IDZ Ryabushko 3.2 Option 12

No. 1 For a triangle ∆ABC with vertices A(-4;2), B(8;-6) and C(2;6), it is necessary to find: a) the equation of side AB; b) height equation CH; c) equation of the median AM; 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) the distance from point C to line AB.

a) Equation of side AB: To find the equation of side AB, it is necessary to calculate the coordinates of the vector AB and then use it to construct the equation of the line passing through A and B. Coordinates of the vector AB: (8-(-4);-6-2) = (12;-8) Equation of the straight line passing through A and B: y-2 = (-8/12)(x-(-4)), that is, y = (-2/3)x + (2/3 ).

b) Equation of height CH: Height CH is perpendicular to side AB and passes through vertex C. So the equation of height can be found using the coordinates of points C and the perpendicular vector to AB (in this case it is the vector (-8,-12)) . The equation of a line passing through C and perpendicular to AB: -12(x-2) + (-8)(y-6) = 0, that is, 12x + 8y - 96 = 0.

c) Equation of median AM: Median AM is the line segment connecting vertex A and the midpoint of side BC. Let's find the coordinates of the middle of side BC: Coordinates of the middle of BC: ((8+2)/2;(-6+6)/2) = (5;0) Equation of the line passing through A and (5;0): y-2 = (0-2/5)(x-(-4)), that is, y = (-2/5)x + (6/5).

d) Point N of the intersection of the median AM and the height CH: To find point N, it is necessary to solve a system of equations consisting of the equation of the median AM and the equation of the height CH. Having solved the system, we obtain the intersection point N(1;-2).

e) Equation of a line passing through vertex C and parallel to side AB: A line parallel to side AB and passing through C has the same direction vector as AB. Thus, the equation of the line can be found using the coordinates of point C and the direction vector AB. The equation of a line passing through C and parallel to AB is: y-6 = (-2/3)(x-2), that is, y = (-2/3)x + (14/3).

e) Distance from point C to line AB: The distance between point C and line AB is equal to the length of the projection of vector AC onto vector AB, divided by the length of vector AB. Length of vector AB: √((8-(-4))^2+(-6-2)^2) = √(144+64) = √208. Projection of vector AC onto vector AB: |AC|cos(θ), where θ is the angle between vectors AC and AB. cos(θ) = ((-4-2)(8-2)+(2-6)(-6+2))/|AC||AB| = (-36-18)/√208*10 = -3/√208. Thus, the distance from point C to line AB is equal to |AC|cos(θ)/√208 = 3√13/13.

IDZ Ryabushko 3.2 Option 12

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Sorry, but you asked two different questions. The first question requires a description of the product “IDZ Ryabushko 3.2 Option 12”, and the second question requires a solution to the problem of finding the equations of the other two sides of a triangle using the known equations of one side and height.

Regarding the second question, to find the equations of the other two sides of the triangle, you can use the properties of the triangle and the system of equations.

a) The equation for side AB is already given: 4x + y = 12. b) The height equation BH passes through vertex B and is perpendicular to side AB. This means that its equation can be written as y - (-6) = (4/5)(x - 8). Transforming, we get y = (4/5)x - 22/5. c) The median AM also passes through vertex B and the midpoint of side AC. Let's find the coordinates of point M, which is the middle of side AC: ((-4+2)/2, (2+6)/2) = (-1, 4). The equation for the median AM can be written as y - (-6) = (-4/5)(x - 8). Transforming, we get y = (-4/5)x + 26/5. d) Point N is the intersection point of the median AM and the height BH. To find it, we solve a system of equations consisting of the equations for the median AM and the height BH. Let's substitute the equations of the median and height into the system and bring it to a convenient form:

x + y = 6 4x - 5y = -2

Having solved the system, we obtain the coordinates of point N: x = 2, y = 4. Thus, point N has coordinates (2, 4). e) The equation of side BC can be found by knowing the coordinates of points B and C. Let’s find them:

B(8, -6) C(2, 6)

Side direction vector BC: (-6 - 6, 8 - 2) = (-12, 6). A line parallel to side AB and passing through C will have an equation of the form y - 6 = (1/2)(x - 2), or y = (1/2)x + 5. f) The distance from point C to line AB can be found using the formula: d = |ax0 + by0 + c| / sqrt(a^2 + b^2), where (x0, y0) are the coordinates of point C, a and b are the coefficients of the equation of straight line AB (4x + y = 12), c = -12. Substituting the values, we get: d = |42 + 16 - 12| / sqrt(4^2 + 1^2) = 2 / sqrt(17).

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IDZ Ryabushko 3.2 Option 12 is a task for solving geometric problems related to triangle ABC, defined by vertices A(–4;2), B(8;–6) and C(2;6).

In the task you need to find:

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

Also in the task the equations of side AB, altitude BN and median AM are given, and you need to find the equations of the other two sides of triangle ABC.

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