Given the vertices of a triangle find the length of the height. How to learn to solve problems in analytical geometry? Typical problem with a triangle on a plane
An example of solving some tasks from the typical work "Analytical geometry on a plane"
Vertices are given,
,
triangle ABC. Find:
Equations of all sides of a triangle;
A system of linear inequalities defining a triangle ABC;
Equations for the height, median, and bisector of a triangle drawn from a vertex BUT;
The point of intersection of the heights of the triangle;
The point of intersection of the medians of the triangle;
The length of the height lowered to the side AB;
Corner BUT;
Make a drawing.
Let the vertices of the triangle have coordinates: BUT (1; 4), AT (5; 3), FROM(3; 6). Let's draw a drawing:
1. To write out the equations of all sides of the 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 BUT, and instead of ( x 1 , y 1 ) point coordinates AT, we get the equation of a straight line AB:
The resulting equation will be the equation of a straight line AB written in general form. Similarly, we find the equation of a straight line AC:
And also the equation of a straight line sun:
2. Note that the set of points of the triangle ABC is 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
set the points lying along different sides from straight 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 proceed similarly with the straight line BC, its equation
. As a test, we use point A (1, 1):
so the desired inequality is:
.
If we check the line AC (trial point B), we get:
so the desired inequality will be of the form
Finally, we obtain a system of inequalities:
The signs "≤", "≥" mean that the 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 top BUT to the side sun, consider the side equation sun:
. Vector with coordinates
perpendicular to side sun and, therefore, parallel to the height. We write the equation of a straight line passing through a point BUT parallel to the vector
:
This is the equation for the height omitted from t. BUT to the side sun.
b) Find the coordinates of the midpoint of the side sun according to the formulas:
Here
are the coordinates. AT, a
- coordinates t. FROM. Substitute and get:
The line passing through this point and the point BUT is the desired median:
c) We will look for the bisector equation, based on the fact that in an isosceles triangle the height, median and bisector, lowered 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
Similarly, the unit vector
coincides in direction with the vector
Sum of vectors
is a vector that coincides in direction with the angle bisector BUT. Thus, the equation of the desired bisector can be written as:
4) We have already built the equation of one of the heights. Let's construct an equation of one more height, for example, from the top AT. Side AC is given by the equation
So the vector
perpendicular AC, and thus parallel to the desired height. Then the equation of the straight line passing through the vertex AT in the direction of the vector
(i.e. perpendicular AC), has the form:
It is known that the heights of a triangle intersect at one point. In particular, this point is the intersection of the heights found, i.e. solution of the system of equations:
are the coordinates of this point.
5. Middle AB has coordinates
. Let's write the equation of the median to the side AB. This line passes through the points with coordinates (3, 2) and (3, 6), so its equation is:
Note that zero in the denominator of a fraction in the equation of a straight line means that this straight line runs parallel to the y-axis.
To find the point of intersection of the medians, it is enough to solve the system of equations:
The point of intersection of the medians of a triangle has coordinates
.
6. The length of the height lowered to the side AB, equal to the distance from the point FROM to straight AB with the equation
and is given by the formula:
7. Cosine of an angle BUT can be found by 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:
.
Exercise 1
57. Vertices of triangle ABC are given. Find
) length of side AB;
) equations of the sides AB and AC and their slopes;
) internal corner A;
) the equation of the median drawn from vertex B;
) the equation of height CD and its length;
) the equation of a circle for which the height CD is the diameter and the points of intersection of this circle with the side AC;
) the equation of the bisector of the internal angle A;
) the area of triangle ABC;
) a system of linear inequalities that define the triangle ABC.
Make a drawing.
