The straight line is the main geometric object on the plane and in three-dimensional space. It is from the straight lines that many figures are built, for example: a parallelogram, a triangle, a prism, a pyramid, and so on. We consider in the article various ways of defining equations of lines.
Definition of a line and types of equations for its description
Each student is well aware of what kind of geometric object in question. A straight line can be represented as a collection of points, and if we connect each of them in turn with all the others, we will get a set of parallel vectors. In other words, you can get to each point of the line from one of its fixed points, transferring it to some unit vector multiplied by a real number. This definition of a line is used to specify vector equality for its mathematical description both on the plane and in three-dimensional space.
A straight line can be mathematically represented by the following types of equations:
- general;
- vectorial;
- parametric;
- in segments;
- symmetric (canonical).
Next, we consider all of these types and show on examples of solving problems how to work with them.
Vector and parametric description of the line
We start by setting a line through a known vector. Suppose that in space there is a fixed point M (x 0 ; y 0 ; z 0 ). It is known that the line passes through it and is directed along the vector segment v¯ (a; b; c). How to find an arbitrary point on the line from these data? The answer to this question will be given by the following equality:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + λ * (a; b; c)
Where λ is an arbitrary number.
A similar expression can be written for the two-dimensional case, where the coordinates of the vectors and points are represented by a set of two numbers:
(x; y) = (x 0 ; y 0 ) + λ * (a; b)
The written equations are called vector equations, and the directed segment v¯ is the directing vector for the line.
From the written expressions the corresponding parametric equations are obtained simply, it is enough just to rewrite them in explicit form. For example, for the case in space, we obtain the following equation:
x = x 0 + λ * a;
y = y 0 + λ * b;
z = z 0 + λ * c
It is convenient to work with parametric equations if it is necessary to analyze the behavior of each coordinate. Note that although the parameter λ can take arbitrary values, in all three equalities it should be the same.
General equation
Another way of defining a straight line, which is often used to work with a geometric object in question, is to use a general equation. For the two-dimensional case, it has the form:
A * x + B * y + C = 0
Here, capital letters represent specific numerical values. The convenience of this equality in solving problems lies in the fact that it explicitly contains a vector that is perpendicular to the line. If we denote it by n¯, then we can write:
n¯ = [A; B]
In addition, the expression is conveniently used to determine the distance from a straight line to a certain point P (x 1 ; y 1 ). The formula for the distance d has the form:
d = | A * x 1 + B * y 1 + C | / √ (A 2 + B 2 )
It is easy to show that if we explicitly express the variable y from the general equation, we get the following well-known form of writing the line:
y = k * x + b
Where k and b are uniquely determined by the numbers A, B, C.
The equation in segments and canonical
The equation in segments is easiest to get from a general view. We show how this can be done.
Suppose that there is the following line:
A * x + B * y + C = 0
We transfer the free term to the right side of the equality, then divide the whole equation into it, we get:
A * x + B * y = -C;
x / (-C / A) + y / (-C / B) = 1;
x / q + y / p = 1, where q = -C / A, p = -C / B
We got the so-called equation in segments. It got its name due to the fact that the denominator into which each variable is divided shows the value of the coordinate of the intersection of the line with the corresponding axis. This fact is convenient to use for the image of a straight line in the coordinate system, as well as for the analysis of its relative position in relation to other geometric objects (straight lines, points).
Now we turn to obtaining the canonical equation. This is easier to do if you consider the parametric option. For the case on the plane, we have:
x = x 0 + λ * a;
y = y 0 + λ * b
We express the parameter λ in each equality, then we equate them, we obtain:
λ = (x - x 0 ) / a;
λ = (y - y 0 ) / b;
(x - x 0 ) / a = (y - y 0 ) / b
This is the desired equation, written in symmetric form. Like a vector expression, it explicitly contains the coordinates of the guide vector and the coordinates of one of the points that belongs to the line.
You may notice that in this section we have given equations for the two-dimensional case. Similarly, you can make the equation of a straight line in space. It should be noted here that if the canonical notation and expression in segments are of the same form, then the general equation in space for a straight line is represented by a system of two equations for intersecting planes.
The problem of constructing the equation of a line
From geometry, every student knows that through two points you can draw a single line. Assume that the following points are specified in the coordinate plane:
M 1 (1; 2);
M 2 (-1; 3)
It is necessary to find the equation of the line to which both points belong, in segments, in vector, canonical and in general form.
First we get the vector equation. To do this, determine for the direct guide vector M 1 M 2 ¯:
M 1 M 2 ¯ = (-1; 3) - (1; 2) = (-2; 1)
Now you can make a vector equation by taking one of two points specified in the problem condition, for example, M 2 :
(x; y) = (-1; 3) + λ * (-2; 1)
To obtain the canonical equation, it is enough to convert the found equality into a parametric form and exclude the parameter λ. We have:
x = -1 - 2 * λ, therefore, λ = x + 1 / (-2);
y = 3 + λ, then we obtain λ = y - 3;
x + 1 / (-2) = (y - 3) / 1
The remaining two equations (general and in segments) can be found from the canonical, transforming it as follows:
x + 1 = -2 * y + 6;
general equation: x + 2 * y - 5 = 0;
in segments the equation: x / 5 + y / 2,5 = 1
The obtained equations show that the vector (1; 2) should be perpendicular to the line. Indeed, if we find its scalar product with a directing vector, then it will be equal to zero. The equation in the segments says that the line intersects the x axis at the point (5; 0), and the y axis at the point (2,5; 0).
The problem of determining the point of intersection of lines
On the plane, two lines are given by the following equations:
2 * x + y -1 = 0;
(x; y) = (0; -1) + λ * (-1; 3)
It is necessary to determine the coordinates of the point at which these lines intersect.
There are two ways to solve the problem:
- Convert a vector equation into a general form, then solve a system of two linear equations.
- Do not perform any transformations, but simply substitute the coordinate of the intersection point, expressed in terms of parameter λ, into the first equation. Then find the value of the parameter.
We will act in the second way. We have:
x = -λ;
y = -1 + 3 * λ;
2 * (-λ) + (-1) + 3 * λ - 1 = 0;
λ = 2
Substitute the resulting number in the vector equation:
(x; y) = (0; -1) + 2 * (-1; 3) = (-2; 5)
Thus, the only point that belongs to both lines is the point with coordinates (-2; 5). The lines intersect in it.