Problem
Points A,B and C are collinear , and AB: BC=1:4 A is located at (-5,-3), B is located at (-2,0) and C is located at (x,y), on the directed line segment AC. what are the values of x and y?
Solution
For this case we know the coordinates of A and B given by:
A=(-5,-3)
B=(-2,0)
We also know that C is located at:
C=(x,y)
We also know the following ratio:
AB/BC= 1/4
We can find the distance between A and B with this formula:
[tex]AB=\sqrt[]{(-2+5)^2+(0+3)^2}=\sqrt[]{18}=3\sqrt[]{2}[/tex]
We can find the distance between B and C on this way:
[tex]BC=\sqrt[]{(-2-x)^2+(y-0)^2}[/tex]
We knwo that AB is 4 times bigger than BC so then if we check the possible options for x and y we can selct the correct one:
(10,12)
And the reason is because if x=10 and y= 12 we have:
[tex]BC=\sqrt[]{(-2-10)^2+(12)^2}=12\sqrt[]{2}[/tex]
And when we divide AB/BC we got:
AB/BC= 3 sqrt(2)/ 12 sqrt(2) = 3/12 = 1/4