A(7, 9); B(-2, -3); C(-7, 7)
Solution:
1)
Find the length of the vector
= (x b -x a )2+ (y b -y a )2 = ((-2)-7)2 + (-3 - 9)2 = 92 + 122 = 225
= \u003d 15 - length of side AB 2)
Let's find the equation of the side AB
Equation of a straight line passing through points Oh a ; at in ) and B(x a ; at in ) in general Substitute the coordinates of points A and B into this equation of a straight line =
=
=
S AB = (- 3, - 4) is called the directing vector of the line AB. This vector is parallel to line AB. 4(x - 7) = - 3(y - 9) 4x + 28 = - 3y + 27 4x + 3y + 1 \u003d 0 - equation of the straight line AB If the equation is written as: y = X - then its slope can be distinguished: k 1 =4/3
Vector N AB = (-4, 3) is called the normal vector of the line AB. Vector N AB = (-4, 3) is perpendicular to line AB. Similarly, we find the equation of the side AC =
=
=
S AS = (- 7, - 1) - direction vector of AC side (x - 7) = - 7 (y - 9) x + 7 = - 7y + 63 x + 7y - 56 = 0 - equation of side AC y= = x + 8 whence slope k 2 = 1/7
Vector N AC = (- 1, 7) is the normal vector of the line AC. Vector N AC = (- 1, 7) is perpendicular to line AC. 3)
Let's find the angle A
We write the formula for the scalar product of vectors and * = *cos∟A To find angle A, it is enough to find the cosine of this angle. From the previous formula, we write the expression for the cosine of angle A cos∟A = Finding the scalar product of vectors and = (x in - X a ; at in - at a ) = (- 2 - 7; - 3 - 9) = (-9, -12)
= (x With - X a ; at With - at a ) = (- 7 - 7; 7 - 9) = (-14; -2)
9*(-14) + (-12)*(-2) = 150
Vector length = 15 (found earlier) Find the length of the vector = (x FROM -x a )2+ (y With -y a )2 = (-14)2 + (-2)2 = 200
= \u003d 14.14 - length of side AC Then cos∟A = = 0,7072
∟A = 45 0
4)
Find the equation for the median BE drawn from point B to side AC
General median equation Now you need to find the direction vector of the straight line BE. We complete the triangle ABC to the parallelogram ABCD, so that the side AC is its diagonal. The diagonals in a parallelogram are divided in half, i.e. AE = EC. Therefore, the point E lies on the line BF. As the direction vector of the straight line BE, one can take the vector , which we will find. = +
= (x c - X b ; at c - at b ) = (- 7- (-2); 7 - (-3)) = (-5. 10)
= + = (-5 + 9; 10 + 12) = (4; 22)
Substitute into the equation Substitute the coordinates of point C (-7; 7) (x + 7) = 2(y - 7) x + 77 = 2y - 14 x - 2y + 91 = 0 - BE median equation Since point E is the midpoint of side AC, then its coordinates X e = (x a + x With )/2 = (7 - 7)/2 = 0
at e = (y a + y With )/2 = (9 + 7)/2 = 8
Point E coordinates (0; 8) 5)
Find the equation for the height of CD and its length
General equation It is necessary to find the direction vector of the straight line CD Line CD is perpendicular to line AB, therefore, the direction vector of line CD is parallel to the normal vector of line AB CD ‖AB That is, as the directing vector of the straight line CD, you can take the normal vector of the straight line AB Vector AB found earlier: AB (-4, 3)
Substitute the coordinates of point C, (- 7; 7) (x + 7) = - 4 (y - 7) x + 21 = - 4y + 28 x + 4y - 7 \u003d 0 - height equation C D Point D coordinates: The point D belongs to the line AB, therefore, the coordinates of the point D(x d . y d ) must satisfy the equation of the straight line AB found earlier The point D belongs to the line CD; therefore, the coordinates of the point D(x d . y d ) must satisfy the straight CD equation, Let us compose a system of equations based on this D(1; 1) coordinates Find the length of the line CD = (x d -x c )2+ (y d -y c )2 = (1 + 7)2 + (1 - 7)2 = 64 +36 = 100
= \u003d 10 - length of straight line CD 6)
Find the equation for a circle with diameter CD
Obviously, the straight line CD passes through the origin of coordinates, since its equation is -3x - 4y \u003d 0, therefore, the circle equation can be written as (x - a) 2 + (y - b) 2= R 2- equation of a circle centered at a point (a; b) Here R \u003d CD / 2 \u003d 10 / 2 \u003d 5 (x - a) 2 + (y - b) 2 = 25
The center of the circle O (a; b) lies in the middle of the segment CD. Let's find its coordinates: X 0=a= = = - 3;
y 0=b= = = 4
Circle equation: (x + 3) 2 + (y - 4) 2 = 25
Find the intersection of this circle with side AC: point K belongs to both the circle and the line AC x + 7y - 56 \u003d 0 - the equation of the straight line AC found earlier. Let's make a system Thus, we got the quadratic equation at 2- 750y +2800 = 0 at 2- 15y + 56 = 0 =
at 1 = 8
at 2= 7 - point corresponding to point C hence the coordinates of the point H: x = 7*8 - 56 = 0
