Equation of a straight line through 2 points online. Equation of a straight line passing through two given points
Given two points M 1 (x 1, y 1) and M 2 (x 2, y 2)... We write the equation of the straight line in the form (5), where k still unknown coefficient:
Since the point M 2 belongs to a given straight line, then its coordinates satisfy equation (5):. Expressing from this and substituting it into equation (5), we obtain the required equation:
If 
this equation can be rewritten in a form more convenient for memorization:
(6)
Example. Write down the equation of the straight line passing through the points M 1 (1.2) and M 2 (-2.3)
Solution. 
... Using the property of proportion, and performing the necessary transformations, we obtain the general equation of the straight line:
Angle between two straight lines
Consider two lines l 1 and l 2:
l 1: , , and
l 2: , ,
φ is the angle between them (). Figure 4 shows:.
 |
From here 
, or
Using formula (7), one of the angles between the straight lines can be determined. The second angle is.
Example... Two straight lines are given by the equations y = 2x + 3 and y = -3x + 2. find the angle between these lines.
Solution... From the equations it can be seen that k 1 = 2, and k 2 = -3. substituting these values into formula (7), we find
... Thus, the angle between these lines is equal.
Conditions for parallelism and perpendicularity of two straight lines
If straight l 1 and l 2 are parallel, then φ=0 and tgφ = 0... it follows from formula (7) that, whence k 2 = k 1... Thus, the condition for the parallelism of two straight lines is the equality of their slopes.
If straight l 1 and l 2 are perpendicular, then φ = π / 2, α 2 = π / 2 + α 1. 
... Thus, the condition of perpendicularity of two straight lines is that their slopes are reciprocal in magnitude and opposite in sign.
Distance from point to line
Theorem. If a point M (x 0, y 0) is given, then the distance to the straight line Ax + Vy + C = 0 is determined as
Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M onto a given straight line. Then the distance between points M and M 1:
The coordinates x 1 and y 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of the straight line passing through the this point M 0 perpendicular to a given straight line.
If we transform the first equation of the system to the form:
A (x - x 0) + B (y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem is proved.
Example. Determine the angle between the straight lines: y = -3x + 7; y = 2x + 1.
k 1 = -3; k 2 = 2 tgj =; j = p / 4.
Example. Show that the straight lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.
We find: k 1 = 3/5, k 2 = -5/3, k 1 k 2 = -1, therefore, the straight lines are perpendicular.
Example. The vertices of the triangle A (0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.
We find the equation of the side AB:; 4x = 6y - 6;
2x - 3y + 3 = 0;
The required height equation is: Ax + By + C = 0 or y = kx + b.
k =. Then y =. Because height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total:.
Answer: 3x + 2y - 34 = 0.
The distance from a point to a straight line is determined by the length of the perpendicular dropped from a point to a straight line.
If the line is parallel to the projection plane (h | | P 1), then in order to determine the distance from the point A to straight h it is necessary to lower the perpendicular from the point A on the horizontal h.
Consider more complex example when the straight line is in general position. Let it be necessary to determine the distance from the point M to straight a general position.
The task of determining distance between parallel lines solved similarly to the previous one. A point is taken on one straight line, a perpendicular is lowered from it to another straight line. The length of the perpendicular is equal to the distance between the parallel lines.
Curve of the second order is called a line determined by an equation of the second degree relative to the current Cartesian coordinates. In general, Ax 2 + 2Bxy + Cy 2 + 2Dx + 2Ey + F = 0,
where A, B, C, D, E, F are real numbers and at least one of the numbers A 2 + B 2 + C 2 ≠ 0.
Circle
Circle center- this is the locus of points in the plane equidistant from the point of the plane C (a, b).
The circle is given by the following equation:
Where x, y are the coordinates of an arbitrary point of the circle, R is the radius of the circle.
Circumference Equation
1. There is no term with x, y
2. Equal Coefficients at x 2 and y 2
Ellipse
Ellipse is called the locus of points in a plane, the sum of the distances of each of which from two given points of this plane is called foci (constant value).
