 General equation of a straight line on a plane, in space

In geometry, after a point, a straight line is perhaps the simplest element. It is used in the construction of any complex figures on the plane and in three-dimensional space. In this article, we will consider the general equation of a straight line and solve a couple of problems using it. Let's get started!

## Straight line in geometry Everyone knows that shapes such as rectangle, triangle, prism, cube and so on are formed by intersecting straight lines. A straight line in geometry is a one-dimensional object that 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 + λu¯.

Here λ is an arbitrary number that can be positive or negative. If equalitywrite above in terms of coordinates, then we get the following equation of a straight line:

(x, y, z)=(x0, y0, z0) + λ(a, b, c).

This equality is called the equation of a straight line in vector form. And the vector u¯ is called a guide.

## General equation of a straight line in a plane

Every student can write it down without any difficulty. But most often the equation is written like this:

y=kx + b.

Where k and b are arbitrary numbers. The number b is called the free member. The parameter k is equal to the tangent of the angle formed by the intersection of the straight line with the x-axis.

The above equation is expressed with respect to the variable y. If we present it in a more general form, then we get the following notation:

Ax + By + C=0.

It is easy to show that this form of writing the general equation of a straight line on a plane is easily transformed into the previous form. To do this, the left and right parts should be divided by the factor B and expressed y.

The figure above shows a straight line passing through two points.

## A line in 3D space

Let's continue our study. We considered the question of how the equation of a straight line in a general form is given on a plane. If we apply the notation given in the previous paragraph of the article for the spatial case, what will we get? Everything is simple - no longer a straight line, but a plane. Indeed, the following expression describes a plane that is parallel to the z-axis:

Ax + By + C=0.

If C=0, then such a plane passesthrough the z-axis. This is an important feature.

How to be then with the general equation of a straight line in space? To understand how to ask it, you need to remember something. Two planes intersect along 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. Let's write this system:

• A1x + B1y + C1z + D 1=0;
• A2x + B2y + C2z + D 2=0.

This system is the general equation of a straight line in space. Note that the planes must not be parallel to each other, that is, their normal vectors must be inclined at some angle relative to each other. Otherwise, the system will have no solutions.

Above we gave the vector form of the equation for a straight line. It is convenient to use when solving this system. To do this, you first need to find the vector product of the normals of these planes. The result of this operation will be a direction vector of a straight line. Then, any point belonging to the line should be calculated. To do this, you need to set 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 actual problem that can arise if you need to write the general equation of a straight line using the known coordinates of two points.Let us show how this problem is solved with an example. Let the coordinates of two points be known:

• P=(x1, y1);
• Q=(x2, y2).

Equation in vector form is quite easy to compose. The direction vector coordinates are:

PQ=(x2-x1, y2-y 1).

Note that there is no difference if the Q coordinates 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 down the vector equation:

(x, y)=(x1, y1) + λ(x2 -x1, y2-y1).

To write the general equation of a straight line, the parameter λ should be expressed in both cases. And then compare the results. We have:

x=x1 + λ(x2-x1)=> λ=(x-x1)/(x2-x1);

y=y1 + λ(y2-y1)=> λ=(y-y1)/(y2-y1)=>

(x-x1)/(x2-x1)=(y-y 1)/(y2-y1).

It remains only to open the brackets and transfer all the terms of the equation to one side of the equation in order to obtain a general expression for a straight 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.

It is necessary to make a general equationa straight line that intersects the x-axis at (-3, 0) and is parallel to the y-axis.

Let's start solving the problem by writing the equation in vector form. Since the line is parallel to the y-axis, then the directing vector for it will be the following:

u¯=(0, 1).

Then the desired line will be written as follows:

(x, y)=(-3, 0) + λ(0, 1).

Now let's translate this expression into a general form, for this we express the parameter λ:

• x=-3;
• y=λ.

Thus, any value of the variable y belongs to the line, however, only the single value of the variable x corresponds to it. Therefore, the general equation will take the form:

x + 3=0.

## Problem with a straight line in space It is known that two intersecting planes are given by the following equations:

• 2x + y - z=0;
• x - 2y + 3=0.

It is necessary to find the vector equation of the straight line along which these planes intersect. Let's get started.

As it was said, the general equation of a straight line in three-dimensional space is already given in the form of a system of two with three unknowns. First of all, we determine the direction vector along which the planes intersect. Multiplying the vector 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 reverses its direction, we can write:

u¯=-1(-2, -1, -5)=(2, 1, 5).

Toto find a vector expression for a straight line, in addition to the direction vector, one should know some point of this straight line. Find since its coordinates must satisfy the system of equations in the condition of the problem, then we will find them. For example, let's put x=0, then we get:

y=z;

y=3/2=1, 5.

Thus, the point belonging to the desired straight line 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 look like:

(x, y, z)=(0, 1, 5, 1, 5) + λ(2, 1, 5).

The correctness of the solution can be easily checked. To do this, you need to choose an arbitrary value of the parameter λ and substitute the obtained coordinates of the point of the straight line into both equations for the planes, you will get an identity in both cases.