The calculation of the coordinates of some point C, which divides the given segment AB in a certain ratio, can be performed using the formulas:
хС = (хА + λхВ) / (1 + λ), уС = (уА + λуВ) / (1 + λ),
where (xA; yA) and (xB; yB) are the coordinates of the ends of the given segment AB; the number λ \u003d AC / CB is the ratio in which the segment AB is divided by the point C, which has coordinates (xC; yC).
If the segment AB is divided by the point C in half, then the number λ \u003d 1 and the formulas for xC and yC will take the form:
xC = (xA + xB)/2, yC = (yA + yB)/2.
It must be borne in mind that in tasks λ is the ratio of the lengths of the segments, and therefore the numbers included in this ratio are not the lengths of the segments themselves in a given unit of measurement. For example, AC = 12 cm, CB = 16 cm: λ = AC/CB = 12 cm / 16 cm = 3/4.
1. Search for the coordinates of the middle of a certain segment, according to the given coordinates of its ends
Example 1
Points A (-2; 3) and B (6; -9) are the ends of the segment AB. Find point C, which is the midpoint of segment AB.
Decision.
In the condition of the problem, it is specified that xA = -2; xB = 6; yA = 3 and yB = -9. It is required to find C(xC; yC).
Applying the formulas xC = (xA + xB)/2, yC = (yA + yB)/2, we get:
xC \u003d (-2 + 6) / 2 \u003d 2, yC \u003d (3 + (-9)) / 2 \u003d -3.
Thus, point C, which is the midpoint of segment AB, has coordinates (-2; 3) (Fig. 1).
2. Calculation of the coordinates of the end of a certain segment, knowing the coordinates of its middle and other end
Example 2
One end of the segment AB is point A, with coordinates (-3; -5), and its midpoint is point C (3; -2). Calculate the coordinates of the second end of the segment - point B.
Decision.
According to the condition of the problem, it becomes clear that xA = -3; yA = -5; xC = 3 and yC = -2.
Substituting these values into the formulas xC = (xA + xB)/2, yC = (yA + yB)/2, we get:
3 = (-3 + xB)/2 and
2 \u003d (-5 + uV) / 2.
Solving the first equation for xB and the second for yB, we find: xB = 9 and yB = 1, it turns out that the desired point B will be given by the coordinates (9; 1) (Fig. 2).
3. Calculation of the coordinates of the vertices of a certain triangle according to the given coordinates of the midpoints of its sides
Example 3
The midpoints of the sides of the triangle ABC are the points D(1; 3), E(-1; -2) and F(4; -1). Find the coordinates of the vertices A, B and C of the given triangle.
Decision.
Let point D be the midpoint of side AB, point E be the midpoint of BC, and point F be the midpoint of side AC (Fig. 3). Find points A, B and C.
We denote the vertices of the triangle as A (xA; yA), B (xB; yB) and C (xC; yC) and knowing the coordinates of the points D, E and F, according to the formulas xC \u003d (xA + xB) / 2, yC \u003d (yA + uV)/2 we get:
(1 = (xA + xB)/2, 
(-1 \u003d (xB + xC) / 2,
(4 \u003d (xA + xC) / 2,
(3 \u003d (uA + uB) / 2,
(-2 \u003d (uV + uS) / 2,
(-1 \u003d (yA + yC) / 2.
We bring the equations to an integer form:
(xA + xB = 2,
(xB + xC = -2,
(xA + xC = 8,
(uA + uB = 6,
(uV + yC = -4,
(uA + yC = -2.
Solving the systems, we get:
xA = 6; xB = -4; xC = 2.
yA = 4; uV = 2; yC = -6.
Points A (6; 4), B (-4; 2) and C (2; -6) are the necessary vertices of the triangle.
4. Calculation of the coordinates of points that divide the segment in a certain ratio, according to the given coordinates of the ends of this segment
Example 4
Segment AB is divided by point C in a ratio of 3: 5 (counting from point A to point B). The ends of segment AB are points A(2; 3) and B(10; 11). Find point C.
Decision.
