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# What is the smallest possible distance between the point (0, 5) and an

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SVP
Status: Preparing GMAT
Joined: 02 Nov 2016
Posts: 2033
Location: Pakistan
GPA: 3.39
What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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Updated on: 06 Jan 2019, 19:16
8
00:00

Difficulty:

45% (medium)

Question Stats:

60% (00:57) correct 40% (01:15) wrong based on 82 sessions

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What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

Source: Experts Global GMAT
Difficulty Level: 600

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Last edited by SajjadAhmad on 06 Jan 2019, 19:16, edited 4 times in total.
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Concentration: Entrepreneurship, Operations
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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09 Oct 2018, 12:36
3
What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

EG

Method 01 :

Distance from a pt (x,y) from line am+bn+c=0 is $$\frac{|ax+by+c|}{\sqrt{(a^2 +b^2)}}$$
Eqn: x+y-3=0 Hence Distance = $$\frac{|0*1 +5*1-3|}{\sqrt{(1^2 +1^2)}}$$ = $$\sqrt{2}$$

Or Alternatively , Method 02: Picture
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Manager
Joined: 24 Jun 2013
Posts: 132
Location: India
Schools: ISB '20
Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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10 Oct 2018, 21:25
1
1
What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

EG

smallest possible distance is always when line from that point meets at 90deg to other line.

From above we know the both lines will be perpendicular to each other.

if two lines are perpendicular to each other then their slope are negative reciprocal.

let L1 = y = -x + 3 ==> m1 (slope) = -1
let L2 (line from point 0,5 to L1) = y=m2 x + C ; we also know from above that m2 = 1

so the equation of L2 becomes = y = x + 5 (C=5, we can substitute (0,5) in L2 equation to get value of C)

now both line meet at (-1,4) you could get this point by equating both line equation to find intersecting x,y co ordinates.

now the distance between (0,5) and (-1,4) will be the smallest distance and can be calculated by using distance formula

$$\sqrt{(x2-x1)^2 + (y2-y1)^2}$$ = $$\sqrt{2}$$
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 00:44
u1983, doomedcat I keep getting 1 as answer using method 1 stated by u1983. Pls what am I doing wrong? Or is the method wrong?
From the diagram posted by u1983 point (0,5) is on the x-axis instead of the y axis.. sorry I asking all these questions, this is because I am not an expert and just need clarification on whether the explanation is accurate. Thanks.
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Concentration: Entrepreneurship, Operations
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 01:08
Kem12 wrote:
u1983, doomedcat I keep getting 1 as answer using method 1 stated by u1983. Pls what am I doing wrong? Or is the method wrong?
From the diagram posted by u1983 point (0,5) is on the x-axis instead of the y axis.. sorry I asking all these questions, this is because I am not an expert and just need clarification on whether the explanation is accurate. Thanks.
Posted from my mobile device

Hi Kem12 ...please post the calculation you are doing. We will try to find the confusion point.
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 02:38
u1983 (|0*1+5*1-3|)/(1^2+1^2) this equals 1.

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Concentration: Entrepreneurship, Operations
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 09:19
1
Kem12 wrote:
u1983 (|0*1+5*1-3|)/(1^2+1^2) this equals 1.

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Hello Kem12 ...Thanks.

The mistake is at my end. I missed out on the sq root at the denominator while typing the soln. Sorry for taking things granted .
I have corrected it now.
SO, now following the formula you should get $$\sqrt{2}$$.
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Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 10:01
1
Please the graph doesn’t make any sense.

When I graph I see 2 points between (0,5) and (0,3) and the perpendicular from (0,5) hits the line at x=-2, giving a 2,2,2root2 90degree triangle between 0,5 and 0,3(on the line).

Where am I wrong?

Or should I just memorize the formula and move on

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Manager
Joined: 14 Feb 2016
Posts: 73
Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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06 Jan 2019, 10:35
1
u1983 wrote:
What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

EG

Method 01 :

Distance from a pt (x,y) from line am+bn+c=0 is $$\frac{|ax+by+c|}{\sqrt{(a^2 +b^2)}}$$
Eqn: x+y-3=0 Hence Distance = $$\frac{|0*1 +5*1-3|}{\sqrt{(1^2 +1^2)}}$$ = $$\sqrt{2}$$

Or Alternatively , Method 02: Picture

I am a bit confused by your graph. If the coordinates are (0,5);(x1, y1). Then why is the coordinate (0,5) not posted on the y-axis?
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Intern
Joined: 18 Nov 2018
Posts: 2
Re: What is the smallest possible distance between the point (0, 5) and an  [#permalink]

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07 Jan 2019, 11:06
Please the graph doesn’t make any sense.

When I graph I see 2 points between (0,5) and (0,3) and the perpendicular from (0,5) hits the line at x=-2, giving a 2,2,2root2 90degree triangle between 0,5 and 0,3(on the line).

Where am I wrong?

Or should I just memorize the formula and move on

u1983 wrote:
What is the smallest possible distance between the point (0, 5) and any point on the line y = -x + 3?

A. 0
B. $$\frac{1}{\sqrt{2}}$$
C. 1
D. $$\sqrt{2}$$
E. 2

EG

Method 01 :

Distance from a pt (x,y) from line am+bn+c=0 is $$\frac{|ax+by+c|}{\sqrt{(a^2 +b^2)}}$$
Eqn: x+y-3=0 Hence Distance = $$\frac{|0*1 +5*1-3|}{\sqrt{(1^2 +1^2)}}$$ = $$\sqrt{2}$$

Or Alternatively , Method 02: Picture

Posted from my mobile device
Re: What is the smallest possible distance between the point (0, 5) and an &nbs [#permalink] 07 Jan 2019, 11:06
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