Equation of a line passing through 2. Equation of a line that passes through two given points: examples, solutions
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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 - slope
straight.
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
js-scriptExample 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
b) perpendicular to the straight line 2x+3y -7 = 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. 
;
.
. Since the lines are parallel, the equation of the desired line is 2x + 5y + C = 0. The area of a 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). The equation parabolas is. There are several options for constructing this equation. It all depends on what parameters are presented in the problem statement.
Instructions
A parabola is a curve that resembles an arc in its shape and is a graph power function. Regardless of the characteristics of a parabola, this one is even. Such a function is called even; for all values of the argument from the definition, when the sign of the argument changes, the value does not change: f (-x) = f (x) Start with the simplest function: y = x^2. From its appearance we can conclude that it is both positive and negative negative values argument x. The point at which x=0, and at the same time, y =0 is considered a point.
Below are all the main options for constructing this function and its . As a first example, below we consider a function of the form: f(x)=x^2+a, where a is an integer. In order to construct a graph of this function, it is necessary to shift the graph of the function f(x) by a units. An example is the function y=x^2+3, where along the y-axis the function is shifted by two units. If a function with the opposite sign is given, for example y=x^2-3, then its graph is shifted down along the y-axis.
Another type of function that can be given a parabola is f(x)=(x +a)^2. In such cases, the graph, on the contrary, shifts along the abscissa axis (x axis) by a units. For example, we can consider the functions: y=(x +4)^2 and y=(x-4)^2. In the first case, where there is a function with a plus sign, the graph is shifted along the x-axis to the left, and in the second case - to the right. All these cases are shown in the figure.
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, we could simply 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 coordinate plane there are two points A(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 sharp corner(the first sign of similarity 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 via 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.
Let the line pass through the points M 1 (x 1; y 1) and M 2 (x 2; y 2). The equation of a straight line passing through 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 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 is x = x 1 .
If y 2 = y I, then the equation of the line can be written as y = y 1, the straight line M 1 M 2 is parallel to the abscissa axis.
Equation of a line in segments
Let the straight line intersect the Ox axis at point M 1 (a;0), and the Oy axis at point M 2 (0;b). The equation will take the form: 
those. 
. This equation is called equation of a straight line in segments, because numbers a and b indicate which segments the line cuts off on the coordinate axes.
Equation of a line passing through a given point perpendicular to a given vector
Let's find the equation of the straight line passing through this point Mo (x O; y o) is perpendicular to the given non-zero vector n = (A; B).
Let's take an arbitrary point M(x; y) on the 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 equal to zero: that is
A(x - xo) + B(y - yo) = 0. (10.8)
Equation (10.8) is called equation of a straight line passing through a given point perpendicular to a given vector .
Vector n= (A; B), perpendicular to the line, is called normal normal vector of this line .
Equation (10.8) can be rewritten as Ah + Wu + C = 0 , (10.9)
where A and B are the coordinates of the normal vector, C = -Ax o - Vu o is the free term. Equation (10.9) There is general equation straight(see Fig. 2).
Fig.1 Fig.2
Canonical equations of the line
,
Where 
- coordinates of the point through which the line passes, and 
- 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.
Canonical equation of a circle of radius
R centered at a point 
:
In particular, if the center of the stake coincides with the origin of coordinates, 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, is a constant quantity 
, greater than the distance between foci 
.
The canonical equation of an ellipse whose foci lie on the Ox axis, and the origin of coordinates in the middle between the foci has the form 
G 
de a semi-major axis length; b – length of the semi-minor axis (Fig. 2).
Let two points be given M(X 1 ,U 1) and N(X 2,y 2). Let's find the equation of the line passing through these points.
Since this line passes through the point M, then according to formula (1.13) its equation has the form
U – Y 1 = K(X–x 1),
Where K– unknown angular coefficient.
The value of this coefficient is determined from the condition that the desired straight line passes through the point N, which means its coordinates satisfy equation (1.13)
Y 2 – Y 1 = K(X 2 – X 1),
From here you can find the slope of this line:
,
Or after conversion
(1.14)
Formula (1.14) determines Equation of a line passing through two points M(X 1, Y 1) and N(X 2, Y 2).
In the special case when points M(A, 0), N(0, B), A ¹ 0, B¹ 0, lie on the coordinate axes, equation (1.14) will take a simpler form
Equation (1.15) called 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. Write an equation for a line passing through the points M(1, 2) and B(3, –1).
. According to (1.14), the equation of the desired line has the form
2(Y – 2) = -3(X – 1).
Transferring all terms to the left side, we finally obtain the desired equation
3X + 2Y – 7 = 0.
Example 1.11. Write an equation for a line passing through a point M(2, 1) and the point of intersection of the lines X+ Y – 1 = 0, X – y+ 2 = 0.
. We will find the coordinates of the point of intersection of the lines by solving these equations together
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 U:
Now let’s write the equation of the straight line passing through the points (2, 1) and:
or .
Hence or –5( Y – 1) = X – 2.
We finally obtain the equation of the desired line in the form X + 5Y – 7 = 0.
Example 1.12. Find the equation of the 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. From the conditions of the problem it is clear that the abscissas of both points have the same value. This means that the desired straight line is parallel to the axis OY and its equation is: x = 2.
Comment . If, when writing the equation of a line using 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.
Let's consider other ways to define a line on a plane.
1. Let a non-zero vector be perpendicular to the given line L, and point M 0(X 0, Y 0) lies on this line (Figure 1.7).
Figure 1.7
Let's denote M(X, Y) any point on a line L. Vectors and 
Orthogonal. Using the conditions of orthogonality of these vectors, we obtain or A(X – X 0) + B(Y – Y 0) = 0.
We have obtained the equation of a line passing through a point M 0 is perpendicular to the vector. This vector is called Normal vector to a straight line L. The resulting equation can be rewritten as
Oh + Wu + WITH= 0, where WITH = –(AX 0 + By 0), (1.16),
Where A And IN– coordinates of the normal vector.
We obtain the general equation of the line in parametric form.
2. A straight line on a plane can be defined as follows: let a non-zero vector be parallel to the given straight line L and period M 0(X 0, Y 0) lies on this line. Let's take an arbitrary point again M(X, y) on a straight line (Figure 1.8).
Figure 1.8
Vectors and 
collinear.
Let us write down the condition for the collinearity of these vectors: , where T– an arbitrary number called a parameter. Let's write this equality in coordinates:
These equations are called Parametric equations Straight. Let us exclude the parameter from these equations T:
These equations can otherwise be written as
. (1.18)
The resulting equation is called The canonical equation of the line. The vector is called The directing vector is straight .
Comment . It is easy to see that if is the normal vector to the line L, then its direction vector can be the vector since , i.e. .
Example 1.13. Write the equation of a line passing through a point M 0(1, 1) parallel to line 3 X + 2U– 8 = 0.
b) perpendicular to the straight line 2x+3y -7 = 0. . The vector is the normal vector to the given and desired lines. Let's use the equation of a line passing through a point M 0 with a given normal vector 3( X –1) + 2(U– 1) = 0 or 3 X + 2у– 5 = 0. We obtained the equation of the desired line.