IDZ Ryabushko 3.2 Option 24

№1

Given vertices ∆АВС: А(–2,–6); B(–3;5); C(4;0). Need to find:

a) Equation of side AB: First, let’s find the coordinates of vector AB: AB = B - A = (-3 - (-2); 5 - (-6)) = (-1; 11) Then the equation of straight line AB can be written in form: y + 6 = 11/1(x + 2)

b) Equation of the height CH: Let's find the coordinates of the vector AB and AC: AB = (-1; 11) AC = (4 - (-2); 0 - (-6)) = (6; 6) Since the height CH is drawn from vertex C is perpendicular to side AB, then it is parallel to vector AB. This means that the coordinates of the vector CH coincide with the coordinates of the vector AC projected onto the vector AB: CH = (AC * AB/|AB|^2) * AB = (6; 6) * (-1/122) * (-1; 11) = (6/61; -66/61) Now the equation of straight line CH can be written as: y = (-66/61)x + 24/61

c) Equation of the median AM: Let's find the coordinates of the vector AM: AM = M - A = ((-2 - 3)/2; (-6 + 0)/2) = (-5/2; -3) Since the median AM is a line passing through vertex A and the middle of side BC, then its direction vector is equal to half of vector BC: BC = C - B = (4 - (-3); 0 - 5) = (7; -5) Median AM passes through point M((-2 + 4)/2; (-6 + 0)/2) = (1; -3) and has a direction vector AM, so its equation can be written as: y + 3 = (-3/ -5)(x - 1)

d) Point N of intersection of the median AM and height CH: In order to find the point of intersection of the median AM and height CH, you need to solve the system of equations: y = (-66/61)x + 24/61 y + 3 = (-3/ -5)(x - 1) Having solved it, we get point N(23/61; -144/61).

e) Equation of a line passing through vertex C and parallel to side AB: Since the line passes through point C and is parallel to side AB, its direction vector coincides with vector AB: y - 0 = 11/1(x - 4)

e) Distance from point C to line AB: First, let’s find the equation of line AB: y + 6 = 11/1(x + 2) Then the distance from point C to line AB can be found using the formula: d = |(y2 - y1 )x0 - (x2 - x1)y0 + x2y1 - y2x1| / √((y2 - y1)^2 + (x2 - x1)^2) where (x0, y0) are the coordinates of point C, (x1, y1) and (x2, y2) are the coordinates of any two points lying on the line AB. Let's choose points A and B: d = |(5 - (-6))3 - ((-3)№1

Given vertices ∆АВС: А(–2,–6); B(–3;5); C(4;0). Need to find:

a) Equation of side AB: Let's start by finding the coordinates of vector AB: AB = B - A = (-3 - (-2); 5 - (-6)) = (-1; 11) Then the equation of straight line AB can be written as : y + 6 = 11/1(x + 2)

b) Equation of the CH height: Let us find the coordinates of the vectors AB and AC: AB = (-1; 11) AC = (4 - (-2); 0 - (-6)) = (6; 6) Since the CH height is drawn from vertex C is perpendicular to side AB, then it is parallel to vector AB. This means that the coordinates of the vector CH coincide with the coordinates of the vector AC projected onto the vector AB: CH = (AC * AB/|AB|^2) * AB = (6; 6) * (-1/122) * (-1; 11 ) = (6/61; -66/61) Now the equation of straight line CH can be written as: y = (-66/61)x + 24/61

c) Equation of the median AM: Let's find the coordinates of the vector AM: AM = M - A = ((-2 - 3)/2; (-6 + 0)/2) = (-5/2; -3) Since the median AM is a line passing through vertex A and the middle of side BC, then its direction vector is equal to half of vector BC: BC = C - B = (4 - (-3); 0 - 5) = (7; -5) Median AM passes through point M((-2 + 4)/2; (-6 + 0)/2) = (1; -3) and has a direction vector AM, so its equation can be written as: y + 3 = (-3/ -5)(x - 1)

d) Point N of intersection of the median AM and height CH: In order to find the point of intersection of the median AM and height CH, you need to solve the system of equations: y = (-66/61)x + 24/61 y + 3 = (-3/ -5)(x - 1) Having solved it, we get point N(23/61; -144/61).

e) Equation of a line passing through vertex C and parallel to side AB: Since the line passes through point C and is parallel to side AB, its direction vector coincides with vector AB: y - 0 = 11/1(x - 4)

f) Distance from point C to line AB: First, let’s find the equation of line AB: y + 6 = 11/1(x + 2) Then the distance from point C to line AB can be found using the formula: d = |(y2 - y1 )x0 - (x2 - x1)y0 + x2y1 - y2x1| / √((y2 - y1)^2 + (x2 - x1)^2) where (x0, y0) are the coordinates of point C, (x1, y1) and (x2, y2) are the coordinates of any two points lying on the line AB. Let's choose points A and B: d = |(5 - (-6))3 - ((-3) - (-

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IDZ Ryabushko 3.2 Option 24 is a task for solving geometric problems on finding the equations of the sides, heights, medians of a triangle, points of intersection of medians and heights, as well as finding the equation of a line passing through the vertex of a triangle and parallel to one of its sides.

The task gives the vertices of the triangle ∆ABC: ​​A(–2,–6); B(–3;5); C(4;0). It is necessary to find the equation of the side AB, the equation of the height CH, the equation of the median AM, the point of intersection of the median AM and the height CH, the equation of the line passing through the vertex C and parallel to the side AB, as well as the distance from the point C to the line AB.

To solve the task, you must use knowledge from geometry and algebra, as well as the ability to work with the coordinates of points on the coordinate plane.

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