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Distance in Cartesian Plane

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Re: Distance in Cartesian Plane

by Vemuri » Tue May 05, 2009 8:47 am
The question stem provides us with 2 points (1,2) & (0,5/2). Using the slope formula ==> ((5/2)-2)/(0-1) ==> -1/2

IMO B
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Re: Distance in Cartesian Plane

by dtweah » Tue May 05, 2009 3:07 pm
Vemuri wrote:The question stem provides us with 2 points (1,2) & (0,5/2). Using the slope formula ==> ((5/2)-2)/(0-1) ==> -1/2

IMO B
Give it another shot vemuri.
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Re: Distance in Cartesian Plane

by Vemuri » Tue May 05, 2009 7:40 pm
dtweah wrote:
Vemuri wrote:The question stem provides us with 2 points (1,2) & (0,5/2). Using the slope formula ==> ((5/2)-2)/(0-1) ==> -1/2

IMO B
Give it another shot vemuri.
Ok, the slope should be determined by using a point on the line & the y intercept. But, how are we to determine the point? I think I am stuck here.

I am not sure I understand how to use the info that (1,2) is (3(2)^0.5)/4 away from the line on the XY plane.
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by dumb.doofus » Tue May 05, 2009 8:31 pm
Phew!!! Man.. such problems come on GMAT??? I am not sure..
anyways, the solution is in the picture below... login to see..
I think the answer should be 1 i.e. option D


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by Vemuri » Tue May 05, 2009 8:46 pm
Damn !!! Unless there is a easier way to solve this problem, I bet this is not a GMAT question.
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Re: Distance in Cartesian Plane

by dtweah » Wed May 06, 2009 2:03 am
dtweah wrote:The distance between the point (1,2) and a line in the X-Y plane is (3(2)^0.5)/4. If the y intercept of the line is 5/2 find the slope of the line?

A. -1

B -1/2

C. 1/2

D. 1

E. 2
Altough you could use the distance to a line formula and solve you can also solve without it. Put eqn into y=mx+b y=mx+5/2 or 2y=mx+5. Plug point in (1,2) you can find slope m.
2(2)=m(1)+5
m=1
Choose D.

Using Distance Formula:
Rationalize distance. 3/(2)2^.5
Denominator equal square root of 8=4+4 or 2^2+2^2
So A=+-2 and B=+-2



|Ax +B +C|=|3| From y intercept info you know C=5 and A =2
|2(1)+B(2)+5|=3
|7+2B|=3

The def of absolute value has:
7+2B= 3, for B=-2, this solves.

7+2B=-3, since distance is never negative we ignore this part

The eqn is 2x-2y=5 from which the slope can be determined to be 1.
Choose D.

The first method solved under 1 min. The second about midway over 1.
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by avenus » Wed May 06, 2009 3:27 am
I believe there's stg wrong with your reasoning:
Altough you could use the distance to a line formula and solve you can also solve without it. Put eqn into y=mx+b y=mx+5/2 or 2y=mx+5. Plug point in (1,2) you can find slope m.
2(2)=m(1)+5
m=1
you're assuming the line will cross (1,2). Why? Distance would then be zero....plus result would be m=-1

The eqn is 2x-2y=5 from which the slope can be determined to be 1.
Choose D.
This eq yields a y-intercept of -5/2, which is different from the 5/2 stated in the problem
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by dtweah » Wed May 06, 2009 4:47 am
avenus wrote:I believe there's stg wrong with your reasoning:
Altough you could use the distance to a line formula and solve you can also solve without it. Put eqn into y=mx+b y=mx+5/2 or 2y=mx+5. Plug point in (1,2) you can find slope m.
2(2)=m(1)+5
m=1
you're assuming the line will cross (1,2). Why? Distance would then be zero....plus result would be m=-1

The eqn is 2x-2y=5 from which the slope can be determined to be 1.
Choose D.
This eq yields a y-intercept of -5/2, which is different from the 5/2 stated in the problem
The latter is a transpositional error. Ax +BY +C= 0 so 2x-2y+5=0. Which has slope 1. The former is a serious error and you are right. (1,2) does not pass through 2x-2y+5=0(1,7/2) does. So the only valid solution is method 2. I spotted the error in V's response when I told him to try again but committed the same error! I think where the error comes from is that in using the formula for finding the distance between a point and a line, we use the point in the equation but this does not mean the point is on the line! Good observation. Thanks.
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