Equation of side ab. Given the coordinates of the vertices of the triangle

Problem 1. The coordinates of the vertices of triangle ABC are given: A(4; 3), B(16;-6), C(20; 16). Find: 1) the length of side AB; 2) equations of sides AB and BC and their angular coefficients; 3) angle B in radians with an accuracy of two digits; 4) equation of height CD and its length; 5) the equation of the median AE and the coordinates of the point K of the intersection of this median with the height CD; 6) the equation of a straight line passing through point K parallel to side AB; 7) coordinates of point M, located symmetrically to point A relative to straight line CD.

Solution:

1. The distance d between points A(x 1 ,y 1) and B(x 2 ,y 2) is determined by the formula

Applying (1), we find the length of side AB:

2. The equation of the line passing through the points A(x 1 ,y 1) and B(x 2 ,y 2) has the form

(2)

Substituting the coordinates of points A and B into (2), we obtain the equation of side AB:

Having solved the last equation for y, we find the equation of side AB in the form of a straight line equation with an angular coefficient:

where

Substituting the coordinates of points B and C into (2), we obtain the equation of straight line BC:

Or

3. It is known that the tangent of the angle between two straight lines, the angular coefficients of which are respectively equal, is calculated by the formula

(3)

The desired angle B is formed by straight lines AB and BC, the angular coefficients of which are found: Applying (3), we obtain

Or glad.

4. The equation of a straight line passing through a given point in a given direction has the form

(4)

The height CD is perpendicular to side AB. To find the slope of the height CD, we use the condition of perpendicularity of the lines. Since then Substituting into (4) the coordinates of point C and the found angular coefficient of height, we obtain

To find the length of the height CD, we first determine the coordinates of point D - the point of intersection of straight lines AB and CD. Solving the system together:

we find i.e. D(8;0).

Using formula (1) we find the length of the height CD:

5. To find the equation of the median AE, we first determine the coordinates of point E, which is the middle of side BC, using the formulas for dividing a segment into two equal parts:

(5)

Hence,

Substituting the coordinates of points A and E into (2), we find the equation for the median:

To find the coordinates of the point of intersection of the height CD and the median AE, we solve together the system of equations

We find.

6. Since the desired straight line is parallel to side AB, its angular coefficient will be equal to the angular coefficient of straight line AB. Substituting into (4) the coordinates of the found point K and the angular coefficient we obtain

3x + 4y – 49 = 0 (KF)

7. Since the straight line AB is perpendicular to the straight line CD, the desired point M, located symmetrically to the point A relative to the straight line CD, lies on the straight line AB. In addition, point D is the midpoint of segment AM. Using formulas (5), we find the coordinates of the desired point M:

Triangle ABC, height CD, median AE, straight line KF and point M are constructed in the xOy coordinate system in Fig. 1.

Task 2. Create an equation for the locus of points whose distances to a given point A(4; 0) and to a given line x=1 are equal to 2.

Solution:

In the xOy coordinate system, we construct the point A(4;0) and the straight line x = 1. Let M(x;y) be an arbitrary point of the desired geometric location of points. Let us lower the perpendicular MB to the given line x = 1 and determine the coordinates of point B. Since point B lies on the given line, its abscissa is equal to 1. The ordinate of point B is equal to the ordinate of point M. Therefore, B(1;y) (Fig. 2 ).

According to the conditions of the problem |MA|: |MV| = 2. Distances |MA| and |MB| we find from formula (1) of problem 1:

Squaring the left and right sides, we get

The resulting equation is a hyperbola in which the real semi-axis is a = 2, and the imaginary half-axis is

Let's define the foci of a hyperbola. For a hyperbola, the equality is satisfied. Therefore, and – hyperbole tricks. As you can see, the given point A(4;0) is the right focus of the hyperbola.

Let us determine the eccentricity of the resulting hyperbola:

The equations of the hyperbola asymptotes have the form and . Therefore, or and are asymptotes of a hyperbola. Before constructing a hyperbola, we construct its asymptotes.

Problem 3. Create an equation for the locus of points equidistant from the point A(4; 3) and the straight line y = 1. Reduce the resulting equation to its simplest form.

