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# ps-coordinate geometry

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VP
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ps-coordinate geometry [#permalink]  31 May 2005, 01:25
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What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
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Director
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[#permalink]  31 May 2005, 03:47
x^2 + y^2 = 1 is a circle with centre at (0,0) and y = 3/4*x - 3 is a line which doesnt pass through origin . So make x = 0 we get y = -3 which is 2 units away from the circle which intercepts the X -axis at -1 . So the distance is 2.

Also check with y=0 we get x=4 which is 3units away from the circle.

So E is the answer
Current Student
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[#permalink]  31 May 2005, 05:33
E it is...

just draw out the 2 and you will see....
VP
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[#permalink]  31 May 2005, 05:56
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4
VP
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[#permalink]  31 May 2005, 06:11
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4

baner, it seems to me that you know for almost every problem the right formula. do you have a list of formulas ?
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VP
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[#permalink]  31 May 2005, 06:46
christoph wrote:
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4

baner, it seems to me that you know for almost every problem the right formula. do you have a list of formulas ?

Altho I don't have a list, but I think in GMAT coordinate geometry, u need to know only three formulas.

1. Finding the slope of a line with two points (x1,y1) (x2,y2) =
(y2-y1)/(x2-x1)

2. Distance of a point (x1,y1) from a line (ax+by+c = 0) :
(ax1+by1+c)/sqrt(a^2+b^2)

3. Distance between two points = sqrt{(x1-x2)^2 + (y1-y2)^2}
Intern
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[#permalink]  31 May 2005, 11:12
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4

bravo, baner... really good approach. Agree on A.
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Re: ps-coordinate geometry [#permalink]  31 May 2005, 13:45
christoph wrote:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0

Draw the circle and the triangle on paper.

The triangle is a 3-4-5 triangle. The height from right angle to hypotenuse = 3*4/5
=2.4

Final Ans is 2.4 -radius = 1.4

This formula might be helpful for right angle triangles.
Height = side1*side2 / hypotenuse
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Director
Joined: 18 Feb 2005
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[#permalink]  31 May 2005, 19:17
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4

Good one Baner....Forgot the concept...Thanks for reminding
[#permalink] 31 May 2005, 19:17
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# ps-coordinate geometry

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