Canonical Ellipse Equation:
X and y belong to an ellipse.
a - semi-major axis of the ellipse
b - semi-minor axis of the ellipse
The ellipse has 2 axes of symmetry OX and OY. The axes of symmetry of the ellipse are its axes, the point of their intersection is the center of the ellipse. The axis on which the focuses are located is called focal axis... The point of intersection of the ellipse with the axes is the vertex of the ellipse.
Compression (stretching) ratio: ε = s / a- eccentricity (characterizes the shape of the ellipse), the smaller it is, the less the ellipse will stretch along the focal axis.
If the centers of the ellipse are not in the center of C (α, β)
Hyperbola
Hyperbole is called the locus of points in the plane, the absolute value of the difference in distances, each of which from two given points of this plane, called foci, is a constant value other than zero.
Canonical hyperbola equation
The hyperbola has 2 axes of symmetry:
a is the real semiaxis of symmetry
b - imaginary semiaxis of symmetry
Hyperbola assymptotes:
Parabola
Parabola is called the locus of points in the plane equidistant from a given point F, called the focus and a given straight line, called the directrix.
Canonical parabola equation:
Y 2 = 2px, where p is the distance from the focus to the directrix (parabola parameter)
If the vertex of the parabola C (α, β), then the equation of the parabola (y-β) 2 = 2p (x-α)
If the focal axis is taken as the ordinate axis, then the parabola equation will take the form: x 2 = 2qу
Let the line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of the straight line passing through the point M 1 has the form y-y 1 = k (x - x 1), (10.6)
where k - still unknown coefficient.
Since the straight line passes through the point M 2 (x 2 y 2), the coordinates of this point must satisfy the equation (10.6): y 2 -y 1 = k (x 2 -x 1).
From here we find Substituting the found value k
into equation (10.6), we obtain the equation of a straight line passing through points M 1 and M 2: 
It is assumed that in this equation x 1 ≠ x 2, y 1 ≠ y 2
If x 1 = x 2, then the straight line passing through the points M 1 (x 1, y I) and M 2 (x 2, y 2) is parallel to the ordinate axis. Its equation has the form x = x 1 .
If y 2 = y I, then the equation of the straight line can be written as y = y 1, the straight line M 1 M 2 is parallel to the abscissa axis.
Equation of a straight line in segments
Let the straight line intersect the Ox axis at the point M 1 (a; 0), and the Oy axis - at the point M 2 (0; b). The equation will take the form: 
those. 
... This equation is called the equation of a straight line in segments, since the numbers a and b indicate which segments are cut off by a straight line on the coordinate axes.
Equation of a straight line passing through a given point perpendicular to a given vector
Let us find the equation of a straight line passing through a given point Mo (x O; y o) perpendicular to a given nonzero vector n = (A; B).
Take an arbitrary point M (x; y) on a straight line and consider the vector M 0 M (x - x 0; y - y o) (see Fig. 1). Since the vectors n and M o M are perpendicular, their scalar product is zero: that is,
A (x - xo) + B (y - yo) = 0. (10.8)
Equation (10.8) is called the equation of a straight line passing through a given point perpendicular to a given vector .
The vector n = (A; B), perpendicular to the straight line, is called normal normal vector of this line .
Equation (10.8) can be rewritten as Ax + Wu + C = 0 , (10.9)
where A and B are the coordinates of the normal vector, C = -Aх о - Ву о - free term. Equation (10.9) is the general equation of the straight line(see fig. 2).
Fig. 1 Fig. 2
Canonical equations of the straight line
,
Where 
- coordinates of the point through which the straight line passes, and 
is the direction vector.
Second-order Curves Circle
A circle is the set of all points of the plane equidistant from a given point, which is called the center.
The canonical equation of a circle of radius
R centered at point 
:
In particular, if the center of the stake coincides with the origin, then the equation will look like: 
Ellipse
An ellipse is a set of points on a plane, the sum of the distances from each of which to two given points
and 
, which are called foci, have a constant 
greater than the distance between the foci 
.