Solution: Let M(x; y) be one of the points of the desired geometric locus of points. Let us drop the perpendicular MB from point M to this straight line y = 1 (Fig. 3). Let us determine the coordinates of point B. Obviously, the abscissa of point B is equal to the abscissa of point M, and the ordinate of point B is equal to 1, i.e. B(x; 1). According to the conditions of the problem |MA|=|MV|. Consequently, for any point M(x;y) belonging to the desired geometric locus of points, the following equality is true:

The resulting equation defines a parabola with a vertex at the point. To bring the parabola equation to its simplest form, let us set and y + 2 = Y, then the parabola equation takes the form:

An example of solving some tasks from the standard work “Analytical geometry on a plane”

The vertices are given,
,
triangle ABC. Find:

Equations of all sides of a triangle;

System of linear inequalities defining a triangle ABC;

Equations of altitude, median and bisector of a triangle drawn from the vertex A;

The intersection point of the triangle's altitudes;

The intersection point of the triangle's medians;

Length of the height lowered to the side AB;

Corner A;

Make a drawing.

Let the vertices of the triangle have coordinates: A (1; 4), IN (5; 3), WITH(3; 6). Let's draw a drawing right away:

1. To write down the equations of all sides of a triangle, we use the equation of a straight line passing through two given points with coordinates ( x 0 , y 0 ) And ( x 1 , y 1 ):

=

Thus, substituting instead of ( x 0 , y 0 ) point coordinates A, and instead of ( x 1 , y 1 ) point coordinates IN, we get the equation of the line AB:

The resulting equation will be the equation of the straight line AB, written in general form. Similarly, we find the equation of the straight line AC:

And also the equation of the straight line Sun:

2. Note that the set of points of the triangle ABC represents the intersection of three half-planes, and each half-plane can be defined using a linear inequality. If we take the equation of either side ∆ ABC, For example AB, then the inequalities

And

define points lying on opposite sides of a line AB. We need to choose the half-plane where point C lies. Let’s substitute its coordinates into both inequalities:

The second inequality will be correct, which means that the required points are determined by the inequality

.

We do the same with straight line BC, its equation
. We use point A (1, 1) as a test point:

This means that the required inequality has the form:

.

If we check straight line AC (test point B), we get:

This means that the required inequality will have the form

We finally obtain a system of inequalities:

The signs “≤”, “≥” mean that points lying on the sides of the triangle are also included in the set of points that make up the triangle ABC.

3. a) In order to find the equation for the height dropped from the vertex A to the side Sun, consider the equation of the side Sun:
. Vector with coordinates
perpendicular to the side Sun and therefore parallel to the height. Let us write down the equation of a straight line passing through a point A parallel to the vector
:

This is the equation for the height omitted from t. A to the side Sun.

b) Find the coordinates of the middle of the side Sun according to the formulas:

Here
– these are the coordinates of t. IN, A
– coordinates t. WITH. Let's substitute and get:

The straight line passing through this point and the point A is the required median:

c) We will look for the equation of the bisector based on the fact that in an isosceles triangle the height, median and bisector descended from one vertex to the base of the triangle are equal. Let's find two vectors
And
and their lengths:

Then the vector
has the same direction as the vector
, and its length
Likewise, the unit vector
coincides in direction with the vector
Vector sum

there is a vector that coincides in direction with the bisector of the angle A. Thus, the equation of the desired bisector can be written as:

4) We have already constructed the equation for one of the heights. Let's construct an equation for another height, for example, from the vertex IN. Side AC given by the equation
So the vector
perpendicular AC, and thus parallel to the desired height. Then the equation of the line passing through the vertex IN in the direction of the vector
(i.e. perpendicular AC), has the form:

It is known that the altitudes of a triangle intersect at one point. In particular, this point is the intersection of the found heights, i.e. solving the system of equations:

- coordinates of this point.

5. Middle AB has coordinates
. Let us write the equation of the median to the side AB. This line passes through points with coordinates (3, 2) and (3, 6), which means its equation has the form:

Note that a zero in the denominator of a fraction in the equation of a line means that this line runs parallel to the ordinate axis.

To find the intersection point of the medians, it is enough to solve the system of equations:

The intersection point of the medians of a triangle has coordinates
.

