; b) of the steering vector:
a3 - b=0=>
b=3a
Now write the vector equation:
(x; y)=(3, 5; 0, 5) + λ(a; 3a)=>
(x; y)=(3, 5; 0, 5) + β(1; 3)
Here we have replaced the product aλ with a new parameter β.
Thus, we have made the equation of a straight line passing through the center of the segment.