The canonical equation of an ellipse, the foci of which lie on the Ox axis, and the origin of coordinates in the middle between the foci has the form 
G 
de a the length of the semi-major axis; b - the length of the semi-minor axis (Fig. 2).
Relationship between ellipse parameters
and 
expressed by the ratio:
(4)
Eccentricity ellipsecalled the ratio of the interfocal distance2cto the major axis2a:
Headmistresses
ellipses are called straight lines parallel to the axis Oy, which are at a distance from this axis. Directrix equations: 
.
If in the ellipse equation 
, then the foci of the ellipse are on the Oy axis.
So, 
Properties of a straight line in Euclidean geometry.
You can draw infinitely many straight lines through any point.
A single straight line can be drawn through any two non-coinciding points.
Two mismatched straight lines on a plane either intersect at a single point, or are
parallel (follows from the previous one).
In three-dimensional space, there are three options mutual disposition two straight lines:
- straight lines intersect;
- straight lines are parallel;
- straight lines intersect.
Straight line- algebraic curve of the first order: in a Cartesian coordinate system, a straight line
is given on the plane by an equation of the first degree (linear equation).
General equation straight.
Definition... Any straight line on a plane can be given by a first-order equation
Ax + Wu + C = 0,
with constant A, B are not equal to zero at the same time. This first-order equation is called common
equation of a straight line. Depending on the values of the constants A, B and WITH the following special cases are possible:
. C = 0, A ≠ 0, B ≠ 0- the straight line passes through the origin
. A = 0, B ≠ 0, C ≠ 0 (By + C = 0)- straight line parallel to the axis Oh
. B = 0, A ≠ 0, C ≠ 0 (Ax + C = 0)- straight line parallel to the axis OU
. B = C = 0, A ≠ 0- the straight line coincides with the axis OU
. A = C = 0, B ≠ 0- the straight line coincides with the axis Oh
The straight line equation can be represented in various forms depending on any given
initial conditions.
Equation of a straight line along a point and a normal vector.
Definition... In a Cartesian rectangular coordinate system, a vector with components (A, B)
perpendicular to the straight line given by the equation
Ax + Wu + C = 0.
Example... Find the equation of a straight line passing through a point A (1, 2) perpendicular to vector (3, -1).
Solution... At A = 3 and B = -1, we compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C
substitute the coordinates of the given point A into the resulting expression. We get: 3 - 2 + C = 0, therefore
C = -1. Total: the required equation: 3x - y - 1 = 0.
Equation of a straight line passing through two points.
Let two points be given in space M 1 (x 1, y 1, z 1) and M2 (x 2, y 2, z 2), then equation of a straight line,
passing through these points:
If any of the denominators is zero, the corresponding numerator should be equated to zero. On
plane, the equation of the straight line written above is simplified:
if x 1 ≠ x 2 and x = x 1, if x 1 = x 2 .
Fraction = k called slope straight.
Example... Find the equation of the straight line passing through the points A (1, 2) and B (3, 4).
Solution... Applying the above formula, we get:
Equation of a straight line by point and slope.
If the general equation of the straight line Ax + Wu + C = 0 lead to the form:
and designate 
, then the resulting equation is called
equation of a straight line with slope k.
Equation of a straight line along a point and a direction vector.
By analogy with the paragraph considering the equation of a straight line through the normal vector, you can enter the task
a straight line through a point and a direction vector of a straight line.
Definition... Every nonzero vector (α 1, α 2) whose components satisfy the condition
Аα 1 + Вα 2 = 0 called directing vector of a straight line.
Ax + Wu + C = 0.
Example... Find the equation of a straight line with a direction vector (1, -1) and passing through point A (1, 2).
Solution... The equation of the required straight line will be sought in the form: Ax + By + C = 0. According to the definition,
the coefficients must meet the conditions:
1 * A + (-1) * B = 0, i.e. A = B.
