Option 13 IDZ 3.2

IDZ - 3.2 No. 1.13

The vertices ∆АВС are given: А(–5;2); B(0;–4); C(5;7).

Find:

1. Equation of side AB;
2. CH height equation;
3. AM the equation media;
4. Point N of intersection of the median AM and height CH;
5. Equation of a line passing through vertex C and parallel to side AB;
6. Distance from point C to straight line AB.

Answer

1. Equation of side AB:

Let's find the coordinates of vector AB:

AB = B - A = (0 - (-5); -4 - 2) = (5; -6)

Then the equation of line AB has the form:

(y - 2) / (-6) = (x + 5) / 5

or

5y + 30 = -6x - 30

or

6x + 5y + 60 = 0

CH height equation:

Let's find the equation of a line passing through C and perpendicular to AB:

Since AB is given by the equation 6x + 5y + 60 = 0, the equation of the line perpendicular to AB has the form:

5h - 6u + S1 = 0,

where C1 is an unknown coefficient that needs to be found by substituting the coordinates of point C:

5 * 5 - 6 * 7 + C1 = 0

S1 = 11

Then the equation for the height of the CH has the form:

5h - 6u + 11 = 0

The equation for media is:

Let's find the coordinates of point M, which is the middle of side AC:

M = ((-5 + 5) / 2; (2 + 7) / 2) = (0; 4.5)

Then the equation of the media AM has the form:

y = -9/5 * x + 13.5

Point N of intersection of the median AM and height CH:

Let's find the coordinates of point N, the intersection of the median AM and the height CH:

Let's solve the system of equations:

5h - 6u + 11 = 0

y = -9/5 * x + 13.5

Let's substitute the equation of the second equation into the first:

5x - 6 * (-9/5 * x + 13.5) + 11 = 0

Solving the equation, we get:

x = 2

u = 4

The current intersection point of the median AM and the height CH is equal to N(2; 4).

Equation of a line passing through vertex C and parallel to side AB:

Since side AB is given by the equation 6x + 5y + 60 = 0, then the equation of a straight line parallel to AB has the form:

6h + 5u + S2 = 0,

where C2 is an unknown coefficient that needs to be found by substituting the coordinates of point C:

6 * 5 + 5 * 7 + S2 = 0

S2 = -65

Then the equation of the line passing through vertex C and parallel to side AB has the form:

6x + 5y - 65 = 0

Distance from point C to straight line AB:

The distance from point C to line AB is equal to the distance from point C to its projection onto line AB. Let's find the coordinates of the projection of point C onto line AB:

Let us find the equation of a line passing through C and perpendicular to AB:

Since AB is given by the equation 6x + 5y + 60 = 0, the equation of the line perpendicular to AB has the form:

5h - 6u + S3 = 0,

where C3 is an unknown coefficient that needs to be found by substituting the coordinates of point C:

5 * 5 - 6 * 7 + C3 = 0

C3 = 11

Then the equation of the line passing through C and perpendicular to AB has the form:

5h - 6u + 11 = 0

Let us find the point of intersection of the line passing through C and perpendicular to AB, and the straight line AB:

Let's solve the system of equations:

6x + 5y + 60 = 0

5h - 6u + 11 = 0

Let's substitute the equation of the second equation into the first:

6x + 5 * (-9/5 * x + 13.5) + 60 = 0

Solving the equation, we get:

x = -1

u = 3

Then the coordinates of the projection of point C onto line AB are equal to S(-1; 3).

The distance from point C to line AB is equal to the distance between points C and S:

d = √[(5 - (-1))^2 + (7 - 3)^2] = √[36 + 16] = √52 = 2√13

IDZ - 3.2 No. 2.13

Given two vertices of triangle ABC: A(–6;2); AT 2

Product description

Option 13 IDZ 3.2

This is a digital product presented in a digital goods store. This product contains solutions to problems from the geometry section related to triangles. In particular, it presents the solution to problem No. 1.13 and problem No. 2.13 from option 13 of IDZ 3.2.

