IDZ Ryabushko 3.2 Option 4

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No. 1 Given vertices ∆ABC: A(1;0); B(–1;4); C(9;5). Find: a) Equation of side AB; b) Equation of CH height; 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) Distance from point C to straight line AB.

a) Equation of the side AB: Find the coordinate vector AB: AB = B - A = (-1 - 1; 4 - 0) = (-2; 4) Then the equation of the line AB has the form: y - yA = (yB - yA) / (xB - xA) * (x - xA) y - 0 = 4 / (-1 - 1) * (x - 1) y = -2x + 2

b) Equation of the height CH: Let us find the coordinates of the vectors AB and AC: AB = (-2; 4) AC = (8; 5) Since the height is drawn from the vertex C, it is perpendicular to the side AB and, therefore, parallel to the vector AB. Then the CH height equation has the form: y - yC = (xB - xC) / (yB - yC) * (x - xC) y - 5 = (-1 - 9) / (4 - 5) * (x - 9) y = -8x + 77

c) Equation of the median AM: We find the coordinates of the vectors AB and AC: AB = (-2; 4) AC = (8; 5) + xC) / 2; (yB + yC) / 2) = ((-1 + 9) / 2; (4 + 5) / 2) = (4; 4.5) The equation of the median AM has the form: y - yA = ( yM - yA) / (xM - xA) * (x - xA) y - 0 = (4.5 - 0) / (4 - 1) * (x - 1) y = 1.5x + 1.5

d) Point N of intersection of the median AM and the height CH: The median AM and the height CH intersect at point N. Let's find the coordinates of point N by solving the system of equations for the height CH and the median AM: y = -8x + 77 y = 1.5x + 1.5 -8x + 77 = 1.5x + 1.5 x = 8 y = -61 Point N has coordinates (8; -61).

e) Equation of a line passing through vertex C and parallel to side AB: The equation of a line passing through vertex C and parallel to side AB is: y - yC = (yB - yA) / (xB - xA) * (x - xC) y - 5 = 4 / (-1 - 1) * (x - 9) y = -2x + 23

f) Distance from point C to straight AB: The distance from point C to straight AB is equal to the distance from point C to its projection onto straight AB. Let's find the coordinates of the projection of point C onto line AB. To do this, we find the equation of a line perpendicular to AB and passing through C: Find the coordinates of the vector AB: AB = B - A = (-1 - 1; 4 - 0) = (-2; 4) Then the coordinates of the vector perpendicular to AB are equal to ( 4; 2). The equation of a line passing through C and perpendicular to AB is: y - yC = (2 / 4) * (x - xC) y - 5 = (1 / 2) * (x - 9) y = (1 / 2) * x - 1.5 Let us find the point of intersection of this line with line AB, which is the projection of point C onto line AB. Let's solve the system of equations: y = (1 / 2) * x - 1.5 y = -2x + 2 (1 / 2) * x - 1.5 = -2x + 2 x = 2 y = -1 The intersection point has coordinates (2; - 1). The distance from point C to line AB is equal to the distance between point C and its projection onto line AB: d = √((xC - x)^2 + (yC - y)^2) = √((9 - 2)^2 + (5 - (-1))^2) = √(49 + 36) = √85.

No. 2 Find the equation of a line that cuts off a segment equal to 2 on the ordinate axis and runs parallel to the line 2y - x = 3.

Let us transfer the equation of the straight line 2y - x = 3 to a general form: x + 2y = 3 Then the coefficients of the equation of the straight line parallel to the given one and passing through the point (0; 2) are equal to (1; 2). The equation of this line has the form: y - y0 = k * (x - x0) y - 2 = 2 * (x - 0) y = 2x + 2 To find the point of intersection of this line with the ordinate axis, substitute x = 0: y = 2 * 0 + 2 = 2 The intersection point has coordinates (0; 2). Since the segment cut off on the ordinate axis is equal to 2, the second point of the segment has coordinates (0; 4). Then the equation of the desired line has the form: y - y0 = k * (x - x0) y - 2 = 2 * (x - 0) y = 2x + 2.

IDZ Ryabushko 3.2 Option 4

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IDZ Ryabushko 3.2 Option 4 is a set of tasks in mathematics, which includes the following tasks:

No. 2 Find the equation of a line that cuts off a segment equal to 2 on the ordinate axis and runs parallel to the line 2y - x = 3.

To solve problems from Ryabushko IDZ 3.2 Option 4, you must apply knowledge in the field of geometry and algebra.

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