Equation of a line that passes through two points. Equation of a line that passes through two given points: examples, solutions

Equation of a line passing through two points. In the article" " I promised you to look at the second way to solve the presented problems of finding the derivative, given a graph of a function and a tangent to this graph. We will discuss this method in , do not miss! Why in the next one?

The fact is that the formula for the equation of a straight line will be used there. Of course, you could just show this formula and advise you to learn it. But it’s better to explain where it comes from (how it is derived). It's necessary! If you forget it, you can quickly restore itwill not be difficult. Everything is outlined below in detail. So, we have two points A on the coordinate plane(x 1;y 1) and B(x 2;y 2), a straight line is drawn through the indicated points:

Here is the direct formula itself:

*That is, when substituting specific coordinates of points, we get an equation of the form y=kx+b.

**If you simply “memorize” this formula, then there is a high probability of getting confused with the indices when X. In addition, indices can be designated in different ways, for example:

That's why it's important to understand the meaning.

Now the derivation of this formula. Everything is very simple!

Triangles ABE and ACF are similar in acute angle (the first sign of similarity of right triangles). It follows from this that the ratios of the corresponding elements are equal, that is:

Now we simply express these segments through the difference in the coordinates of the points:

Of course, there will be no error if you write the relationships of the elements in a different order (the main thing is to maintain consistency):

The result will be the same equation of the line. This is all!

That is, no matter how the points themselves (and their coordinates) are designated, by understanding this formula you will always find the equation of a straight line.

The formula can be derived using the properties of vectors, but the principle of derivation will be the same, since we will be talking about the proportionality of their coordinates. In this case, the same similarity of right triangles works. In my opinion, the conclusion described above is more clear)).

View output using vector coordinates >>>

Let a straight line be constructed on the coordinate plane passing through two given points A(x 1;y 1) and B(x 2;y 2). Let us mark an arbitrary point C on the line with coordinates ( x; y). We also denote two vectors:

It is known that for vectors lying on parallel lines (or on the same line), their corresponding coordinates are proportional, that is:

— we write down the equality of the ratios of the corresponding coordinates:

Let's look at an example:

Find the equation of a straight line passing through two points with coordinates (2;5) and (7:3).

You don’t even have to build the straight line itself. We apply the formula:

It is important that you grasp the correspondence when drawing up the ratio. You can't go wrong if you write:

Answer: y=-2/5x+29/5 go y=-0.4x+5.8

In order to make sure that the resulting equation is found correctly, be sure to check - substitute the coordinates of the data in the condition of the points into it. The equations should be correct.

That's all. I hope the material was useful to you.

Sincerely, Alexander.

P.S: I would be grateful if you tell me about the site on social networks.

Definition. Any straight line on the plane can be specified by a first-order equation

Ax + Wu + C = 0,

Moreover, the constants A and B are not equal to zero at the same time. This first order equation is called general equation of a straight line. Depending on the values constant A, B and C 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 Ox axis

B = 0, A ≠0, C ≠ 0 (Ax + C = 0) – straight line parallel to the Oy axis

B = C = 0, A ≠0 – the straight line coincides with the Oy axis

A = C = 0, B ≠0 – the straight line coincides with the Ox axis

The equation of a straight line can be represented in in various forms depending on any given initial conditions.

Equation of a straight line from a point and normal vector

Definition. In Cartesian rectangular system coordinate vector with components (A, B) is perpendicular to the line, given by the equation Ax + Wu + C = 0.

Example. Find the equation of the line passing through the point A(1, 2) perpendicular to (3, -1).

Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x – y + C = 0. To find the coefficient C, we 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 line passing through two points

Let two points M 1 (x 1, y 1, z 1) and M 2 (x 2, y 2, z 2) be given in space, then the equation of the line passing through these points is:

If any of the denominators is equal to zero, the corresponding numerator should be equal to zero. On the plane, the equation of the line written above is simplified:

if x 1 ≠ x 2 and x = x 1, if x 1 = x 2.

The fraction = k is called slope straight.

Example. Find the equation of the line passing through points A(1, 2) and B(3, 4).

