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The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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13 Jul 2011, 09:50
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The equation of line n is y = 4/3*x  100. What is the smallest possible distance in the xyplane from the point with coordinates (0, 0) to any point on line n? A. 48 B. 50 C. 60 D. 75 E. 100
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Last edited by Bunuel on 08 Dec 2013, 06:24, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Distance from the origin [#permalink]
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13 Jul 2011, 10:10
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kannn wrote: The equation of line n is y = \(\frac{4}{3}\) x – 100. What is the smallest possible distance in the xyplane from the point with coordinates (0, 0) to any point on line n ?
a) 48 b) 50 c) 60 d) 75 e) 100 The shortest line is a perpendicular, which goes through \((0;0)\) We know that lines (\(y=kx+b\)) are perpendicular when \(k1*k2=1\). So, \((4/3)*k2=1\) k2=3/4 Line has point (0;0) also. So, \(y=(3/4)x\) This line intersects with \(y=\frac{4}{3}x100\). Let's find the intersection point. \(\frac{4}{3}x100=\frac{3}{4}\) \(x=48\) \(y=\frac{4}{3}*48100\) \(y=36\) \((48;36)\) The shortest distance from (0;0) to above mentioned point is \(\sqrt{48*48+36*36}\)=\(60\) (C)



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Re: Distance from the origin [#permalink]
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07 Dec 2013, 09:13
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A shortcut to find the shortest distance between a point  P(m,n) and a line that doesn't contain it:
1. Rearrange equation of the line such that its in the form ax+by+c=0 2. Put the "point in the line", i.e substitute the a and the b in the line's equation with m and n respectively. 3. The distance is = absolute value of [a(m)+b(n)+c]/[(a^2+b^2)^1/2]
Taking the example of the problem at hand, Your point is P(0,0) and line is 4x3y300=0
So the distance between (0,0) and the line is: [4(0)3(0)300]/[(4^2+3^2)^1/2]
= 300/5 = 60 units



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Re: Distance from the origin [#permalink]
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08 Dec 2013, 06:36
Plotting the line on the graph gives us a right triangle with, say base=75 and height =100. Now area on this triangle will be 1/2 *b*h= 1/2*75*100. Now, the shortest distance from the origin will be an altitude (say, x) drawn to the hypotenuse (whose length is 125; we can calculate because we already have the base and height) of the same triangle. Now 1/2*125*x= 1/2*75*100. Hence, x= 60



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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28 Dec 2013, 06:27
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kannn wrote: The equation of line n is y = 4/3*x  100. What is the smallest possible distance in the xyplane from the point with coordinates (0, 0) to any point on line n?
A. 48 B. 50 C. 60 D. 75 E. 100 1. Draw your line first 2. Note the pythagorean triple, that is divide by 25 to get your 3,4,5 3. Given proportions Hypothenuse is 125 4. Now area is either b*h or leg*leg (both are divided by 2 but in this case we can cancel them out) 5. 75(100) = 125 (x). 'x' represents the height which is perpendicular to the base and is the smallest distance from the origin 6. x = 60 Answer is C Bonus method: If y'all want to save some time in there's a formula for distance from point to line that has been mentioned in the posts above but anyways this is how it would work for this problem d = 100 / sqrt (1^2 + (4/3)^2) = 100 / 5/3 = 60 Sanity check, answer still 60 Hope this gives a hand Hope it helps Cheers! J
Last edited by jlgdr on 31 Mar 2014, 07:58, edited 1 time in total.



