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Optimization problem: Find the point on the line $−x + 2y − 1 = 0$ that is closest to the point $(1, 2)$.

Find the closest point to the originTangent line Optimization HomeworkFind the point on the line $x + 4y − 7 = 0$ which is closest to the point $(-2, -2)$Closest points on two line segmentsOptimization, point on parabola closest to another pointfinding the closest distance between a point a curveClosest line to point after non-linear mapFinding the point on a line closest to another point not on the lineOptimization-Coordinate closest to originPoint in $e^x$ that is closest to a line

1

$begingroup$

Find the point on the line $−x + 2y − 1 = 0$ that is closest to the point $(1, 2)$.

I solved the optimization and got $x=14/10$ and $y = 1.7$ but my $y$ coordinate is not correct. can anyone explain why it's wrong. I used $x$ coordinate and solved for $y$ using $y=(x+1)/2$.

calculus optimization

share|cite|improve this question

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(1.4+1)/2=1.2ne1.7$
  $endgroup$
  – Shubham Johri
  Mar 21 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh shoot how did I do that, thank you
  $endgroup$
  – McA
  Mar 21 at 19:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nevermind, happens with all of us :)
  $endgroup$
  – Shubham Johri
  Mar 21 at 19:46
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Find the point on the line $−x + 2y − 1 = 0$ that is closest to the point $(1, 2)$.

I solved the optimization and got $x=14/10$ and $y = 1.7$ but my $y$ coordinate is not correct. can anyone explain why it's wrong. I used $x$ coordinate and solved for $y$ using $y=(x+1)/2$.

calculus optimization

share|cite|improve this question

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $(1.4+1)/2=1.2ne1.7$
  $endgroup$
  – Shubham Johri
  Mar 21 at 19:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oh shoot how did I do that, thank you
  $endgroup$
  – McA
  Mar 21 at 19:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nevermind, happens with all of us :)
  $endgroup$
  – Shubham Johri
  Mar 21 at 19:46
  
  

  
  
  
  

add a comment | 

$begingroup$

Find the point on the line $−x + 2y − 1 = 0$ that is closest to the point $(1, 2)$.

I solved the optimization and got $x=14/10$ and $y = 1.7$ but my $y$ coordinate is not correct. can anyone explain why it's wrong. I used $x$ coordinate and solved for $y$ using $y=(x+1)/2$.

calculus optimization

share|cite|improve this question

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

$endgroup$

share|cite|improve this question

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

share|cite|improve this question

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

edited Mar 21 at 19:43

Shubham Johri

5,500818

edited Mar 21 at 19:43

Shubham Johri

5,500818

asked Mar 21 at 19:38

McAMcA

1

asked Mar 21 at 19:38

McAMcA

1

2 Answers
2

active

oldest

votes

0

$begingroup$

A point on that line is given by $left(x,yright)=left(x,fracx+12right)$. Using the equation for the distance between two points, the distance between a point on this line and $left(1,2right)$ is given by $sqrtleft(x-1right)^2+left(fracx+12-2right)^2$. You already know this is an optimization problem, so you just need to solve the equation
$$fractextdtextdxsqrtleft(x-1right)^2+left(fracx+12-2right)^2=0.$$

share|cite|improve this answer

answered Mar 21 at 19:53

JakeJake

566314

$endgroup$

add a comment | 

0

$begingroup$

Any point on the line $(2k-1,k)$

Now the point will be on the perpendicular line as well

So, the product of the gradients

$$dfrac12cdotdfrack-22k-1-1=-1$$

Alternatively, if $d$ is the distance,

$d^2=(2k-1-1)^2+(k-2)^2=5k^2-12k+20=5(k-6/5)^2+20-5(6/5)^2$

The equality occurs if $k-6/5=0$

share|cite|improve this answer

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

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0

$begingroup$

A point on that line is given by $left(x,yright)=left(x,fracx+12right)$. Using the equation for the distance between two points, the distance between a point on this line and $left(1,2right)$ is given by $sqrtleft(x-1right)^2+left(fracx+12-2right)^2$. You already know this is an optimization problem, so you just need to solve the equation
$$fractextdtextdxsqrtleft(x-1right)^2+left(fracx+12-2right)^2=0.$$

share|cite|improve this answer

answered Mar 21 at 19:53

JakeJake

566314

$endgroup$

add a comment | 

0

$begingroup$

A point on that line is given by $left(x,yright)=left(x,fracx+12right)$. Using the equation for the distance between two points, the distance between a point on this line and $left(1,2right)$ is given by $sqrtleft(x-1right)^2+left(fracx+12-2right)^2$. You already know this is an optimization problem, so you just need to solve the equation
$$fractextdtextdxsqrtleft(x-1right)^2+left(fracx+12-2right)^2=0.$$

share|cite|improve this answer

answered Mar 21 at 19:53

JakeJake

566314

$endgroup$

add a comment | 

$begingroup$

A point on that line is given by $left(x,yright)=left(x,fracx+12right)$. Using the equation for the distance between two points, the distance between a point on this line and $left(1,2right)$ is given by $sqrtleft(x-1right)^2+left(fracx+12-2right)^2$. You already know this is an optimization problem, so you just need to solve the equation
$$fractextdtextdxsqrtleft(x-1right)^2+left(fracx+12-2right)^2=0.$$

share|cite|improve this answer

answered Mar 21 at 19:53

JakeJake

566314

$endgroup$

share|cite|improve this answer

answered Mar 21 at 19:53

JakeJake

566314

answered Mar 21 at 19:53

JakeJake

566314

answered Mar 21 at 19:53

JakeJake

566314

0

$begingroup$

Any point on the line $(2k-1,k)$

Now the point will be on the perpendicular line as well

So, the product of the gradients

$$dfrac12cdotdfrack-22k-1-1=-1$$

Alternatively, if $d$ is the distance,

$d^2=(2k-1-1)^2+(k-2)^2=5k^2-12k+20=5(k-6/5)^2+20-5(6/5)^2$

The equality occurs if $k-6/5=0$

share|cite|improve this answer

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

0

$begingroup$

Any point on the line $(2k-1,k)$

Now the point will be on the perpendicular line as well

So, the product of the gradients

$$dfrac12cdotdfrack-22k-1-1=-1$$

Alternatively, if $d$ is the distance,

$d^2=(2k-1-1)^2+(k-2)^2=5k^2-12k+20=5(k-6/5)^2+20-5(6/5)^2$

The equality occurs if $k-6/5=0$

share|cite|improve this answer

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

add a comment | 

$begingroup$

Any point on the line $(2k-1,k)$

Now the point will be on the perpendicular line as well

So, the product of the gradients

$$dfrac12cdotdfrack-22k-1-1=-1$$

Alternatively, if $d$ is the distance,

$d^2=(2k-1-1)^2+(k-2)^2=5k^2-12k+20=5(k-6/5)^2+20-5(6/5)^2$

The equality occurs if $k-6/5=0$

share|cite|improve this answer

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

$endgroup$

share|cite|improve this answer

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279

answered Mar 22 at 1:25

lab bhattacharjeelab bhattacharjee

228k15158279