Find the other two vertices of that square whose opposite vertices are (1,1) and (1,8).
First find the length of the diagonal from (1,1) to (1,8)
you can do this in your head , and it is 7, and half of it is 3.5
so the centre of your square is (1, 4.5)
notice since the first diagonal is vertical, the second one must be horizontal.
The diagonals of the square bisect each other, and the diagonals are equal, so just add 3.5 to the 1 to get (4.5, 4.5) as one of the other points, then subtract 3.5 from 1 to get
(-2.5, 4.5) for the last point.
To find the other two vertices of the square, we can start by finding the distance between the given opposite vertices. Let's call those points A(1, 1) and B(1, 8).
The distance between two points in a coordinate plane can be calculated using the distance formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Using the distance formula, let's calculate the distance between points A and B:
Distance = √((1 - 1)^2 + (8 - 1)^2)
= √(0^2 + 7^2)
= √(0 + 49)
= √49
= 7
Since we know that all sides of a square are equal, the length of each side of the square is 7 units. From the given opposite vertices, we already have two points: A(1, 1) and B(1, 8). Let's use this information to find the other two vertices.
For a square, if we go 7 units horizontally from a point and then 7 units vertically, we will reach the next vertex. Similarly, if we go 7 units vertically and then 7 units horizontally, we will reach the opposite vertex.
Starting from point A(1, 1), let's find the other two vertices:
Vertex 1: Going 7 units horizontally from A(1, 1):
x-coordinate of Vertex 1 = 1 + 7 = 8
y-coordinate stays the same = 1
So, Vertex 1 = (8, 1)
Vertex 2: Going 7 units vertically from A(1, 1):
x-coordinate stays the same = 1
y-coordinate of Vertex 2 = 1 + 7 = 8
So, Vertex 2 = (1, 8)
Therefore, the other two vertices of the square with opposite vertices (1, 1) and (1, 8) are:
Vertex 1 = (8, 1)
Vertex 2 = (1, 8)