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,,
Geometry
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- goyalsau
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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
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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- goyalsau
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I am not able to understand one thing, this line will also pass from the origin ( 0 , 0 )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
If p , q are at origin then........
Saurabh Goyal
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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.goyalsau wrote:I am not able to understand one thing, this line will also pass from the origin ( 0 , 0 )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
If p , q are at origin then........
Rahul Lakhani
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- MAAJ
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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
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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