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Given two points A(-3,4) and B(2,5) find the coordinates of one point P on the line and passing por A and B. Look that the point P is two times more distant from A than B.

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Why don't you use the Two point form to obtain the equation for the line? – Pragabhava Oct 12 '12 at 0:20
Also, if you change the title to something more descriptive, like "problem involving points on a line" or something like that, your question will get a lot of attention. – Pragabhava Oct 12 '12 at 0:24
Using the two point form i got x-5y+23=0 – Vinicius L. Beserra Oct 12 '12 at 3:45
I see you've solved it. Congrats! – Pragabhava Oct 12 '12 at 14:13
Thanks Pragabhava for the help. – Vinicius L. Beserra Oct 12 '12 at 16:34

1) Find the equation of the line connecting the two points. P must fulfill this equation since it is on the line.

2) Let's assume that d(x,y) is the distance between x and y, then your second constraint "the distance of P from A is twice the distance from P to B" can be written as: d(A,P)=2*d(B,P). You should know how to calculate d(x,y), so just insert into both sides of the equation.

3) Using the equations from (1) and (2), you have two equations with two variables, solve this simple system of equations and you will find P.

Good luck!

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So I can use this too. AP=2PB, so, P-A=2[B-P]; (x,y)-(3,4)=2[(2,5)-(x,y); So, i got (x+3, y-4)=(4-2x,10-2y); x+3=4-2x => x+2x=4-3 => 3x=1=> x=1/3 – Vinicius L. Beserra Oct 12 '12 at 4:17
And I have found y= 14/3. – Vinicius L. Beserra Oct 12 '12 at 4:18
@ViniciusL.Beserra good job! – Bitwise Oct 12 '12 at 13:51

Hint: Since your point P lies on $AB$, this just finding an inner point formula: $$AP:PB=2:1$$

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thanks everyone, the points are (1/3,14/3). – Vinicius L. Beserra Oct 12 '12 at 4:07

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