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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09 Jan 2014, 13:51
distance from (0,0) to any point on given line is Sqrt( (x0)^2 + (y0)^2)
For minimum distance this can be differentiated and equated to zero.
Distance (D)^2 = x^2 + y^2  (1) put y= 4x/3 100
(1) becomes 25x^2/9 800x/3 +10000
differentiate this equation and equate it to zero, x= 48 so y=36.
Distance then is sqrt(48^2 + 36^2) = 60
(C)
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Re: Distance from the origin [#permalink]
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12 Jan 2014, 09:01
reetskaur wrote: Plotting the line on the graph gives us a right triangle with, say base=75 and height =100. Now area on this triangle will be 1/2 *b*h= 1/2*75*100. Now, the shortest distance from the origin will be an altitude (say, x) drawn to the hypotenuse (whose length is 125; we can calculate because we already have the base and height) of the same triangle. Now 1/2*125*x= 1/2*75*100. Hence, x= 60 Can u please explain this in more detail. Thank you



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Re: Distance from the origin [#permalink]
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17 Feb 2014, 09:44
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Vidhi1 wrote: reetskaur wrote: Plotting the line on the graph gives us a right triangle with, say base=75 and height =100. Now area on this triangle will be 1/2 *b*h= 1/2*75*100. Now, the shortest distance from the origin will be an altitude (say, x) drawn to the hypotenuse (whose length is 125; we can calculate because we already have the base and height) of the same triangle. Now 1/2*125*x= 1/2*75*100. Hence, x= 60 Can u please explain this in more detail. Thank you In the equation of the line put X=0 you would get Y=100 and putting Y=0 would give you X=75. So the line intercepts the X and Y axis at 75 and 100 respectively. This creates a right angle triangle. We can use the Pythagoras theorem to calculate the Hypotnuse. Notice that the trip let we have here should be 3(25), 4(25), 5(25) (we have arms as 75 and 100 so the hypotnuse should be 125). Now the area of the traingle formed would be 1/2 * 75 * 100 = 1/2 * 125 * H (This H would be the closest distnace of line from the origin) This gives H=60. Hope this helps!!!



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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17 Feb 2014, 15:00
Is the goal as described as the attachment. I follow the area of the triangle is (100*75)/2, but why are you setting the area of the triangle to (125*H)/2. I dont understand the second formula logic. Please explain.
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Book1.xlsx [10.95 KiB]
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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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17 Feb 2014, 15:02
lalania1 wrote: Is the goal as described as the attachment.
I follow the area of the triangle is (100*75)/2, but why are you setting the area of the triangle to (125*H)/2. I dont understand the second formula logic. Please explain. Yes buddy that's correct. Now remember that the area of a triangle is b*h/2. Therefore one can use both legs or the base and height to find the area. So basically you can equal both areas and solve for the distance you correctly mentioned in the excel. Hope this clarifies Cheers J



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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23 Feb 2014, 07:06
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This can be solve in two steps and without any complex calculation.
Given : equation of line as y=(4/3)x 100. So the line intercept the axes at (0,100) and (75,0). This can be considered a right angle triangle with right angle at (0,0) . So Base=100 , Height=75 and Hypotenuse =125 (By Pythagoras triplet)
So a perpendicular from the (0,0) to hypotenuse will be the answer.
Area of triangle= 0.5*100*75=0.5*125* x => x=60;
SO answer is 60



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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23 Feb 2014, 11:28
Good step by step solution jlgdr!



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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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23 Feb 2014, 17:59
agreed, nice approach jlgdr
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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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23 Feb 2014, 18:23
Thanks guys. Let me know if anything remains unclear ok?
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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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12 Mar 2014, 14:10
I know calculus isn't something expected to be known on the GMAT, but you can also use calculus to quickly solve this problem:
\(D^2 = x^2 + ((4x/3)100)^2\)
\(D^2 = x^2 + ((16/9)x^2  (8/3)(100)x +100^2)\)
\((D^2)' = 2x + 2*(16/9)*x  (8/3)(100) = 0\)
to find the minimum value we then solve for x when \((D^2)' = 0\)
\(0 = 2x + 32x/9  (8/3)*100
(8/3)*(100) = 50x/9
x = (8*100*9)/(3*50)\)
Canceling out terms from the numerator and denominator we get \(x = 48\)
This x value will give us the minimum distance. Plugging this back in, we get:
\(D^2 = 48^2 + ((4/3)(48)  100)^2\)
\(D^2 = 48^2 + (64  100)^2 = 48^2 + 36^2\)
Noticing that 6 is a factor of both 48 and 36;
\(D^2 = 6^2*(8^2 + 6^2)\)
\(D^2 = 36*(100)^2\)
\(D = 6*10 = 60.\)
Answer: C
This isn't math that will be tested on the GMAT, but maybe it could still provide shortcuts in problems like these.



