The coordinates of the midpoint of the line AB are (1, 2). The length of the line AB is 10 units.
If the gradient of AB is 0, find the coordinates of A and B
The line AB would be the horizontal diameter of a circle with centre (1,2) and radius 5
So just go 5 units to the left and to the right of (1,2)
A could be (-4,2) and B could be (6,2)
or B(-4,2) and A(6,2)
notice AB = √(6+4)^2 + 0^2) = 10
To find the coordinates of points A and B, we can use the midpoint formula and the length formula.
Let's denote the coordinates of A as (x1, y1) and the coordinates of B as (x2, y2).
According to the information given, the midpoint of line AB is (1, 2). The midpoint formula is:
Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)
We can plug in the coordinates of the midpoint (1, 2) to find:
(1, 2) = ((x1 + x2) / 2, (y1 + y2) / 2)
Simplifying this equation, we get:
1 = (x1 + x2) / 2 ---- (1) and 2 = (y1 + y2) / 2 ---- (2)
Now, we are told that the gradient (or slope ) of line AB is 0. A line with a gradient of 0 is a horizontal line. A horizontal line means that the y-coordinates of both points on the line are the same. So, from equation (2), we can conclude that:
y1 + y2 = 2 * 2 = 4 ---- (3)
Next, we are given that the length of line AB is 10 units. The length formula for the distance between two points (x1, y1) and (x2, y2) is:
Length = sqrt((x2 - x1)^2 + (y2 - y1)^2)
We need to solve this equation for the coordinates of A and B, so we can square both sides:
Length^2 = (x2 - x1)^2 + (y2 - y1)^2
Since the gradient is 0, and the y-coordinates are the same, (y2 - y1) would be 0. Thus, equation (4) becomes:
Length^2 = (x2 - x1)^2
Plugging in the given length of 10 units, we have:
10^2 = (x2 - x1)^2
Simplifying further, we get:
100 = (x2 - x1)^2 ---- (4)
We now have three equations:
1) 1 = (x1 + x2) / 2 ---- (1)
2) 4 = y1 + y2 ---- (3)
3) 100 = (x2 - x1)^2 ---- (4)
We can solve this system of equations to find the coordinates of A and B. Is there any additional information given?