A(-3, 1), B(4, 5)
coordinates of P on AB such that AP/PB = 5/2
AP/PB = 5/2 . . . . . desired result
2AP = 5PB . . . . . . multiply by 2PB
2(P-A) = 5(B-P) . . . meaning of the above
2P -2A = 5B -5P . . eliminate parentheses
7P = 2A +5B . . . . . collect P terms
P = (2A +5B)/7 . . . .divide by the coefficient of P
P = (2(-3, 1) +5(4, 5))/7 . . . . substitute the given points
P = (-6+20, 2+25)/7 . . . . . . simplify
P = (2, 3 6/7)
Firstly, let's look at this ratio 5:2. Despite what you may think, 5:2 is not equal to 5/2, but rather it is equal to "5 equal parts to 2 equal parts", which make up a total of 7 equal parts. In short, they want us to place point P at 5/7 of line AB.
Firstly, how far is -3 to 4 (the x coordinates)? That would be 7 units. Multiply 5/7 by 7:
[tex] \frac{5}{7}\times \frac{7}{1}=\frac{35}{7}=5 [/tex]
Next, how far is 1 from 5 (the y-coordinates)? That would be 4 units. Multiply 5/7 by 4:
[tex] \frac{5}{7}\times \frac{4}{1}=\frac{20}{7}=2\frac{6}{7} [/tex]
Next, since from -3 to 4 you are increasing, add 5 to -3, which gets you a sum of 2. 2 is the x-coordinate of point P.
Next, since from 1 to 5 you are increasing, add 2 6/7 to 1, which gets you a sum of 3 6/7. 3 6/7 is the y coordinate of point P.
Putting it together, point P is at [tex] (2,3\frac{6}{7}) [/tex]
