In geometry, after a point, a straight line is perhaps the simplest element. It is used in the construction of any complex figures on a plane and in three-dimensional space. In this article, we consider the general equation of the line and solve a couple of problems using it. Let's get started!
Straight line in geometry
Everyone knows that shapes such as a rectangle, a triangle, a prism, a cube, and so on, are formed by intersecting straight lines. A straight line in geometry is a one-dimensional object, which can be obtained by transferring a certain point to a vector having the same or opposite direction. To better understand this definition, imagine that there is some point P in space. Take an arbitrary vector u¯ in this space. Then any point Q of the line can be obtained as a result of the following mathematical operations:
Q = P + λ * u¯.
Here λ is an arbitrary number that can be positive and negative. If the above equality is written through the coordinates, then we get the following equation of the line:
(x, y, z) = (x 0 , y 0 , z 0 ) + λ * (a, b, c).
This equality is called the equation of the line in vector form. And the vector u¯ is called the guideline.
The general equation of a line on a plane
Each student will be able to write it down without any difficulties. But most often the equation is written like this:
y = k * x + b.
Where k and b are arbitrary numbers. The number b is called the free term. The parameter k is equal to the tangent of the angle formed by the intersection of the line with the abscissa.
The above equation is expressed with respect to the variable y. If it is presented in a more general form, then we obtain the following form of notation:
A * x + B * y + C = 0.
It is easy to show that this form of writing the general equation of a straight line on a plane is easily converted to the previous form. To do this, the left and right parts should be divided by coefficient B and express y.
The figure above shows a line passing through two points.
Direct in three-dimensional space
We continue our study. We examined the question of how the equation of the line in general form is set on the plane. If we use the notation form given in the previous paragraph of the article for the spatial case, what will we do? Everything is simple - it is no longer a straight line, but a plane. Indeed, the following expression describes a plane that is parallel to the z axis:
A * x + B * y + C = 0.
If C = 0, then such a plane passes through the z axis. This is an important feature.
Then what about the general equation of a straight line in space? To understand how to ask it, you need to remember something. Two planes intersect in a certain straight line. What does this mean? Only that the general equation is the result of solving a system of two equations for planes. We write this system:
- A 1 * x + B 1 * y + C 1 * z + D 1 = 0;
- A 2 * x + B 2 * y + C 2 * z + D 2 = 0.
This system is a general equation of a straight line in space. Note that the planes should not be parallel to each other, that is, their normal vectors should be tilted at a certain angle relative to each other. Otherwise, the system will not have solutions.
Above, we presented a vector form for writing the equation for a straight line. It is convenient to use when solving this system. To do this, first find the vector product of the normals of these planes. The result of this operation will be the directing vector of the line. Then, you should calculate any point that belongs to the line. To do this, put any of the variables equal to a certain value, the two remaining variables can be found by solving the reduced system.
How to translate a vector equation into a general one? Nuances
This is an urgent task that may arise if one should write the general equation of a line along the known coordinates of two points. We show how this problem is solved by an example. Let the coordinates of two points be known:
- P = (x 1 , y 1 );
- Q = (x 2 , y 2 ).
Equation in vector form is quite simple. The coordinates of the guide vector are:
PQ = (x 2 -x 1 , y 2 -y 1 ).
Note that there is no difference, if the coordinates of Q are subtracted from the coordinates of the point P, the vector will only change its direction to the opposite. Now you should take any point and write the vector equation:
(x, y) = (x 1 , y 1 ) + λ * (x 2 -x 1 , y 2 -y 1 ).
To write the general equation of the line, in both cases the parameter λ should be expressed. And then equate the results. We have:
x = x 1 + λ * (x 2 -x 1 ) => λ = (xx 1 ) / (x 2 -x 1 );
y = y 1 + λ * (y 2 -y 1 ) => λ = (yy 1 ) / (y 2 -y 1 ) =>
(xx 1 ) / (x 2 -x 1 ) = (yy 1 ) / (y 2 -y 1 ).
It remains only to open the brackets and transfer all the terms of the equation to one side of the equality in order to obtain a general expression for the line passing through two known points.
In the case of a three-dimensional problem, the solution algorithm is preserved, only its result will be a system of two equations for planes.
Task
It is necessary to draw up a general equation for a line that intersects the x axis at (-3,0), and which is parallel to the y axis.
We start solving the problem by writing an equation in vector form. Since the line is parallel to the ordinate axis, the following vector will be the guiding vector for it:
u¯ = (0, 1).
Then the desired line can be written as follows:
(x, y) = (-3,0) + λ * (0, 1).
Now we translate this expression into general form, for this we express the parameter λ:
Thus, any value of the variable y belongs to the line, however, only the only value of the variable x corresponds to it. Therefore, the general equation will take the form:
x + 3 = 0.
The problem with a straight line in space
It is known that two intersecting planes are given by the following equations:
- 2 * x + y - z = 0;
- x - 2 * y + 3 = 0.
It is necessary to find the vector equation of the line along which these planes intersect. Let's get started.
As was said, the general equation of a straight line in three-dimensional space has already been set in the form of a system of two with three unknowns. First of all, we define a direction vector along which the planes intersect. Vector multiplying the coordinates of the normals to the planes, we get:
u¯ = [(2, 1, -1) * (1, -2, 0)] = (-2, -1, -5).
Since multiplying a vector by a negative number changes its direction to the opposite, we can write:
u¯ = -1 * (- 2, -1, -5) = (2, 1, 5).
To find a vector expression for a line, in addition to the directing vector, you need to know some point of this line. Find since its coordinates must satisfy the system of equations in the condition of the problem, we find them. For example, put x = 0, then we get:
y = z;
y = 3/2 = 1.5.
Thus, the line that belongs to the desired point has the coordinates:
P = (0, 1.5, 1.5).
Then we get the answer to this problem, the vector equation of the desired line will have the form:
(x, y, z) = (0, 1.5, 1.5) + λ * (2, 1, 5).
The correctness of the solution can be easily checked. For this, it is necessary to choose an arbitrary value of the parameter λ and substitute the obtained coordinates of the point of the straight line in both equations for the planes, we obtain the identity in both cases.