The condition of the problem says that xA = 2; xB = 10; yA = 3; uV = 11; λ = AC/CB = 3/5. Find C(xC; yC) (Fig. 4).
according to the formulas xC = (xA + λxB) / (1 + λ), yC = (yA + λyB) / (1 + λ) we get:
xC = (2 + 3/5 10) / (1 + 3/5) = 5 and yC = (3 + 3/5 11) / (1 + 3/5) = 6. Thus, we have C( 5; 6).
Let's check: AC = 3√2, CB = 5√2, λ = AC/CB = 3√2/5√2 = 3/5.
Comment. The condition of the problem states that the division of the segment is carried out in a given ratio from point A to point B. If this were not specified, then the problem would have two solutions. Second solution: division of the segment from point B to point A. 
Example 5
Some segment AB is divided in the ratio 2: 3: 5 (counting from point A to point B), its ends are points with coordinates A (-11; 1) and B (9; 11). Find the division points of the given segment.
Decision.
Let us denote the division points of the segment from A to B through C and D. In the condition of the problem, it is given that
xA = -11; xB = 9; yA = 1; yB = 11. Find C(xC; yC) and D(xD; yD) if AC: CD: DB = 2: 3: 5.
Point C divides segment AB in relation to λ = AC/CB = 2/(3 + 5) = 2/8 = 1/4.
According to the formulas xC = (xA + λxB) / (1 + λ), yC = (yA + λyB) / (1 + λ) we get:
xC = (-11 + ¼ 9) / (1 + 1/4) = -7 and yC = (1 + ¼ 11) / (1 + 1/4) = 3.
Thus, C(-7; 3).
Point D is the midpoint of segment AB. Applying the formulas xD = (xA + xB)/2, yD = (yA + yB)/2, we find:
xD \u003d (-11 + 9) / 2 \u003d -1, yD \u003d (1 + 11) / 2 \u003d 6. Hence, D has coordinates (-1; 6).
5. Calculation of the coordinates of the points that divide the segment, if the coordinates of the ends of this segment and the number of parts into which this segment is divided are given
Example 6
The ends of the segment are points A(-8; -5) and B(10; 4). Find points C and D that divide this segment into three equal parts.
Decision.
From the condition of the problem it is known that xA = -8; xB = 10; yA = -5; yB = 4 and n = 3. Find C(xC; yC) and D(xD; yD) (Fig. 5).
Let's find the point C. It divides the segment AB with respect to λ = 1/2. We divide from point A to point B. According to the formulas xC = (xA + λxB) / (1 + λ), yC = (yA + λyB) / (1 + λ) we have:
xC = (-8 + ½ 10) / (1 + 1/2) = -2 and yC = (-5 + ½ 4) / (1 + 1/2) = -2. So C(-2; -2).
The division of the segment CB is performed in a ratio of 1: 1, so we use the formulas
xD = (xA + xB)/2, yD = (yA + yB)/2:
xD \u003d (-2 + 10) / 2 \u003d 4, yD \u003d (-2 + 4) / 2 \u003d 1. Thus, D (4; 1).
Division points C(-2; -2) and D(4; 1).
Note: Point D can be found by dividing segment AB in relation to 2: 1. In this case, it will be necessary to apply the formulas xD = (xA + λxB) / (1 + λ), yD = (yA + λyB) / (1 + λ).
Example 7
Points A(5; -6) and B(-5; 9) are the ends of the segment. Find the points that divide the given segment into five equal parts.
Decision.
Let the consecutive division points from A to B be C(xC; yC), D(xD; yD), E(xE; yE), and F(xF; yF). The conditions of the problem say that xA = 5; xB = -5; yA = -6; yB = 9 and n = 5.
Using the formulas xC = (xA + λxB) / (1 + λ), yC = (yA + λyB) / (1 + λ) point C. It divides the segment AB in relation to λ = 1/4:
xC = (5 + 1/4 (-5)) / (1 + 1/4) = 3 and yC = (-6 + 1/4 9) / (1 + 1/4) = -3, we get that point C has coordinates (3; -3).