Solution. Applying the formula written above, we get:

Equation of a straight line from a point and slope

If the total Ax + Bu + C = 0, lead to the form:

and designate , then the resulting equation is called equation of a straight line with slopek.

Equation of a straight line from a point and a direction vector

By analogy with the point considering the equation of a straight line through a normal vector, you can enter the definition of a straight line through a point and the directing vector of the straight line.

Definition. Every nonzero vector(α 1, α 2), the components of which satisfy the condition A α 1 + B α 2 = 0 is called the directing vector of the line

Ax + Wu + C = 0.

Example. Find the equation of a straight line with a direction vector (1, -1) and passing through the point A(1, 2).

Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. In accordance with the definition, the coefficients must satisfy 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. for x = 1, y = 2 we obtain C/ A = -3, i.e. required equation:

Equation of a line in segments

If in the general equation of the straight line Ах + Ву + С = 0 С≠0, then, dividing by –С, we get: or

The geometric meaning of the coefficients is that the coefficient A is the coordinate of the point of intersection of the line with the Ox axis, and b– the coordinate of the point of intersection of the straight line with the Oy axis.

Example. Set general equation line x – y + 1 = 0. Find the equation of this line in segments.

C = 1, , a = -1, b = 1.

Normal equation of a line

If both sides of the equation Ax + By + C = 0 are multiplied by the number which is called normalizing factor, then we get

xcosφ + ysinφ - p = 0 –

normal equation of a line. The sign ± of the normalizing factor must be chosen so that μ * C< 0. р – длина перпендикуляра, опущенного из начала координат на прямую, а φ - угол, образованный этим перпендикуляром с положительным направлением оси Ох.

Example. Given the general equation of the straight line 12x – 5y – 65 = 0. You need to write Various types equations of this line.

equation of this line in segments:

equation of this line with slope: (divide by 5)

; 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 of coordinates.

Example. The straight line cuts off equal positive segments on the coordinate axes. Write an equation for a straight line if the area of ​​the triangle formed by these segments is 8 cm 2.

Solution. The equation of the straight line has the form: , ab /2 = 8; ab=16; a=4, a=-4. a = -4< 0 не подходит по условию задачи. Итого: или х + у – 4 = 0.

Example. Write an equation for a straight line passing through point A(-2, -3) and the origin.

Solution. The equation of the straight line is: , where x 1 = y 1 = 0; x 2 = -2; y 2 = -3.

Angle between straight lines on a plane

Definition. If two lines are given y = k 1 x + b 1, y = k 2 x + b 2, then the 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. The lines Ax + Bу + C = 0 and A 1 x + B 1 y + C 1 = 0 are parallel when the coefficients A 1 = λA, B 1 = λB are proportional. If also C 1 = λC, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.

Equation of a line passing through a given point perpendicular to a given line

Definition. A straight line passing through the point M 1 (x 1, y 1) and perpendicular to the straight line y = kx + b is represented by the equation:

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Bу + C = 0 is determined as

.

Proof. Let point M 1 (x 1, y 1) be the base of the perpendicular dropped from point M to a given straight line. Then the distance between points M and M 1:

(1)

The coordinates x 1 and y 1 can be found by solving the system of equations:

The second equation of the system is the equation of a line passing through a given point M 0 perpendicular to a given 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 proven.

Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.

k 1 = -3; k 2 = 2; tgφ = ; φ= π /4.

Example. Show that the lines 3x – 5y + 7 = 0 and 10x + 6y – 3 = 0 are perpendicular.

Solution. We find: k 1 = 3/5, k 2 = -5/3, k 1* k 2 = -1, therefore, the lines are perpendicular.

Example. Given are the vertices of the triangle A(0; 1), B (6; 5), C (12; -1). Find the equation of the height drawn from vertex C.

Solution. We find the equation of side AB: ; 4 x = 6 y – 6;

2 x – 3 y + 3 = 0;

The required height equation has the form: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: from where b = 17. Total: .

Answer: 3 x + 2 y – 34 = 0.

Properties of a straight line in Euclidean geometry.

An infinite number of straight lines can be drawn through any point.

