the coordinates of the endpoints A and B of segment AB are 6 and 21, respestively. find the coordinate of point C between A and B such that AC= 2/3(CB)
The separation of the points A and B is 21-6 = 15. C is located 2/5 of the way from A to B, and 3/5 of the AB distance remains from B to C. That pus C 6 units beyond point A, at a coordinate value of 6 + 6 = 12.
AC = 2/3 CB
so AC:CB = 2:3
6----------X------------------21
A..........C...................B
-----2----- ---------3---------
then (X-6)/2 = (21-X)/3
cross multiply
...
...
X=12
To find the coordinate of point C between A and B such that AC is two-thirds of CB, we can use the concept of linear interpolation.
Let's assume the coordinate of point C is represented by x.
According to the given condition, AC = 2/3(CB).
We know that the distance between two points can be calculated using the formula:
Distance = Square root of ((x2 - x1)^2 + (y2 - y1)^2)
In this case, we have the coordinates of A and B: A(6, ?) and B(21, ?). We are only interested in the x-coordinates since we want to find the x-coordinate of point C.
The distance between A and B is therefore:
AB = |21 - 6| = 15
Since AC is two-thirds of CB, we have:
AC = 2/3(CB)
Let's substitute the known values:
AC = 2/3(AB)
AC = 2/3(15)
AC = 10
Now, let's set up an equation to solve for x.
Since C lies on the segment AB, the x-coordinate of C is between 6 and 21.
We can find the ratio of the distances from A to C and C to B. The ratio must be the same as the ratio of the difference in coordinates:
(x - 6)/(21 - x) = 10/15
Cross-multiply to get:
15(x - 6) = 10(21 - x)
Simplify:
15x - 90 = 210 - 10x
Combine like terms:
15x + 10x = 210 + 90
25x = 300
Divide both sides by 25:
x = 300/25
x = 12
Therefore, the coordinate of point C between A and B, such that AC is two-thirds of CB, is (12, ?).