Each problem contains a step-by-step solution with detailed calculations and answers. For ease of viewing and reading, the text is formatted as an html page using appropriate html tags. Beautiful page design makes browsing even more pleasant and convenient.

This digital product is intended for students studying geometry and solving problems related to triangles. Solutions to problems in this product will help you better understand the material and prepare for exams, and can also be used as additional material for independent study of geometry.

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IDZ 3.2 № 1.13

Need to find:

a) Equation of side AB;

b) Equation of CH height;

(c) AM the equational media;

d) Point N of intersection of the median AM and height CH;

e) Equation of a line passing through vertex C and parallel to side AB;

e) Distance from point C to straight line AB.

There are vertices of the triangle ∆ABC: ​​A(–5;2); B(0;–4); C(5;7).

a) Construct vectors AB and BC, then find their coordinates:

AB = (0 - (-5), -4 - 2) = (5, -6) BC = (5 - 0, 7 - (-4)) = (5, 11)

The equation for side AB is: 5x - 6y + c = 0

To find the value of the constant c, we substitute the coordinates of point A: 5*(-5) - 62 + c = 0 c = 55 + 6*2 = 35

Answer: equation of side AB: 5x - 6y + 35 = 0.

b) Let's find the equations of straight lines containing side AB and height CH passing through vertex C. To do this, find the coordinates of point H, the intersection of height CH and side AB. First, find the length of the sides of the triangle:

AB = √(5^2 + (-6)^2) = √61 AND = √(10^2 + 5^2) = √125 BC = √(5^2 + 11^2) = √146

Triangle semi-perimeter p = (AB + AC + BC) / 2 = (√61 + √125 + √146) / 2 ≈ 12.776

Area of ​​the triangle S = √(p(p-AB)(p-AC)(p-BC)) ≈ 30.5

The height dropped from vertex C is equal to h = 2S/AB ≈ 10

Point H lies on side AB and the distance from it to vertex C is equal to h. This means that the coordinates of point H can be found by solving the system of equations:

5x - 6y + 35 = 0 5x + 11y - 35 = 0 y = 7

Substituting y = 7 into the first equation, we find:

x = -2

This means that the coordinates of point H are equal to (-2, 7).

The CH height equation has the form: x + 6y - 16 = 0

Answer: CH height equation: x + 6y - 16 = 0.

c) Find the midpoint of side AB and the coordinates of point M. The midpoint of side AB has coordinates (x, y), where:

x = (-5 + 0) / 2 = -2.5 y = (2 - 4) / 2 = -1

Point M lies on the side AC and divides it in the ratio AM/MC = 1/1, that is, the coordinates of point M can be found by solving the system of equations:

5x - 6y + 35 = 0 x + y - 3 = 0

Solving this system, we get:

x = -2 y = 1

Answer: the equation of the median AM has the form: 5x - 6y + 35 = 0

d) The intersection point of the median AM and the height CH can be found by solving the system of equations for the equation of the median and height:

5x - 6y + 35 = 0 x + 6y - 16 = 0

Solving this system, we get:

x = -1 y = 3

This means that the intersection point of the median AM and the height CH has coordinates (-1, 3).

Answer: point N(-1, 3).

e) A straight line passing through vertex C and parallel to side AB has the equation:

5x - 6y + c = 0

To find the value of the constant c, we substitute the coordinates of point C: 55 - 67 + c = 0 c = 7

Answer: the equation of a line passing through vertex C and parallel to side AB has the form: 5x - 6y + 7 = 0.

f) The distance from point C to line AB can be found using the formula for the distance from point to line:

d = |5*(-5) - 6*2 + 35| / √(5^2 + (-6)^2) ≈ 4.52

Answer: the distance from point C to straight line AB ≈ 4.52.

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