Question

Точка А(хА; yА) – вершина квадрата АВСD, а його діагональ ВD знаходиться на прямій ах+bу+с = 0. Необхідно знайти рівняння сторін АВ, ВС, СD и АD та координати вершин В, С, и D. Зробіть креслення.

Answer 1

Ответ:

Объяснение:

на фото

Answer 2

Ответ: прошу две минуты для загрузки ответа. т.к. сбой программы.

Объяснение:

Answer 3

#### Problem Analysis We are given that point A (xA, yA) is the vertex of a square ABCD, and its diagonal BD lies on the line ax + by + c = 0. We need to find the equations of the sides AB, BC, CD, and AD, as well as the coordinates of the vertices B, C, and D. We also need to provide a diagram. #### Solution To find the equations of the sides AB, BC, CD, and AD, we need to determine the slope of the line BD. Since BD is a diagonal of a square, it is perpendicular to the sides AB and BC. Therefore, the slopes of AB and BC are negative reciprocals of the slope of BD. Let's start by finding the slope of BD. We can use the fact that BD lies on the line ax + by + c = 0. The slope-intercept form of a line is y = mx + b, where m is the slope. To convert the given equation into slope-intercept form, we solve for y: ax + by + c = 0 by = -ax - c y = (-a/b)x - c/b Comparing this equation with y = mx + b, we can see that the slope of BD is -a/b. Now, let's find the slopes of AB and BC. Since AB and BC are perpendicular to BD, their slopes are negative reciprocals of the slope of BD. Therefore, the slope of AB is b/a, and the slope of BC is -b/a. We know that point A (xA, yA) is the vertex of the square. To find the coordinates of the other vertices, we can use the slopes of AB and BC. The equation of AB can be written as y - yA = (b/a)(x - xA). Simplifying this equation, we get y = (b/a)x + (yA - b/a * xA). This gives us the equation of AB. Similarly, the equation of BC can be written as y - yB = (-b/a)(x - xB), where (xB, yB) are the coordinates of point B. Simplifying this equation, we get y = (-b/a)x + (yB + b/a * xB). This gives us the equation of BC. Since AB and BC are perpendicular, the product of their slopes is -1. Therefore, (b/a) * (-b/a) = -1. Solving this equation, we get b^2 = a^2. Now, let's find the coordinates of the other vertices. Since AB and BC are perpendicular, the distance between A and B is equal to the distance between B and C. Let's denote this distance as d. Using the distance formula, we have: d^2 = (xB - xA)^2 + (yB - yA)^2 d^2 = (xC - xB)^2 + (yC - yB)^2 Since AB and BC have the same length, we can equate these two equations: (xB - xA)^2 + (yB - yA)^2 = (xC - xB)^2 + (yC - yB)^2 Expanding and simplifying this equation, we get: (xB^2 - 2xBxA + xA^2) + (yB^2 - 2yByA + yA^2) = (xC^2 - 2xCxB + xB^2) + (yC^2 - 2yCyB + yB^2) Canceling out the common terms, we get: -2xBxA + xA^2 - 2yByA + yA^2 = -2xCxB + xB^2 - 2yCyB + yB^2 Since the coordinates of A are given as (xA, yA), we can substitute these values into the equation. Similarly, we can substitute the coordinates of B as (xB, yB). This will give us an equation with the coordinates of C. Finally, we can find the coordinates of D by using the fact that D is the midpoint of AC. The midpoint formula is: xD = (xA + xC) / 2 yD = (yA + yC) / 2 Let's summarize the steps to find the equations of the sides AB, BC, CD, and AD, as well as the coordinates of the vertices B, C, and D: 1. Find the slope of BD: -a/b. 2. Find the slopes of AB and BC: b/a and -b/a, respectively. 3. Use the slope-intercept form to find the equations of AB and BC. 4. Use the fact that AB and BC are perpendicular to find b^2 = a^2. 5. Use the distance formula to equate the distances between A and B, and B and C. 6. Simplify the equation to find the coordinates of C. 7. Use the midpoint formula to find the coordinates of D. Let's now find the equations of the sides AB, BC, CD, and AD, as well as the coordinates of the vertices B, C, and D. #### Equations of the Sides The equation of side AB is **y = (b/a)x + (yA - b/a * xA)**. The equation of side BC is **y = (-b/a)x + (yB + b/a * xB)**. The equation of side CD is **y = (b/a)x + (yC - b/a * xC)**. The equation of side AD is **y = (-b/a)x + (yD + b/a * xD)**. #### Coordinates of the Vertices The coordinates of vertex B are **(xB, yB)**. The coordinates of vertex C are **(xC, yC)**. The coordinates of vertex D are **(xD, yD)**. #### Diagram Here is a diagram to help visualize the square ABCD: ``` C _______ B | | | | |_______| A D ``` Please note that the diagram is not to scale and is only for illustrative purposes. I hope this helps! Let me know if you have any further questions.