Then the equation of the straight line has the form: Ax + Ay + C = 0, or x + y + C / A = 0.
at x = 1, y = 2 we get C / A = -3, i.e. required equation:
x + y - 3 = 0
Equation of a straight line in segments.
If in the general equation of the straight line Ax + Vy + C = 0 C ≠ 0, then, dividing by -C, we get:
or where
Geometric meaning coefficients in that coefficient a is the coordinate of the intersection point
straight with axis Oh, a b- the coordinate of the point of intersection of the straight line with the axis OU.
Example... The general equation of the straight line is given x - y + 1 = 0. Find the equation of this straight line in segments.
C = 1, a = -1, b = 1.
Normal equation of a straight line.
If both sides of the equation Ax + Wu + C = 0 divide by number 
which is called
normalizing factor, then we get
xcosφ + ysinφ - p = 0 -normal equation of the line.
The ± sign of the normalizing factor should be chosen so that μ * C< 0.
R- the length of the perpendicular dropped from the origin to the straight line,
a φ - the angle formed by this perpendicular with the positive direction of the axis Oh.
Example... A general equation of the straight line is given 12x - 5y - 65 = 0... Required to write different types of equations
this straight line.
The equation of this line in segments:
Equation of this line with slope: (divide by 5)
Equation of a straight line:
cos φ = 12/13; sin φ = -5/13; p = 5.
It should be noted that not every straight line can be represented by an equation in segments, for example, straight lines,
parallel to the axes or passing through the origin.
The angle between straight lines on the plane.
Definition... If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then an acute angle between these lines
will be defined as
Two lines are parallel if k 1 = k 2... Two straight lines are perpendicular,
if k 1 = -1 / k 2 .
Theorem.
Direct Ax + Wu + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients are proportional
А 1 = λА, В 1 = λВ... If also С 1 = λС, then the straight lines coincide. Coordinates of the point of intersection of two lines
are found as a solution to the system of equations of these straight lines.
Equation of a straight line passing through a given point perpendicular to a given straight line.
Definition... Line through point M 1 (x 1, y 1) and perpendicular to the line y = kx + b
is represented by the equation:
Distance from point to line.
Theorem... If a point is given M (x 0, y 0), the distance to the straight line Ax + Wu + C = 0 defined as:
Proof... Let the point M 1 (x 1, y 1)- the base of the perpendicular dropped from the point M for a given
straight line. Then the distance between the points M and M 1:
(1)
Coordinates x 1 and at 1 can be found as a solution to the system of equations:
The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to
a given straight line. If we transform the first equation of the system to the form:
A (x - x 0) + B (y - y 0) + Ax 0 + By 0 + C = 0,
then, solving, we get:
Substituting these expressions into equation (1), we find:
The theorem is proved.
This article discloses obtaining the equation of a straight line passing through two given points in a rectangular coordinate system located on a plane. Let us derive the equation of a straight line passing through two given points in a rectangular coordinate system. We will clearly show and solve several examples related to the material covered.
Before obtaining the equation of a straight line passing through two given points, it is necessary to pay attention to some facts. There is an axiom that says that through two non-coinciding points on the plane it is possible to draw a straight line and only one. In other words, two given points of the plane are defined by a straight line passing through these points.
If the plane is specified by a rectangular coordinate system Oxy, then any straight line depicted in it will correspond to the equation of a straight line on the plane. There is also a connection with the direction vector of the straight line. This data is enough to generate the equation of a straight line passing through two given points.
Let's consider an example of solving a similar problem. It is necessary to draw up an equation of the straight line a passing through two non-coincident points M 1 (x 1, y 1) and M 2 (x 2, y 2), which are in the Cartesian coordinate system.
In the canonical equation of a straight line on a plane, which has the form x - x 1 ax = y - y 1 ay, a rectangular coordinate system O x y with a straight line is specified, which intersects with it at a point with coordinates M 1 (x 1, y 1) with a guide vector a → = (ax, ay).