6. Length of height lowered to the side AB, equal to the distance from the point WITH to a straight line AB with equation
and is found by the formula:

7. Cosine of angle A can be found using the formula for the cosine of the angle between vectors And , which is equal to the ratio of the scalar product of these vectors to the product of their lengths:

.

In problems 1 - 20 the vertices of triangle ABC are given.
Find: 1) the length of side AB; 2) equations of sides AB and AC and their angular coefficients; 3) Internal angle A in radians with an accuracy of 0.01; 4) equation for the height of CD and its length; 5) the equation of a circle for which the height CD is the diameter; 6) a system of linear inequalities defining triangle ABC.

Length of triangle sides:
|AB| = 15
|AC| = 11.18
|BC| = 14.14
Distance d from point M: d = 10
The coordinates of the vertices of the triangle are given: A(-5,2), B(7,-7), C(5,7).
2) Length of the sides of the triangle
The distance d between points M 1 (x 1 ; y 1) and M 2 (x 2 ; y 2) is determined by the formula:

8) Equation of a line
A straight line passing through points A 1 (x 1 ; y 1) and A 2 (x 2 ; y 2) is represented by the equations:

Equation of line AB


or

or
y = -3 / 4 x -7 / 4 or 4y + 3x +7 = 0
Equation of line AC
Canonical equation of the line:

or

or
y = 1 / 2 x + 9 / 2 or 2y -x - 9 = 0
Equation of line BC
Canonical equation of the line:

or

or
y = -7x + 42 or y + 7x - 42 = 0
3) Angle between straight lines
Equation of straight line AB:y = -3 / 4 x -7 / 4
Line equation AC:y = 1 / 2 x + 9 / 2
The angle φ between two straight lines, given by equations with angular coefficients y = k 1 x + b 1 and y 2 = k 2 x + b 2, is calculated by the formula:

The slopes of these lines are -3/4 and 1/2. Let's use the formula, and take its right-hand side modulo:

tg φ = 2
φ = arctan(2) = 63.44 0 or 1.107 rad.
9) Equation of height through vertex C
The straight line passing through the point N 0 (x 0 ;y 0) and perpendicular to the straight line Ax + By + C = 0 has a direction vector (A;B) and, therefore, is represented by the equations:



This equation can be found in another way. To do this, let's find the slope k 1 of straight line AB.
AB equation: y = -3 / 4 x -7 / 4, i.e. k 1 = -3 / 4
Let's find the angular coefficient k of the perpendicular from the condition of perpendicularity of two straight lines: k 1 *k = -1.
Substituting the slope of this line instead of k 1, we get:
-3 / 4 k = -1, whence k = 4 / 3
Since the perpendicular passes through the point C(5,7) and has k = 4 / 3, we will look for its equation in the form: y-y 0 = k(x-x 0).
Substituting x 0 = 5, k = 4 / 3, y 0 = 7 we get:
y-7 = 4 / 3 (x-5)
or
y = 4 / 3 x + 1 / 3 or 3y -4x - 1 = 0
Let's find the point of intersection with line AB:
We have a system of two equations:
4y + 3x +7 = 0
3y -4x - 1 = 0
From the first equation we express y and substitute it into the second equation.
We get:
x = -1
y=-1
D(-1;-1)
9) Length of the altitude of the triangle drawn from vertex C
The distance d from the point M 1 (x 1 ;y 1) to the straight line Ax + By + C = 0 is equal to the absolute value of the quantity:

Find the distance between point C(5;7) and line AB (4y + 3x +7 = 0)


The length of the height can be calculated using another formula, as the distance between point C(5;7) and point D(-1;-1).
The distance between two points is expressed in terms of coordinates by the formula:

5) the equation of a circle for which the height CD is the diameter;
The equation of a circle of radius R with center at point E(a;b) has the form:
(x-a) 2 + (y-b) 2 = R 2
Since CD is the diameter of the desired circle, its center E is the midpoint of the segment CD. Using the formulas for dividing a segment in half, we get:


Therefore, E(2;3) and R = CD / 2 = 5. Using the formula, we obtain the equation of the desired circle: (x-2) 2 + (y-3) 2 = 25

6) a system of linear inequalities defining triangle ABC.
Equation of line AB: y = -3 / 4 x -7 / 4
Equation of line AC: y = 1 / 2 x + 9 / 2
Equation of line BC: y = -7x + 42