Find the point, M, that divides segment AB into a ratio of 2:3 if A is at (0, 15) and B is at (20, 0).
can you jus say the answer
To find the point, M, that divides segment AB into a ratio of 2:3, we can use the concept of the section formula in coordinate geometry.
The section formula states that if a line segment AB is divided by a point M with ratio m:n, then the coordinates of point M can be found using the following formula:
M(x, y) = ((n * Ax) + (m * Bx)) / (m + n), ((n * Ay) + (m * By)) / (m + n)
Let's substitute the given values into the formula.
Given: A(0, 15), B(20, 0), ratio of 2:3
m = 2 and n = 3
M(x, y) = ((3 * 0) + (2 * 20)) / (2 + 3), ((3 * 15) + (2 * 0)) / (2 + 3)
Simplifying further:
M(x, y) = (40 / 5, 45 / 5)
M(x, y) = (8, 9)
Therefore, the point, M, that divides segment AB into a ratio of 2:3 is (8, 9).
Hint:
the ratio is 2:3, so from A to B, it is 2/(2+3)=2/5.
The coordinates of M are:
M(xa+(xb-xa)*2/5, ya+(yb-ya)*2/5)