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
Distance in Cartesian Plane
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Vemuri
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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.dtweah wrote:Give it another shot vemuri.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
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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dumb.doofus
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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
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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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.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
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.
I believe there's stg wrong with your reasoning:
you're assuming the line will cross (1,2). Why? Distance would then be zero....plus result would be m=-1Altough 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
This eq yields a y-intercept of -5/2, which is different from the 5/2 stated in the problemThe eqn is 2x-2y=5 from which the slope can be determined to be 1.
Choose D.
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.avenus wrote:I believe there's stg wrong with your reasoning:
you're assuming the line will cross (1,2). Why? Distance would then be zero....plus result would be m=-1Altough 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
This eq yields a y-intercept of -5/2, which is different from the 5/2 stated in the problemThe eqn is 2x-2y=5 from which the slope can be determined to be 1.
Choose D.