1. The equation of the sides AB and BC and their slopes.
The task gives the coordinates of the points through which these lines pass, so we will use the equation of a straight line passing through two given points $$\frac(x-x_1)(x_2-x_1)=\frac(y-y_1)(y_2-y_1)$ $ substitute and get the equations
equation of line AB $$\frac(x+6)(6+6)=\frac(y-8)(-1-8) => y = -\frac(3)(4)x + \frac(7 )(2)$$ the slope of line AB is \(k_(AB) = -\frac(3)(4)\)
equation of line BC $$\frac(x-4)(6-4)=\frac(y-13)(-1-13) => y = -7x + 41$$ slope of line BC is \(k_( BC) = -7\)
2. Angle B in radians to two decimal places
Angle B - the angle between lines AB and BC, which is calculated by the formula $$tg\phi=|\frac(k_2-k_1)(1+k_2*k_1)|$$substitute the slope coefficients of these lines and get $$tg\ phi=|\frac(-7+\frac(3)(4))(1+7*\frac(3)(4))| = 1 => \phi = \frac(\pi)(4) \approx 0.79$$
3.Length of side AB
The length of the side AB is calculated as the distance between the points and is equal to \(d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)\) => $$d_(AB) = \sqrt((6+ 6)^2+(-1-8)^2) = 15$$
4. Equation of CD height and its length.
We will find the height equation by the formula of a straight line passing through given point C(4;13) in the given direction - perpendicular to the line AB according to the formula \(y-y_0=k(x-x_0)\). Find the slope of the height \(k_(CD)\) using the property of perpendicular lines \(k_1=-\frac(1)(k_2)\) we get $$k_(CD)= -\frac(1)(k_(AB) ) = -\frac(1)(-\frac(3)(4)) = \frac(4)(3)$$ (x-4) => y = \frac(4)(3)x+\frac(23)(3)$$ = \frac(Ax_0+By_0+C)(\sqrt(A^2+B^2))$$ in the numerator is the equation of the line AB, we bring it to this form \(y = -\frac(3)(4)x + \frac(7)(2) => 4y+3x-14 = 0\) , substitute the resulting equation and point coordinates into the formula $$d = \frac(4*13+3*4-14 )(\sqrt( 4^2+3^2)) = \frac(50)(5) =10$$
5. The equation of the median AE and the coordinates of the point K, the intersection of this median with the height CD.
We will look for the median equation as the equation of a straight line passing through two given points A(-6;8) and E , where point E is the midpoint between points B and C and its coordinates are found by the formula \(E(\frac(x_2+x_1) (2);\frac(y_2+y_1)(2))\) substitute the coordinates of the points \(E(\frac(6+4)(2);\frac(-1+13)(2))\) = > \(E(5; 6)\), then the equation for median AE is $$\frac(x+6)(5+6)=\frac(y-8)(6-8) => y = - \frac(2)(11)x + \frac(76)(11)$$Find the coordinates of the point of intersection of the heights and the median, i.e. find their common point To do this, compose the system equation $$\begin(cases)y = -\frac(2)(11)x + \frac(76)(11)\\y = \frac(4)(3)x+ \frac(23)(3)\end(cases)=>\begin(cases)11y = -2x +76\\3y = 4x+23\end(cases)=>$$$$\begin(cases)22y = -4x +152\\3y = 4x+23\end(cases)=> \begin(cases)25y =175\\3y = 4x+23\end(cases)=> $$$$\begin(cases) y =7\\ x=-\frac(1)(2)\end(cases)$$ Intersection coordinates \(K(-\frac(1)(2);7)\)
6. Equation of a straight line that passes through point K parallel to side AB.
If the lines are parallel, then their slopes are equal, i.e. \(k_(AB)=k_(K) = -\frac(3)(4)\) , the coordinates of the point \(K(-\frac(1)(2);7)\) are also known, i.e. . to find the equation of a straight line, we apply the formula for the equation of a straight line passing through a given point in a given direction \(y - y_0=k(x-x_0)\), we substitute the data and get $$y - 7= -\frac(3)(4) (x-\frac(1)(2)) => y = -\frac(3)(4)x + \frac(53)(8)$$
