If the adjacent vertices of a square ABCD are A ( -2 , 0 ) , find
- The equation of the two diagonals AC and BD
- The point of intersection O of the two diagonals AC and BD
- The equation of the altitude from O to the side CD
- The centroid of the triangle OCD by finding any two of its medians
- The perimeter of the square ABCD
To find the answers to these questions, let's go through each one step by step.
1. Equation of the two diagonals AC and BD:
The equation of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by:
(y - y₁) = (y₂ - y₁) / (x₂ - x₁) * (x - x₁)
For the diagonal AC, we have points A(-2, 0) and C(2, 0):
(y - 0) = (0 - 0) / (2 - (-2)) * (x - (-2))
y = 0
Similarly, for the diagonal BD, we have points B(0, 2) and D(0, -2):
(y - 2) = (-2 - 2) / (0 - 0) * (x - 0)
x = 0
So, the equations of the diagonals AC and BD are y = 0 and x = 0, respectively.
2. Point of intersection O of the diagonals AC and BD:
Since the equations of the diagonals AC and BD are y = 0 and x = 0, respectively, the point of intersection O is (0, 0).
3. Equation of the altitude from O to the side CD:
Since O(0, 0) is the midpoint of CD, the slope of the altitude from O to the side CD will be the negative reciprocal of the slope of CD.
To find the slope of CD:
m = (y₂ - y₁) / (x₂ - x₁) = (0 - (-2)) / (0 - 2) = 2/2 = 1
Now, the slope of the altitude will be -1/1 = -1.
Since O(0, 0) lies on the altitude, the equation becomes:
(y - 0) = -1*(x - 0)
Simplifying the equation:
y = -x
So, the equation of the altitude from O to the side CD is y = -x.
4. Centroid of the triangle OCD using any two medians:
The medians of a triangle intersect at its centroid. Let's take O(0, 0) as one of the medians' endpoints.
The other endpoint, M, can be found by taking the midpoint of OC.
Coordinates of O(0, 0) and C(2, 0):
Midpoint M = ((0 + 2)/2, (0 + 0)/2)
M = (1, 0)
The coordinates of the centroid G will be the average of the coordinates of O and M:
G = ((0 + 1)/2, (0 + 0)/2)
G = (0.5, 0)
So, the centroid of the triangle OCD is G(0.5, 0).
5. Perimeter of the square ABCD:
To find the perimeter of a square, we need to calculate the sum of the lengths of all four sides.
Considering the adjacent vertices of the square ABCD as A(-2, 0), B(0, 2), C(2, 0), and D(0, -2), the lengths of the sides are as follows:
AB = √((x₂ - x₁)² + (y₂ - y₁)²) = √((0 - (-2))² + (2 - 0)²) = √(2² + 2²) = √8 = 2√2
BC = √((x₂ - x₁)² + (y₂ - y₁)²) = √((2 - 0)² + (0 - 2)²) = √(2² + 2²) = √8 = 2√2
CD = √((x₂ - x₁)² + (y₂ - y₁)²) = √((0 - 2)² + (-2 - 0)²) = √((-2)² + (-2)²) = √(4 + 4) = √8 = 2√2
DA = √((x₂ - x₁)² + (y₂ - y₁)²) = √((-2 - 0)² + (0 - (-2))²) = √((-2)² + 2²) = √(4 + 4) = √8 = 2√2
Summing up these lengths:
Perimeter of ABCD = AB + BC + CD + DA = 2√2 + 2√2 + 2√2 + 2√2 = 8√2
So, the perimeter of the square ABCD is 8√2 (units).