The segment AB is divided by the point D in the ratio 2: 3 (i.e. λ = 2/3), therefore:
xD = (5 + 2/3 (-5)) / (1 + 2/3) = 1 and yD = (-6 + 2/3 9) / (1 + 2/3) = 0, so D (ten).
Let's find the point E. It divides the segment AB in relation to λ = 2/3:
XE = (5 + 3/2 (-5)) / (1 + 3/2) = -1 and yE = (-6 + 3/2 9) / (1 + 3/2) = 3. Thus Thus, E(-1; 3).
Point F divides the segment AB in relation to λ = 4/1, therefore:
XF = (5 + 4 (-5)) / (1 + 4) = -3 and yF = (-6 + 4 9) / (1 + 4) = 6, F(-3; 6).
Division points С(-2; -2); D(4; 1); E(-1; 3) and F(-3; 6).
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Let the points M 1 , M 2 , M 3 be located on one straight line. It is said that the point M divides the segment M 1 M 2 with respect to λ(λ≠-1) if .
Let the coordinates of the points M 1 and M 2 be known with respect to some coordinate system: M 1 (x 1 , y 1 , z 1), M 2 (x 2 , y 2 , z 2), then the coordinates of the point M(x, y, z ) relative to the same coordinate system are found by the formulas:
If the point M is in the middle of the segment M 1 M 2 , then 
, that is, λ=1 and formulas (*) will take the form:
(**)
Use the following calculator to solve:
- Points are given by two coordinates: A(x 1 ,y 1), B(x 2 ,y 2).
- The points are given by three coordinates: A(x 1 ,y 1 ,z 1), B(x 2 ,y 2 ,z 2).
Example #1. The triangle is given by the coordinates of its vertices A(3, -2, 1), B(3, 1, 5), C(4, 0, 3). Find the coordinates D(x, y, z) - the points of intersection of its medians.
Decision. Denote by M(x 0 , y 0 , z 0) the midpoint of BC, then by formulas (**)
and M(7/2, ½, 4). Point D divides the median AM with respect to λ=2 . Applying formulas (*), we find .
Example #2. Segment AB is divided by point C(4,1) with respect to λ=1/4 , counting from point A . Find the coordinates of A if B(8,5).
Decision. Applying formulas (*), we get: 
, whence we find x=3 , y=0 .
Example #3. Segment AB is divided into three equal parts by points C(3, -1) and D(1,4). Find the coordinates of the ends of the segment.
Decision. Denote A(x 1 , y 1), B(x 2 , y 2). Point C is the midpoint of segment AD, therefore, using the formulas (**) we find: 
whence x 1 = 5, y 1 = -6. Similarly, the coordinates of point B are found: x 2 \u003d -1, y 2 \u003d 9.
When there are conditions for dividing a segment in a certain ratio, it is necessary to be able to determine the coordinates of the point that serves as a separator. We derive a formula for finding these coordinates by setting the problem on the plane.
Initial data: a rectangular coordinate system O x y and two non-coinciding points lying on it with given coordinates A (x A , y A) and B (x B , y B) are given. And also a point C is given, dividing the segment A B with respect to λ (some positive real number). It is necessary to determine the coordinates of point C: x C and y C .
Before proceeding with the solution of the task, let's reveal a little the meaning of the given condition: "point C, dividing the segment A B in relation to λ". Firstly, this expression indicates that point C lies on the segment A B (that is, between points A and B). Secondly, it is clear that according to the given condition, the ratio of the lengths of the segments A C and C B is equal to λ. Those. equality is correct:
In this case, point A is the beginning of the segment, point B is the end of the segment. If it were given that the point C divides the segment B A in a given ratio, then the equality would be true: .
Well, it’s a completely obvious fact that if λ = 1, then point C is the midpoint of the segment A B.
Let's solve the problem with the help of vectors. Let's arbitrarily display points A, B and point C on the segment A B in some rectangular coordinate system. Let's construct the radius vectors of these points, as well as the vectors A C → and C B → . According to the conditions of the problem, point C divides the segment A B in relation to λ.
The coordinates of the radius vector of the point are equal to the coordinates of the point, then the equalities are true: O A → = (x A , y A) and O B → = (x B , y B) .