Through any two non-coinciding points a single straight line can be drawn.

Two divergent lines in a plane either intersect at a single point or are

parallel (follows from the previous one).

IN three-dimensional space there are three options relative position two straight lines:

• lines intersect;

• lines are parallel;

• straight lines intersect.

Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system

is given on the plane by an equation of the first degree (linear equation).

General equation of a straight line.

Definition. Any straight line on the plane can be specified by a first-order equation

Ax + Wu + C = 0,

and constant A, B are not equal to zero at the same time. This first order equation is called general

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- a 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 equation of a straight line can be presented in different forms depending on any given

initial conditions.

Equation of a straight line from a point and a normal vector.

Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)

perpendicular to the line given by the equation

Ax + Wu + C = 0.

Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).

Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C

Let's 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 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 line,

passing through these points:

If any of the denominators is zero, the corresponding numerator should be set equal 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 line passing through points A(1, 2) and B(3, 4).

Solution. Applying the formula written above, we get:

Equation of a straight line using a point and slope.

If the general equation of the line Ax + Wu + C = 0 lead to:

and designate , then the resulting equation is called

equation of a straight line with slope k.

Equation of a straight line from a point and a direction vector.

By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a directing vector of a straight line.

Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition

Aα 1 + Bα 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 the point A(1, 2).

Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,

coefficients must satisfy the following 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 Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:

or where

The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point

straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.

Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.

C = 1, , a = -1, b = 1.

Normal equation of a 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 a line.

The sign ± of the normalizing factor must 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. The general equation of the line is given 12x - 5y - 65 = 0. Required to write different types of equations

this straight line.

The equation of this line in segments:

The equation of this line with the slope: (divide by 5)

Equation of a 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 a plane.

Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, then the acute angle between these lines

will be defined as

Two lines are parallel if k 1 = k 2. Two 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 parallel when the coefficients are proportional

A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines

are found as a solution to the system of equations of these lines.

The equation of a line passing through a given point perpendicular to a given line.

Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b

represented by the equation:

Distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then 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 a perpendicular dropped from a point M for a given

direct. Then the distance between 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 perpendicularly

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 proven.

Properties of a straight line in Euclidean geometry.

An infinite number of straight lines can be drawn through any point.

Through any two non-coinciding points a single straight line can be drawn.

Two divergent lines in a plane either intersect at a single point or are

parallel (follows from the previous one).

In three-dimensional space, there are three options for the relative position of two lines:

• lines intersect;

• lines are parallel;

• straight lines intersect.

Straight line— algebraic curve of the first order: a straight line in the Cartesian coordinate system

is given on the plane by an equation of the first degree (linear equation).

General equation of a straight line.

Definition. Any straight line on the plane can be specified by a first-order equation

Ax + Wu + C = 0,

and constant A, B are not equal to zero at the same time. This first order equation is called general

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- a 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 equation of a straight line can be presented in different forms depending on any given

initial conditions.

Equation of a straight line from a point and a normal vector.

Definition. In a Cartesian rectangular coordinate system, a vector with components (A, B)

perpendicular to the line given by the equation

Ax + Wu + C = 0.

Example. Find the equation of a line passing through a point A(1, 2) perpendicular to the vector (3, -1).

Solution. With A = 3 and B = -1, let’s compose the equation of the straight line: 3x - y + C = 0. To find the coefficient C

Let's 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 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 line,

passing through these points:

If any of the denominators is zero, the corresponding numerator should be set equal 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 line passing through points A(1, 2) and B(3, 4).

Solution. Applying the formula written above, we get:

Equation of a straight line using a point and slope.

If the general equation of the line Ax + Wu + C = 0 lead to:

and designate , then the resulting equation is called

equation of a straight line with slope k.

Equation of a straight line from a point and a direction vector.

By analogy with the point considering the equation of a straight line through the normal vector, you can enter the task

a straight line through a point and a directing vector of a straight line.

Definition. Every non-zero vector (α 1 , α 2), whose components satisfy the condition

Aα 1 + Bα 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 the point A(1, 2).

