Question

Дано координати трьох вершин прямокутника АВСD: А(1;6); В(5;6); С(5;-2). Знайдіть координати точки перетину його діагоналей. У відповідь впишіть суму абсциси і ординати даної точки. ПЛІС ПОМОГІТЕ

Answer 1

To find the coordinates of the intersection point of the diagonals of the rectangle ABCD, we can use the properties of rectangles.

Properties of Rectangles:

1. The diagonals of a rectangle bisect each other. 2. The diagonals of a rectangle are equal in length.

Using these properties, we can find the coordinates of the intersection point as follows:

1. Find the midpoint of the diagonal AC: - The coordinates of A are (1, 6) and the coordinates of C are (5, -2). - To find the midpoint, we can use the formula: - Midpoint_x = (x1 + x2) / 2 - Midpoint_y = (y1 + y2) / 2 - Substituting the values, we get: - Midpoint_x = (1 + 5) / 2 = 3 - Midpoint_y = (6 + -2) / 2 = 2

2. Find the midpoint of the diagonal BD: - The coordinates of B are (5, 6) and the coordinates of D are (5, -2). - Using the same formula as above, we get: - Midpoint_x = (5 + 5) / 2 = 5 - Midpoint_y = (6 + -2) / 2 = 2

3. The intersection point of the diagonals is the same as the midpoint of both diagonals. - Therefore, the coordinates of the intersection point are (3, 2) and (5, 2).

To find the sum of the abscissa and ordinate of the intersection point, we add the x-coordinate (abscissa) and y-coordinate (ordinate) together: - Sum = 3 + 5 + 2 + 2 = 12

Therefore, the sum of the abscissa and ordinate of the intersection point is 12.

Please let me know if there's anything else I can help you with!