Let's determine the coordinates of the vector: they will be equal to the coordinates of the point C, which are required to be found according to the condition of the problem.
Using the operation of vector addition, we write the equalities: O C → = O A → + A C → O B → = O C → + C B → ⇔ C B → = O B → - O C →
According to the condition of the problem, point C divides the segment A B in relation to λ, i.e. the equality A C = λ · C B is true.
The vectors A C → and C B → lie on the same straight line and are codirectional. λ > 0 by the condition of the problem, then, according to the operation of multiplying a vector by a number, we get: A C → = λ · C B → .
Let's transform the expression by substituting into it: C B → = O B → - O C → .
A C → = λ · (O B → - O C →) .
The equality O C → = O A → + A C → can be rewritten as O C → = O A → + λ · (O B → - O C →) .
Using the properties of operations on vectors, the last equality implies: O C → = 1 1 + λ · (O A → + λ · O B →) .
Now it remains for us to directly calculate the coordinates of the vector O C → = 1 1 + λ · O A → + λ · O B → .
Let's perform the necessary operations on the vectors O A → and O B → .
O A → = (x A , y A) and O B → = (x B , y B) , then O A → + λ O B → = (x A + λ x B , y A + λ y B) .
Thus, O C → = 1 1 + λ · (O A → + λ · O B →) = (x A + λ · x B 1 + λ , y A + λ · y B 1 + λ) .
Summarizing: the coordinates of the point C dividing the segment A B in a given ratio λ are determined by the formulas: x C \u003d x A + λ x B 1 + λ and y C \u003d y A + λ y B 1 + λ.
Determining the coordinates of a point dividing a segment in a given ratio in space
Initial data: rectangular coordinate system O x y z , points with given coordinates A (x A , y A , z A) and B (x B , y B , z B) .
Point C divides segment A B with respect to λ. It is necessary to determine the coordinates of point C.
Using the same reasoning scheme as in the case above on the plane, we arrive at the equality:
O C → = 1 1 + λ (O A → + λ O B →)
The vectors and are the radius vectors of the points A and B, which means:
O A → = (x A , y A , z A) and O B → = (x B , y B , z B) , therefore
O C → = 1 1 + λ (O A → + λ O B →) = (x A + λ x B 1 + λ , y A + λ y B 1 + λ , z A + λ z B 1 + λ)
Thus, point C, dividing the segment A B in space in a given ratio λ, has coordinates: (x A + λ x B 1 + λ, y A + λ y B 1 + λ, z A + λ z B 1+λ)
Let's consider the theory on concrete examples.
Example 1
Initial data: point C divides segment A B in a ratio of five to three. The coordinates of points A and B are given by A (11 , 1 , 0) , B (- 9 , 2 , - 4) .
Decision
By the condition of the problem λ = 5 3 . Let's apply the above formulas and get:
x A + λ x B 1 + λ = 11 + 5 3 (- 9) 1 + 5 3 = - 3 2
y A + λ y B 1 + λ = 1 + 5 3 2 1 + 5 3 = 13 8
z A + λ z B 1 + λ = 0 + 5 3 (- 4) 1 + 5 3 = - 5 2
Answer: C (- 3 2 , 13 8 , - 5 2)
Example 2
Initial data: it is necessary to determine the coordinates of the center of gravity of the triangle A B C.
The coordinates of its vertices are given: A (2 , 3 , 1) , B (4 , 1 , - 2) , C (- 5 , - 4 , 8)
Decision
It is known that the center of gravity of any triangle is the point of intersection of its medians (let this be the point M). Each of the medians is divided by point M in a ratio of 2 to 1, counting from the top. Based on this, we find the answer to the question posed.
Assume that A D is the median of triangle A B C. Point M is the intersection point of the medians, has coordinates M (x M, y M, z M) and is the center of gravity of the triangle. M, as the point of intersection of the medians, divides the segment A D in a ratio of 2 to 1, i.e. λ = 2 .