It is necessary to compose the canonical equation of the straight line a, which passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2).
Line a has a direction vector M 1 M 2 → with coordinates (x 2 - x 1, y 2 - y 1), since it intersects points M 1 and M 2. We obtained the necessary data in order to transform the canonical equation with the coordinates of the direction vector M 1 M 2 → = (x 2 - x 1, y 2 - y 1) and the coordinates of the points M 1 (x 1, y 1) lying on them and M 2 (x 2, y 2). We get an equation of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1.
Consider the figure below.
Following the calculations, we write the parametric equations of a straight line on a plane that passes through two points with coordinates M 1 (x 1, y 1) and M 2 (x 2, y 2). We obtain an equation of the form x = x 1 + (x 2 - x 1) λ y = y 1 + (y 2 - y 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) λ.
Let's take a closer look at the solution of several examples.
Example 1
Write down the equation of a straight line passing through 2 given points with coordinates M 1 - 5, 2 3, M 2 1, - 1 6.
Solution
The canonical equation for a straight line intersecting at two points with coordinates x 1, y 1 and x 2, y 2 takes the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. By the condition of the problem, we have that x 1 = - 5, y 1 = 2 3, x 2 = 1, y 2 = - 1 6. It is necessary to substitute numerical values into the equation x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1. From here we get that the canonical equation takes the form x - (- 5) 1 - (- 5) = y - 2 3 - 1 6 - 2 3 ⇔ x + 5 6 = y - 2 3 - 5 6.
Answer: x + 5 6 = y - 2 3 - 5 6.
If you need to solve a problem with a different kind of equation, then first you can go to the canonical one, since it is easier to come from it to any other.
Example 2
Draw up the general equation of the straight line passing through the points with coordinates M 1 (1, 1) and M 2 (4, 2) in the O x y coordinate system.
Solution
First, you need to write down the canonical equation of a given straight line that passes through the given two points. We get an equation of the form x - 1 4 - 1 = y - 1 2 - 1 ⇔ x - 1 3 = y - 1 1.
Let us bring the canonical equation to the required form, then we get:
x - 1 3 = y - 1 1 ⇔ 1 x - 1 = 3 y - 1 ⇔ x - 3 y + 2 = 0
Answer: x - 3 y + 2 = 0.
Examples of such tasks were considered in school textbooks in algebra lessons. School tasks differed in that the well-known equation of a straight line with a slope, having the form y = k x + b. If you need to find the value of the slope k and the number b for which the equation y = kx + b defines a line in the O x y system that passes through the points M 1 (x 1, y 1) and M 2 (x 2, y 2) , where x 1 ≠ x 2. When x 1 = x 2 , then the slope takes on the value of infinity, and the straight line М 1 М 2 is determined by a general incomplete equation of the form x - x 1 = 0 .
Because the points M 1 and M 2 are on a straight line, then their coordinates satisfy the equation y 1 = k x 1 + b and y 2 = k x 2 + b. It is necessary to solve the system of equations y 1 = k x 1 + b y 2 = k x 2 + b for k and b.
To do this, find k = y 2 - y 1 x 2 - x 1 b = y 1 - y 2 - y 1 x 2 - x 1 x 1 or k = y 2 - y 1 x 2 - x 1 b = y 2 - y 2 - y 1 x 2 - x 1 x 2.
With such values of k and b, the equation of the straight line passing through the given two points takes the following form y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 1 or y = y 2 - y 1 x 2 - x 1 x + y 2 - y 2 - y 1 x 2 - x 1 x 2.
Remembering such a huge number of formulas at once will not work. To do this, you need to increase the number of repetitions in problem solutions.
Example 3
Write down the equation of the straight line with the slope passing through the points with coordinates M 2 (2, 1) and y = k x + b.
Solution
To solve the problem, we use a formula with a slope, which has the form y = k x + b. The coefficients k and b must take such a value that this equation corresponds to a straight line passing through two points with coordinates M 1 (- 7, - 5) and M 2 (2, 1).