Solution. We will look for the equation of the desired line in the form: Ax + By + C = 0. According to the definition,

coefficients must satisfy the following 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 Ах + Ву + С = 0 С≠0, then, dividing by -С, we get:

or where

The geometric meaning of the coefficients is that the coefficient a is the coordinate of the intersection point

straight with axis Oh, A b- coordinate of the point of intersection of the line with the axis OU.

Example. The general equation of a straight line is given x - y + 1 = 0. Find the equation of this line in segments.

C = 1, , a = -1, b = 1.

Normal equation of a 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 a line.

The sign ± of the normalizing factor must 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. The general equation of the line is given 12x - 5y - 65 = 0. Required to write different types of equations

this straight line.

The equation of this line in segments:

The equation of this line with the slope: (divide by 5)

Equation of a 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 a plane.

Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2, then the acute angle between these lines

will be defined as

Two lines are parallel if k 1 = k 2. Two 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 parallel when the coefficients are proportional

A 1 = λA, B 1 = λB. If also С 1 = λС, then the lines coincide. Coordinates of the point of intersection of two lines

are found as a solution to the system of equations of these lines.

The equation of a line passing through a given point perpendicular to a given line.

Definition. Line passing through a point M 1 (x 1, y 1) and perpendicular to the line y = kx + b

represented by the equation:

Distance from a point to a line.

Theorem. If a point is given M(x 0, y 0), then 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 a perpendicular dropped from a point M for a given

direct. Then the distance between 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 perpendicularly

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 proven.

The line passing through the point K(x 0 ; y 0) and parallel to the line y = kx + a is found by the formula:

y - y 0 = k(x - x 0) (1)

Where k is the slope of the line.

Alternative formula:
A line passing through the point M 1 (x 1 ; y 1) and parallel to the line Ax+By+C=0 is represented by the equation

A(x-x 1)+B(y-y 1)=0 . (2)

Write an equation for a line passing through point K( ;) parallel to the straight line y = x+ .

Example No. 1. Write an equation for a straight line passing through the point M 0 (-2,1) and at the same time:
a) parallel to the straight line 2x+3y -7 = 0;
b) perpendicular to the straight line 2x+3y -7 = 0.
Solution . Let's imagine the equation with the slope in the form y = kx + a. To do this, move all values ​​except y to the right side: 3y = -2x + 7 . Then divide the right-hand side by a factor of 3. We get: y = -2/3x + 7/3
Let's find the equation NK passing through the point K(-2;1), parallel to the straight line y = -2 / 3 x + 7 / 3
Substituting x 0 = -2, k = -2 / 3, y 0 = 1 we get:
y-1 = -2 / 3 (x-(-2))
or
y = -2 / 3 x - 1 / 3 or 3y + 2x +1 = 0

Example No. 2. Write the equation of a line parallel to the line 2x + 5y = 0 and forming, together with the coordinate axes, a triangle whose area is 5.
Solution . Since the lines are parallel, the equation of the desired line is 2x + 5y + C = 0. Area right triangle, where a and b are its legs. Let's find the intersection points of the desired line with the coordinate axes:
;
.
So, A(-C/2,0), B(0,-C/5). Let's substitute it into the formula for area: . We get two solutions: 2x + 5y + 10 = 0 and 2x + 5y – 10 = 0.

Example No. 3. Write an equation for a line passing through the point (-2; 5) and parallel to the line 5x-7y-4=0.
Solution. This straight line can be represented by the equation y = 5 / 7 x – 4 / 7 (here a = 5 / 7). The equation of the desired line is y – 5 = 5 / 7 (x – (-2)), i.e. 7(y-5)=5(x+2) or 5x-7y+45=0 .

Example No. 4. Having solved example 3 (A=5, B=-7) using formula (2), we find 5(x+2)-7(y-5)=0.

Example No. 5. Write an equation for a line passing through the point (-2;5) and parallel to the line 7x+10=0.
Solution. Here A=7, B=0. Formula (2) gives 7(x+2)=0, i.e. x+2=0. Formula (1) is not applicable, since this equation cannot be resolved with respect to y (this straight line is parallel to the ordinate axis).