Let's find the coordinates of the point D. Since A D is the median, then point D is the midpoint of the segment B C. Then, using the formula for finding the coordinates of the midpoint of the segment, we get:
x D = x B + x C 2 = 4 + (- 5) 2 = - 1 2 y D = y B + y C 2 = 1 + (- 4) 2 = - 3 2 z D = z B + z C 2 = - 2 + 8 2 = 3
Calculate the coordinates of the point M:
x M = x A + λ x D 1 + λ = 2 + 2 (- 1 2) 1 + 2 = 1 3
y M = y A + λ y D 1 + λ = 3 + 2 (- 3 2) 1 + 2 = 0
z M = z A + λ z D 1 + λ = 1 + 2 3 1 + 2 = 7 3
Answer: (1 3 , 0 , 7 3)
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Let a directed line segment AB be given; say dot
M of this line divides the segment AB in a ratio equal to X, where is an arbitrary real number, if
When point M lies between points A and B (i.e., inside the segment
AB), then the vectors AM and MB are directed in the same direction (Fig. 2) and the ratio (1) is positive.
When point M lies outside the segment
AB, then the vectors AM and MB are directed in opposite directions (Fig. 3) and the ratio (1) is negative.
Let's see how the relation (1) changes when the point M runs through the entire line. When point M coincides with point A, then relation (1) is equal to zero; if then the point M runs through the segment AB in the direction from A to B, then the ratio (1) continuously increases, becoming arbitrarily large as the point M approaches B. When , then the fraction (1) loses its meaning, since its denominator turns into a zero vector. With further movement of the point along a straight line in the same direction (in Fig. 3, a to the right of B), ratio (1) becomes negative, and if W is close enough to B, then this ratio has an arbitrarily large absolute value.
Since , then (by virtue of Proposition 8 of § 4) we have
When the point M, moving all the time in the same direction (in our Fig. 3, and from left to right), but goes straight to infinity, then the fraction - tends to zero (since its numerator remains constant, and the denominator increases indefinitely), therefore , ratio , - tends to -1.
Now let M go over to the "left" of the two half-lines into which the point A divides the line (that is, into that half-line which does not contain the segment AB). If, in this case, the point M is far enough from the point A, then, again, arbitrarily small, and, therefore, the ratio of the formula differs arbitrarily little from -1. When point M approaches point A from the left (Fig. 3, b), the ratio (I), remaining negative, continuously decreases in absolute value and finally becomes equal to zero when point M returns to point A.
Note that for any position of the point M on the line, the ratio is not equal to -1. Indeed, the ratio is negative only when the point M lies outside the segment AB. But in this case the segments AM and MB are never equal, i.e.
Now let a coordinate system be established on the line and O be the origin of this system. We denote the coordinate of point A through points B - through , and the variable point M - through . Then and
If the point M (x; y) lies on a straight line passing through two given points M 1 (x 1; y 1), M 2 (x 2; y 2), and the ratio λ \u003d M 1 M / MM 2 is given, in where the point M divides the segment M 1 M 2, then the coordinates of the point M
are determined by the formulas
x = (x 1 + λx 2)/(1 + λ), y = (y 1 + λy 2)/(1 + λ)
If the point M is the midpoint of the segment M 1 M 2, then its coordinates are determined by the formulas
x \u003d (x 1 + x 2) / 2, y \u003d (y 1 + y 2) / 2
86. Given the ends A(3; -5) and 6(-1; 1) of a homogeneous rod. Determine the coordinates of its center of gravity.
87. The center of gravity of a homogeneous rod is at the point M (1; 4), one of its ends is at the point P (-2; 2). Determine the coordinates of the point Q of the other end of this rod
88. Triangle vertices A(1; -3), 6(3; -5) and C(-5; 7) are given. Determine the midpoints of its sides.
89. Two points A(3; - 1) and B(2; 1) are given. Define:
1) coordinates of point M, symmetrical to point A with respect to point B;
2) coordinates of point N, symmetrical to point B with respect to point A.
90. Points M (2; -1), N (-1; 4) and P (-2; 2) are the midpoints of the sides of the triangle. Determine its vertices.
91. Three vertices of a parallelogram A(3; -5), B(5; -3), C(-1; 3) are given. Determine the fourth vertex D, opposite to B.