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Re: Distance from the origin [#permalink]
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17 Jun 2014, 05:43
ulm wrote: kannn wrote: The equation of line n is y = \(\frac{4}{3}\) x – 100. What is the smallest possible distance in the xyplane from the point with coordinates (0, 0) to any point on line n ?
a) 48 b) 50 c) 60 d) 75 e) 100 The shortest line is a perpendicular, which goes through \((0;0)\) We know that lines (\(y=kx+b\)) are perpendicular when \(k1*k2=1\). So, \((4/3)*k2=1\) k2=3/4 Line has point (0;0) also. So, \(y=(3/4)x\) This line intersects with \(y=\frac{4}{3}x100\). Let's find the intersection point. \(\frac{4}{3}x100=\frac{3}{4}\) \(x=48\) \(y=\frac{4}{3}*48100\) \(y=36\) \((48;36)\) The shortest distance from (0;0) to above mentioned point is \(\sqrt{48*48+36*36}\)=\(60\) (C) Hi , can you please explain how K2 = 3/4 , should't it be 3/4 ?



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Re: Distance from the origin [#permalink]
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18 Jun 2014, 05:08
gauravsoni wrote: ulm wrote: kannn wrote: The equation of line n is y = \(\frac{4}{3}\) x – 100. What is the smallest possible distance in the xyplane from the point with coordinates (0, 0) to any point on line n ?
a) 48 b) 50 c) 60 d) 75 e) 100 The shortest line is a perpendicular, which goes through \((0;0)\) We know that lines (\(y=kx+b\)) are perpendicular when \(k1*k2=1\). So, \((4/3)*k2=1\) k2=3/4 Line has point (0;0) also. So, \(y=(3/4)x\) This line intersects with \(y=\frac{4}{3}x100\). Let's find the intersection point. \(\frac{4}{3}x100=\frac{3}{4}\) \(x=48\) \(y=\frac{4}{3}*48100\) \(y=36\) \((48;36)\) The shortest distance from (0;0) to above mentioned point is \(\sqrt{48*48+36*36}\)=\(60\) (C) Hi , can you please explain how K2 = 3/4 , should't it be 3/4 ? For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal \(\frac{1}{m}\). In other words, the two lines are perpendicular if and only if the product of their slopes is \(1\). So, yes, you are right, the slope of a line which is perpendicular to y = 4/3*x  100 is 3/4 (1/(4/3) = 3/4). Hope this helps.
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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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09 Nov 2014, 20:25
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the distance of a point(x1,y2) from a line Ax+by+c=0 is given by.. Modulus{(Ax1+By2+c)/sqrt(A^2+B^2)}
high school formula, I guess.



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The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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10 May 2015, 00:23
There is a formula to solve such questions. The min distance between points (p,q) and the line segment ax+by+c=0 is given by (pa+bq+c)/[sqrt(a^2+b^2)]
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Re: The equation of line n is y = 4/3*x  100. What is the small [#permalink]
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10 May 2015, 14:57
Hi All, This question is LOADED with patternmatching shortcuts. If you can spot these shortcuts, then you can save LOTS of time and avoid much of the math that other Test Takers would need to do to answer this question. The shortcuts are: 1) Draw a quick graph of the line; you should notice that you have a 3/4/5 right triangle with sides 75/100/125 2) You can 'cut' this big triangle into 2 smaller right triangles that are ALSO 3/4/5 right triangles 3) Using the 75 and 100 as reference, you can fill in the missing sides and end up with a 45/60/75 triangle and a 60/80/100 triangle 4) The common side also happens to be the shortest length = 60 GMAT assassins aren't born, they're made, Rich
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Re: The equation of line n is y = 4/3*x  100. What is the small
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