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Math Expert V
Joined: 02 Sep 2009
Posts: 60647
What is the smallest possible distance between origin and any point on  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 45% (01:50) correct 55% (01:50) wrong based on 56 sessions

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What is the smallest possible distance between origin and any point on the line $$y=\frac{1}{2}x+50$$?

A. 25
B. $$20\sqrt{5}$$
C. 50
D. $$50\sqrt{2}$$
E. 100

Are You Up For the Challenge: 700 Level Questions

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Math Expert V
Joined: 02 Aug 2009
Posts: 8336
Re: What is the smallest possible distance between origin and any point on  [#permalink]

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Bunuel wrote:
What is the smallest possible distance between origin and any point on the line $$y=\frac{1}{2}x+50$$?

A. 25
B. $$20\sqrt{5}$$
C. 50
D. $$50\sqrt{2}$$
E. 100

Are You Up For the Challenge: 700 Level Questions

There can be many ways to solve it..

(I) Direct formula....
If the line is ax+by+c=0, then the shortest/perpendicular distance of point (X,Y) is $$\frac{aX+bY+c}{\sqrt{a^2+b^2)}}$$
Now $$y=\frac{1}{2}x+50......2y=x+100......x-2y+100=0$$, so a=1, b=-2, c=100 and (X,Y)=(0,0)
$$\frac{aX+bY+c}{\sqrt{a^2+b^2}}$$=$$\frac{0+0+100}{\sqrt{1^2+(-2)^2}}=$$$$\frac{100}{\sqrt{5}}=20\sqrt{5}$$

(II) Slope
Now slope in $$y=\frac{1}{2}x+50$$ is 1/2, so the slope of the perpendicular will be -2 as the product of slopes of perpendicular lines is -1.
The equation of the line passing through (0,0) and of slope -2 is...$$\frac{y-y_1}{x-x_1}=m.......\frac{y-0}{x-0}=-2.....y=-2x$$
Now where will the two lines meet or the perpendicular from (0,0) meet the line $$y=\frac{1}{2}x+50$$.....$$y=-2x=\frac{1}{2}x+50...x=-20$$
and y=-2x=-2*20=-40..
Distance between (0,0) and (-20,-40)=$$\sqrt{(-20-0)^2+(-40-0)^2}=\sqrt{20^2+40^2}=20\sqrt{1^2+2^2}=20\sqrt{5}$$

(II) Graph..
Draw a free hand graph as shown..
The area of $$\triangle{ABO}=\frac{1}{2}*50*100=2500$$...(i)
The area of $$\triangle{ABO}=\frac{1}{2}*AB*OC$$, when AB is taken as the BASE..
AB=$$\sqrt{100^2+50^2}=50\sqrt{2^2+1^2}=50\sqrt{5}$$, so Area = $$\frac{1}{2}*50\sqrt{5}*OC=25\sqrt{5}*OC$$...(ii)

From (i) and (ii), we get $$25\sqrt{5}*OC=2500....OC=\frac{2500}{25\sqrt{5}}=20\sqrt{5}$$

B
Attachments Untitled00.png [ 9.62 KiB | Viewed 458 times ]

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##### General Discussion
Intern  B
Joined: 24 Jun 2019
Posts: 4
Re: What is the smallest possible distance between origin and any point on  [#permalink]

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3
Equation of a line is y=mx+c
If a line passes from the origin then the equation of line is of the form y=mx

Here the given line is y= (1/2)x + 50, which gives us slope=1/2
So slope of line perpendicular to it is -2 as the product of the slopes =-1 in case of perpendicular lines.

So equation of line from origin is y=-2x

Let's find the point of intersection by solving the two simultaneous equations.
-2x = (1/2)x +50
-2.5x = 50
x = -20

y=-2x=40
So the distance from origin, (0,0) to (-20,40) is
Sqrt( (-20)^2 + (40)^2 )
= 20sqrt(1+4)
=20 sqrt(5)

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Intern  B
Joined: 07 Apr 2019
Posts: 31
Re: What is the smallest possible distance between origin and any point on  [#permalink]

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Is eyeball solution possible?

y intercept is 50... x intercept is -100... the slope is positive so... the shortest distance will be the distance between origin and the y intercept: i.e. 50.....
Intern  B
Joined: 24 Jun 2019
Posts: 4
Re: What is the smallest possible distance between origin and any point on  [#permalink]

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deeeuce wrote:
Is eyeball solution possible?

y intercept is 50... x intercept is -100... the slope is positive so... the shortest distance will be the distance between origin and the y intercept: i.e. 50.....

We are asked about the shortest distance, and it is always a perpendicular line from that point to the line.
so it is not the shortest distance from origin to the y intercept
Director  P
Joined: 07 Mar 2019
Posts: 595
Location: India
GMAT 1: 580 Q43 V27
WE: Sales (Energy and Utilities)
Re: What is the smallest possible distance between origin and any point on  [#permalink]

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Bunuel wrote:
What is the smallest possible distance between origin and any point on the line $$y=\frac{1}{2}x+50$$?

A. 25
B. $$20\sqrt{5}$$
C. 50
D. $$50\sqrt{2}$$
E. 100

Are You Up For the Challenge: 700 Level Questions

The line $$y=\frac{1}{2}x+50$$ intersect y-axis at (0,50) and x-axis at (-100,0) forming a right angled triangle with coordinates (0,0), (0,50) and (-100,0).
Area of this triangle is = $$\frac{1}{2}$$ * 50 * 100 = 2500

Also, length of hypotenuse = $$\sqrt{50^2 + 100^2}$$
So, Area of the triangle can also be = $$\frac{1}{2}$$ * Base(hypotenuse) * perpendicular from base intersecting point of origin
=> $$\frac{1}{2}$$ * $$\sqrt{50^2 + 100^2}$$ * perpendicular = 2500
perpendicular = 2 * $$\frac{2500}{50\sqrt{5}}$$
= $$20\sqrt{5}$$

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GMATPREPSoft1 680(Q48,V35) June 26, 2019 Re: What is the smallest possible distance between origin and any point on   [#permalink] 09 Nov 2019, 23:49
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