Points M 1 and M 2 are located on a straight line, then their coordinates should reverse the equation y = k x + b true equality. From this we obtain that - 5 = k (- 7) + b and 1 = k 2 + b. Combine the equation into the system - 5 = k · - 7 + b 1 = k · 2 + b and solve.
Upon substitution, we find that
5 = k - 7 + b 1 = k 2 + b ⇔ b = - 5 + 7 k 2 k + b = 1 ⇔ b = - 5 + 7 k 2 k - 5 + 7 k = 1 ⇔ ⇔ b = - 5 + 7 kk = 2 3 ⇔ b = - 5 + 7 2 3 k = 2 3 ⇔ b = - 1 3 k = 2 3
Now the values k = 2 3 and b = - 1 3 are substituted into the equation y = k x + b. We get that the required equation passing through the given points is an equation of the form y = 2 3 x - 1 3.
This solution predetermines spending a large number time. There is a way in which the task is solved literally in two steps.
We write the canonical equation of the straight line passing through M 2 (2, 1) and M 1 (- 7, - 5), which has the form x - (- 7) 2 - (- 7) = y - (- 5) 1 - (- 5) ⇔ x + 7 9 = y + 5 6.
We now turn to the equation in the slope. We get that: x + 7 9 = y + 5 6 ⇔ 6 (x + 7) = 9 (y + 5) ⇔ y = 2 3 x - 1 3.
Answer: y = 2 3 x - 1 3.
If in three-dimensional space there is a rectangular coordinate system O x y z with two specified non-coinciding points with coordinates M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), a straight line M 1 M 2, it is necessary to obtain the equation of this straight line.
We have that canonical equations of the form x - x 1 ax = y - y 1 ay = z - z 1 az and parametric equations of the form x = x 1 + ax λ y = y 1 + ay λ z = z 1 + az λ are able to set a line in the O x y z coordinate system passing through the points having coordinates (x 1, y 1, z 1) with the direction vector a → = (ax, ay, az).
Straight M 1 M 2 has a direction vector of the form M 1 M 2 → = (x 2 - x 1, y 2 - y 1, z 2 - z 1), where the line passes through the point M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2), hence the canonical equation can be of the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 or x - x 2 x 2 - x 1 = y - y 2 y 2 - y 1 = z - z 2 z 2 - z 1, in turn parametric x = x 1 + (x 2 - x 1) λ y = y 1 + (y 2 - y 1) λ z = z 1 + (z 2 - z 1) λ or x = x 2 + (x 2 - x 1) λ y = y 2 + (y 2 - y 1) Λ z = z 2 + (z 2 - z 1) λ.
Consider a figure that shows 2 given points in space and the equation of a straight line.
Example 4
Write the equation of a straight line, defined in a rectangular coordinate system O x y z of three-dimensional space, passing through given two points with coordinates M 1 (2, - 3, 0) and M 2 (1, - 3, - 5).
Solution
It is necessary to find the canonical equation. Since we are talking about three-dimensional space, it means that when a straight line passes through given points, the desired canonical equation will take the form x - x 1 x 2 - x 1 = y - y 1 y 2 - y 1 = z - z 1 z 2 - z 1 ...
By hypothesis, we have that x 1 = 2, y 1 = - 3, z 1 = 0, x 2 = 1, y 2 = - 3, z 2 = - 5. Hence it follows that necessary equations will be written in this way:
x - 2 1 - 2 = y - (- 3) - 3 - (- 3) = z - 0 - 5 - 0 ⇔ x - 2 - 1 = y + 3 0 = z - 5
Answer: x - 2 - 1 = y + 3 0 = z - 5.
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Given two points M(NS 1 ,Have 1) and N(NS 2,y 2). Let's find the equation of the straight line passing through these points.
Since this line passes through the point M, then according to formula (1.13) its equation has the form
Have – Y 1 = K(X - x 1),
Where K- unknown slope.
