The coordinates of the midpoint of the line AB are (1, 2). The length of the line AB is 10 units.
(a) If the gradient of AB is 0, find the coordinates of A and B
(b) If the gradient of AB is 34 , find the coordinates of A and B.
I need help with the first question
(a) If the gradient of AB is 0, it means the line is horizontal.
To find the coordinates of A and B, we can use the midpoint formula.
Let the coordinates of A be (x1, y1) and the coordinates of B be (x2, y2).
According to the midpoint formula, the x-coordinate of the midpoint is the average of the x-coordinates of A and B, and the y-coordinate of the midpoint is the average of the y-coordinates of A and B.
Given that the midpoint is (1, 2), we can set up the following equations:
(x1 + x2)/2 = 1 (1)
(y1 + y2)/2 = 2 (2)
Since the line is horizontal and the x-coordinates of A and B should be different, we can deduce that y1 = y2.
Substituting this into equation (2), we get:
(y1 + y1)/2 = 2
Simplifying, we get:
2y1 = 4
y1 = 2
Substituting this value of y1 into equation (1), we get:
(x1 + x2)/2 = 1
Since the x-coordinates of A and B should be different and the midpoint is (1, 2), we can deduce that x1 < x2.
Let's assume x1 = 0, then we can solve for x2:
(0 + x2)/2 = 1
x2/2 = 1
x2 = 2
So the coordinates of A are (0, 2) and the coordinates of B are (2, 2).
(b) If the gradient of AB is 34, it means the line is inclined and has a positive slope.
To find the coordinates of A and B, we can use the gradient formula.
Given that the length of AB is 10 units, we can use the distance formula to set up another equation.
Let the coordinates of A be (x1, y1) and the coordinates of B be (x2, y2).
According to the gradient formula, the gradient (m) is given by:
m = (y2 - y1)/(x2 - x1)
Given that the gradient is 34, we can express this equation as:
34 = (y2 - y1)/(x2 - x1) (3)
According to the distance formula, the length of AB is given by:
√[(x2 - x1)^2 + (y2 - y1)^2] = 10 (4)
Since we have two unknowns (x1, y1) and (x2, y2), we need two equations to solve for them.
From equation (3), we can rearrange it as:
(y2 - y1) = 34(x2 - x1)
Simplifying, we get:
y2 - y1 = 34x2 - 34x1 (5)
Let's rewrite equation (4) as:
[(x2 - x1)^2 + (y2 - y1)^2] = 100
Substituting equation (5) into equation (4), we get:
[(x2 - x1)^2 + (34x2 - 34x1)^2] = 100
Simplifying, we get:
x2^2 - 2x1x2 + x1^2 + 1156x2^2 - 2356x1x2 + 1156x1^2 = 100
Combine like terms:
1157x2^2 - 2358x1x2 + 1157x1^2 = 100
Since we have two unknowns, it is challenging to solve this equation directly. Thus, we need additional information or constraints to solve for x1, y1, x2, and y2.
To find the coordinates of points A and B, we can use the given information about the midpoint and length of the line AB.
(a) If the gradient of AB is 0:
In this case, the line AB is horizontal, meaning the y-coordinates of A and B are the same.
Given that the midpoint of AB is (1, 2), the y-coordinate of both A and B is 2.
Now, we need to find the x-coordinates of A and B. Since the line AB is horizontal, the x-coordinate of the midpoint is the average of the x-coordinates of A and B. So, we have:
(x-coordinate of A + x-coordinate of B) / 2 = 1
(x-coordinate of A + x-coordinate of B) = 2
Since the y-coordinate of A and B is the same (2), we have:
A = (x-coordinate of A, 2)
B = (x-coordinate of B, 2)
So, the coordinates of A and B are (x-coordinate of A, 2) and (x-coordinate of B, 2) respectively.
(b) If the gradient of AB is 34:
In this case, we have the gradient (slope) of the line AB, which tells us the rate of change of y with respect to x.
The gradient is m = (change in y) / (change in x).
Given that the midpoint of AB is (1, 2), we know the y-coordinate of both A and B is 2.
Using the formula for the gradient, we have:
m = (change in y) / (change in x)
34 = (2 - y-coordinate of A) / (x-coordinate of B - x-coordinate of A)
Since the y-coordinate of both A and B is 2, we have:
34 = (2 - 2) / (x-coordinate of B - x-coordinate of A)
Since the numerator is 0, we have:
34 = 0 / (x-coordinate of B - x-coordinate of A)
This equation is true for any value of (x-coordinate of B - x-coordinate of A), as long as the numerator is 0.
In this case, the length of the line AB is given as 10 units. Therefore, the x-coordinate of B - x-coordinate of A is 10.
Substituting this value into the equation, we have:
34 = 0 / 10
This equation is true for any value of the denominator, which means that the gradient 34 is not possible for a line with a length of 10 units crossing the midpoint (1, 2). Therefore, there are no specific coordinates for A and B in this scenario.
The first part I already answered in your previous post.
If the gradiant is 34, the diameter is quite steep.
Lets find its equation:
y = 34x + b
but (1,2) lies on it, so
2 = 34(1) + b
b = -32
y = 34b - 32
the equation of the circle I described in the earlier post is
(x-1)^2 + (y-2)^2 = 25
sub in the equation of the diameter to get your quadratic, solve the quadratic.
You will get the two intersection points A and B