92. Given two adjacent vertices of a parallelogram A(-3; 5), B(1; 7) and the intersection point of its diagonals M(1; 1). Define two other vertices.
93. Three vertices A(2; 3), 6(4; -1) and C(0; 5) of parallelogram ABCD are given. Find its fourth vertex D.
94. Vertices of a triangle A(1; 4), B(3; -9), С(-5; 2) are given. Find the length of its median drawn from vertex B.
95. The segment bounded by points A (1;-3) and B(4; 3) is divided into three equal parts. Determine the coordinates of the division points.
96. Vertices of a triangle A(2; -5), B(1; -2), C(4; 7) are given. Find the point of intersection with the side AC of the bisector of its interior angle at the vertex B.
97. Triangle vertices A(3; -5), B(-3; 3) and C(-1; -2) are given. Determine the length of the bisector of its interior angle at vertex A.
98. Given the vertices of a triangle A(-1; -1), B(3; 5), C(-4; 1). Find the point of intersection with the extension of the side BC of the bisector of its outer angle at the vertex A.
99. Given the vertices of the triangle A (3; -5), B (1; - 3), C (2; -2). Determine the length of the bisector of its exterior angle at vertex B.
100. Given three points A(1; -1), B(3; 3) and C(4; 5) lying on the same straight line. Determine the ratio λ in which each of them divides the segment bounded by the other two.
101. Determine the coordinates of the ends A and B of the segment, which is divided by points P (2; 2) and Q (1; 5) into three equal parts.
102. The straight line passes through the points M 1 (-12; -13) and M 2 (- 2; -5). Find a point on this line whose abscissa is 3.
103. The straight line passes through the points M(2; -3) and N(-6; 5). On this line, find a point whose ordinate is -5.
104. The straight line passes through the points A(7; -3) and B(23;. -6). Find the point of intersection of this line with the x-axis.
105. The line passes through the points A(5; 2) and B(-4; -7). Find the point of intersection of this line with the y-axis.
106. Vertices of the quadrilateral A(-3; 12), B(3; -4), C(5; -4) and D(5; 8) are given. Determine in what ratio its diagonal AC divides diagonal BD.
107. Vertices A(-2; 14), B(4; -2), C(6; -2) and D(6; 10) are given. Find the intersection point of its diagonals AC and BD.
108. Given the vertices of a homogeneous triangular plate A (x 1; y 1), B (x 2; y 2) and C (x 3; y 3). Determine the coordinates of its center of gravity,
Instruction. The center of gravity is at the point of intersection of the medians.
109. The point M of the intersection of the medians of the triangle lies on the abscissa axis, its two vertices are points A (2; -3) and B (-5; 1), the third vertex C lies on the y-axis. Determine the coordinates of points M and C.
110. Given the vertices of a homogeneous triangular plate A (x 1; y 1), B (x 2; y 2) and C (x 3; y 3). If you connect the midpoints of its sides, then a new homogeneous triangular plate is formed. Prove that the centers of gravity of both plates are the same.
Instruction. Use the result of task 108.
111. A homogeneous plate has the shape of a square with a side equal to 12, in which a square cut is made, the cut lines pass through the center of the square, the axes
coordinates are directed along the edges of the plate (Fig. 4). Determine the center of gravity of this plate.
112. A homogeneous plate has the shape of a rectangle with sides equal to a and b, in which a rectangular cut is made; the straight lines of the cut pass through the center, the coordinate axes are directed along the edges of the plate (Fig. 5). Determine the center of gravity of this plate.
113. A homogeneous plate has the shape of a square with a side equal to 2a, from which a triangle is cut off; the cut line connects the midpoints of two adjacent sides, the coordinate axes are directed along the edges of the plate (Fig. 6). Determine the center of gravity of the plate.
114. At the following points A (x 1; y 1), B (x 2; y 2) and C (x 3; y 3) masses m, n and p are concentrated. Determine the coordinates of the center of gravity of this system of three masses.
115. Points A (4; 2), B (7; -2) and C (1; 6) are the vertices of a triangle made of a homogeneous wire. Find the center of gravity of this triangle.