8. Coordinates of point M which is symmetrical to point A with respect to line CD.
The point M lies on the line AB, because CD - height to this side. Find the intersection point of CD and AB. To do this, solve the system of equations $$\begin(cases)y = \frac(4)(3)x+\frac(23)(3)\\y = -\frac(3)(4) x + \frac(7)(2)\end(cases) =>\begin(cases)3y = 4x+23\\4y =-3x + 14\end(cases) => $$$$\begin(cases )12y = 16x+92\\12y =-9x + 42\end(cases) =>
\begin(cases)0= 25x+50\\12y =-9x + 42\end(cases) => $$$$\begin(cases)x=-2\\y=5 \end(cases)$$ Point coordinates D(-2;5). By the condition AD=DK, this distance between points is found by the Pythagorean formula \(d = \sqrt((x_2-x_1)^2+(y_2-y_1)^2)\), where AD and DK are the hypotenuses of equal right triangles, and \(Δx =x_2-x_1\) and \(Δy=y_2-y_1\) are the legs of these triangles, i.e. find the legs and find the coordinates of the point M. \(Δx=x_D-x_A = -2+6=4\), and \(Δy=y_D-y_A = 5-8=-3\), then the coordinates of the point M will be equal to \ (x_M-x_D = Δx => x_D +Δx =-2+4=2 \), and \(y_M-y_D = Δy => y_D +Δy =5-3=2 \), got that the coordinates of the point \( M(2;2)\)
In problems 1 - 20, the vertices of the triangle ABC are given.
Find: 1) the length of the side AB; 2) equations of the sides AB and AC and their slopes; 3) Internal angle A in radians with an accuracy of 0.01; 4) CD height equation and its length; 5) the equation of a circle, for which the height CD is the diameter; 6) a system of linear inequalities that define the triangle ABC.
The length of the sides of the triangle:
|AB| = 15
|AC| = 11.18
|BC| = 14.14
Distance d from point M: d = 10
Given the coordinates of the vertices of the triangle: A(-5,2), B(7,-7), C(5,7).
2) The 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) Straight line equation
The straight line passing through the 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
Line AC equation
Canonical equation of a straight line:
or
or
y = 1 / 2 x + 9 / 2 or 2y -x - 9 = 0
Line BC equation
Canonical equation of a straight line:
or
or
y = -7x + 42 or y + 7x - 42 = 0
3) Angle between straight lines
Straight line equation AB:y = -3 / 4 x -7 / 4
Straight line equation AC:y = 1 / 2 x + 9 / 2
The angle φ between two straight lines given by equations with slope coefficients y \u003d k 1 x + b 1 and y 2 \u003d k 2 x + b 2 is calculated by the formula:
The slopes of these straight lines are -3/4 and 1/2. We use the formula, and we take its right side modulo:
tan φ = 2
φ = arctan(2) = 63.44 0 or 1.107 rad.
9) Height equation through vertex C
The line passing through the point N 0 (x 0; y 0) and perpendicular to the line Ax + By + C = 0 has a direction vector (A; B) and, therefore, is represented by the equations:
This equation can also be found in another way. To do this, we find the slope k 1 of the straight line AB.
Equation AB: y = -3 / 4 x -7 / 4, i.e. k 1 \u003d -3 / 4
Let's find the slope k of the perpendicular from the condition of perpendicularity of two straight lines: k 1 *k = -1.
Substituting instead of k 1 the slope of this straight line, 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 \u003d 5, k \u003d 4 / 3, y 0 \u003d 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 the line AB:
We have a system of two equations:
4y + 3x +7 = 0
3y -4x - 1 = 0
Express y from the first equation and substitute it into the second equation.
We get:
x=-1
y=-1
D(-1;-1)
9) The length of the height of the triangle drawn from the vertex C
The distance d from the point M 1 (x 1; y 1) to the straight line Ax + By + C \u003d 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 also 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 centered at the 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 \u003d CD / 2 \u003d 5. Using the formula, we get the equation of the desired circle: (x-2) 2 + (y-3) 2 \u003d 25
6) a system of linear inequalities that define the triangle ABC.
Straight line AB equation: y = -3 / 4 x -7 / 4
Line AC equation: y = 1 / 2 x + 9 / 2
Straight line BC equation: y = -7x + 42