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Geometry

by goyalsau » Wed Nov 17, 2010 10:38 am
HI! guys,
[spoiler]
I don't Understand How the answer is 2:3, in explanation it is given that points are on the same line of slope 3:2, so the points will be on th ratio of 2:3, but origin is also on the line, which has coordinate 0,0
How it that possible then,
[/spoiler]
Please Guys Help,,
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by Rahul@gurome » Wed Nov 17, 2010 10:58 am
Equation of the line AB is 3x = 2y => x/y = 2/3
Therefore, for any point (x, y) on the line AB, x/y = 2/3
As (p, q) is on AB, p/q = 2/3
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by goyalsau » Fri Nov 19, 2010 7:39 am
Rahul@gurome wrote:Equation of the line AB is 3x = 2y => x/y = 2/3
Therefore, for any point (x, y) on the line AB, x/y = 2/3
As (p, q) is on AB, p/q = 2/3
I am not able to understand one thing, this line will also pass from the origin ( 0 , 0 )

If p , q are at origin then........
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by Rahul@gurome » Fri Nov 19, 2010 7:50 am
goyalsau wrote:
Rahul@gurome wrote:Equation of the line AB is 3x = 2y => x/y = 2/3
Therefore, for any point (x, y) on the line AB, x/y = 2/3
As (p, q) is on AB, p/q = 2/3
I am not able to understand one thing, this line will also pass from the origin ( 0 , 0 )

If p , q are at origin then........
As the point shown in the figure suggests that point B is not origin, the answer remains same as I discussed. But I should have also mentioned that for any point (x, y) on the line AB (except (0, 0)), x/y = 2/3.
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by MAAJ » Fri Nov 19, 2010 11:08 am
My noob way to solve this:

m = (y2-y1)/(x2-x1)
m = (3-0)/(2-0)
m = 3/2

so... if x=0 and y=0

y = 3/2x + b
0 = 3/2(0) + b
0 = 0 + b
0 = b

so... if "b" is always 0

y = 3/2x

so... if x=p and y=q

q=3/2p

then p/q:

p/q
p/(3/2p)
2/3
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