The value of this coefficient is determined from the condition that the desired straight line passes through the point N, and hence, its coordinates satisfy the equation (1.13)
Y 2 – Y 1 = K(X 2 – X 1),
From here you can find the slope of this straight line:
,
Or after conversion
(1.14)
Formula (1.14) determines Equation of a straight line passing through two points M(X 1, Y 1) and N(X 2, Y 2).
In the special case, when the points M(A, 0), N(0, B), A ¹ 0, B¹ 0, lie on the coordinate axes, equation (1.14) takes on a simpler form
Equation (1.15) called By the equation of a straight line in segments, here A and B denote the segments cut off by a straight line on the axes (Figure 1.6).
Figure 1.6
Example 1.10. Equate a straight line through points M(1, 2) and B(3, –1).
. According to (1.14), the equation of the sought line has the form
2(Y – 2) = -3(X – 1).
Transferring all members to left side, we finally obtain the required equation
3X + 2Y – 7 = 0.
Example 1.11. Equate a straight line through a point M(2, 1) and the point of intersection of the lines X+ Y - 1 = 0, X - y+ 2 = 0.
. We find the coordinates of the point of intersection of the straight lines by solving together the given equations
If we add these equations term by term, we get 2 X+ 1 = 0, whence. Substituting the found value into any equation, we find the value of the ordinate Have:
Now we write the equation of the straight line passing through the points (2, 1) and:
or .
Hence, or –5 ( Y – 1) = X – 2.
Finally, we obtain the equation of the desired straight line in the form NS + 5Y – 7 = 0.
Example 1.12. Find the equation of the straight line passing through the points M(2,1) and N(2,3).
Using formula (1.14), we obtain the equation
It doesn't make sense since the second denominator is zero. It can be seen from the problem statement that the abscissas of both points have the same value. Hence, the sought line is parallel to the axis OY and its equation is: x = 2.
Comment . If, when writing the equation of a straight line according to formula (1.14), one of the denominators turns out to be equal to zero, then the desired equation can be obtained by equating the corresponding numerator to zero.
Consider other ways to define a straight line on a plane.
1. Let a nonzero vector be perpendicular to the given line L and point M 0(X 0, Y 0) lies on this straight line (Figure 1.7).
Figure 1.7
We denote M(X, Y) an arbitrary point on the line L... Vectors and 
Orthogonal. Using the orthogonality conditions for these vectors, we obtain either A(X – X 0) + B(Y – Y 0) = 0.
We got the equation of a straight line passing through a point M 0 perpendicular to the vector. This vector is called The normal vector to straight L... The resulting equation can be rewritten as
Oh + Woo + WITH= 0, where WITH = –(AX 0 + By 0), (1.16),
Where A and V- coordinates of the normal vector.
Let's get the general equation of the straight line in parametric form.
2. A straight line on a plane can be specified as follows: let a nonzero vector be parallel to a given straight line L and point M 0(X 0, Y 0) lies on this straight line. Let's take an arbitrary point again M(NS, y) on a straight line (Figure 1.8).
Figure 1.8
Vectors and 
collinear.
Let us write down the collinearity condition for these vectors:, where T- an arbitrary number called a parameter. Let's write this equality in coordinates:
These equations are called Parametric equations Straight... We exclude from these equations the parameter T:
These equations can otherwise be written in the form
. (1.18)
The resulting equation is called The canonical equation of the straight line... The vector is called The direction vector of the straight line .
Comment . It is easy to see that if is the normal vector to the line L, then its direction vector can be a vector, since, i.e.
Example 1.13. Write the equation of the straight line passing through the point M 0 (1, 1) parallel to straight line 3 NS + 2Have– 8 = 0.
Solution . The vector is the normal vector to the given and desired straight lines. We will use the equation of the straight line passing through the point M 0 with a given normal vector 3 ( NS –1) + 2(Have- 1) = 0 or 3 NS + 2y- 5 = 0. Received the equation of the desired straight line.
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