Find the point, M, that divides segment AB into a ratio of 3:2 if A is at (0, 15) and B is at (20, 0).
Let M be (x,y)
Make a sketch , draw two right-angled triangles, with hypotenuse AM = 3 and hypotenuse MB = 2
then use similar triangle ratios.
for the x:
(x-0)/20-x) = 3/2
2x = 60-3x
5x = 60
x = 12
for the y:
(15-y)/(y-0) = 3/2
3y = 30 - 2y
5y = 30
y = 6
M is (12,6)
or:
the x of M is 3/5 of the way to 20 ---> 12
the y of M is 2/5 of the way to 15 ----> 6
To find the point M that divides segment AB into a ratio of 3:2, we can use the section formula.
The section formula states that if we have two points A(x1, y1) and B(x2, y2), and we want to find a point M that divides the line AB into a ratio of p:q, then the coordinates of M can be found using the following formula:
M = ( (px2 + qx1)/(p+q) , (py2 + qy1)/(p+q) )
In this case, A is at (0, 15) and B is at (20, 0). The desired ratio is 3:2, so we have p = 3 and q = 2.
Using the section formula, let's substitute the values into the formula to find the coordinates of M:
M = ( (3*20 + 2*0)/(3+2) , (3*0 + 2*15)/(3+2) )
M = ( (60 + 0)/5 , (0 + 30)/5 )
M = ( 60/5 , 30/5 )
M = ( 12 , 6 )
Therefore, the point M that divides segment AB into a ratio of 3:2 is M(12, 6).