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Determine the slope and write the Cartesian equation of the line.

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2

$begingroup$

Write the equation of the line by the origin of coordinates that has vector director of components (1,2) determine the slope and write the Cartesian equation of the line.

My attempt:

Let $l$ a line such that pass for the origin.

Let $a$ a director vector of $l$. we know as $a$ is a director vector of $l$ then the components $(1,2)$ are in the line.

We have two points of the line.
We know the general equation for the line is $Ax+By+C=0$

Here i'm stuck can someone help me?

geometry

share|cite|improve this question

asked 3 hours ago

Bvss12Bvss12

1,839719

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a ‘director vector’?
  $endgroup$
  – 雨が好きな人
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is a vector that gives the direction of a line and also orients it, that is to say, gives it a certain sense. @雨が好きな人
  $endgroup$
  – Bvss12
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Write the equation of the line by the origin of coordinates that has vector director of components (1,2) determine the slope and write the Cartesian equation of the line.

My attempt:

Let $l$ a line such that pass for the origin.

Let $a$ a director vector of $l$. we know as $a$ is a director vector of $l$ then the components $(1,2)$ are in the line.

We have two points of the line.
We know the general equation for the line is $Ax+By+C=0$

Here i'm stuck can someone help me?

geometry

share|cite|improve this question

asked 3 hours ago

Bvss12Bvss12

1,839719

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a ‘director vector’?
  $endgroup$
  – 雨が好きな人
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is a vector that gives the direction of a line and also orients it, that is to say, gives it a certain sense. @雨が好きな人
  $endgroup$
  – Bvss12
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Write the equation of the line by the origin of coordinates that has vector director of components (1,2) determine the slope and write the Cartesian equation of the line.

My attempt:

Let $l$ a line such that pass for the origin.

Let $a$ a director vector of $l$. we know as $a$ is a director vector of $l$ then the components $(1,2)$ are in the line.

We have two points of the line.
We know the general equation for the line is $Ax+By+C=0$

Here i'm stuck can someone help me?

geometry

share|cite|improve this question

asked 3 hours ago

Bvss12Bvss12

1,839719

$endgroup$

share|cite|improve this question

asked 3 hours ago

Bvss12Bvss12

1,839719

share|cite|improve this question

asked 3 hours ago

Bvss12Bvss12

1,839719

asked 3 hours ago

Bvss12Bvss12

1,839719

asked 3 hours ago

Bvss12Bvss12

1,839719

3 Answers
3

active

oldest

votes

1

$begingroup$

A directing vector of a line is a vector which is collinear to any vector formed by two points on this line.

Now take the point $O(0,0)$ which belongs to the line (as given) and any point $M(x, y) $ on this line. Thus the vector $OM$ is collinear to $(1,2)$; meaning that there exists a real number $k$ such that $OM=$k$a$.

Thus, by comparing their coordinates, we get: $x=k$ and $y=2k$, and thus $y=2x$ is the equation of this line (or $2x-y=0$).

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

user7857462user7857462

1238

$endgroup$

add a comment | 

2

$begingroup$

Assuming that $(1,2)$ is a vector parallel to the line, the line has slope $frac21 = 2$.

From this, we can deduce that the Cartesian equation of the line is

$$y = 2x + c$$

for some constant $c$. We cannot determine $c$ unless we are given more information (such as a particular point on the line).

share|cite|improve this answer

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

$endgroup$

add a comment | 

2

$begingroup$

I take it you want the line through the origin and $(1,2)$. The equation is $y=2x$. This is in slope-intercept form (slope $2$, $y$-intercept $0$). The slope is $m=dfrac Delta yDelta x=dfrac2-01-0=2$.

share|cite|improve this answer

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

$endgroup$

add a comment | 

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1

$begingroup$

A directing vector of a line is a vector which is collinear to any vector formed by two points on this line.

Now take the point $O(0,0)$ which belongs to the line (as given) and any point $M(x, y) $ on this line. Thus the vector $OM$ is collinear to $(1,2)$; meaning that there exists a real number $k$ such that $OM=$k$a$.

Thus, by comparing their coordinates, we get: $x=k$ and $y=2k$, and thus $y=2x$ is the equation of this line (or $2x-y=0$).

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

user7857462user7857462

1238

$endgroup$

add a comment | 

1

$begingroup$

A directing vector of a line is a vector which is collinear to any vector formed by two points on this line.

Now take the point $O(0,0)$ which belongs to the line (as given) and any point $M(x, y) $ on this line. Thus the vector $OM$ is collinear to $(1,2)$; meaning that there exists a real number $k$ such that $OM=$k$a$.

Thus, by comparing their coordinates, we get: $x=k$ and $y=2k$, and thus $y=2x$ is the equation of this line (or $2x-y=0$).

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

user7857462user7857462

1238

$endgroup$

add a comment | 

$begingroup$

A directing vector of a line is a vector which is collinear to any vector formed by two points on this line.

Now take the point $O(0,0)$ which belongs to the line (as given) and any point $M(x, y) $ on this line. Thus the vector $OM$ is collinear to $(1,2)$; meaning that there exists a real number $k$ such that $OM=$k$a$.

Thus, by comparing their coordinates, we get: $x=k$ and $y=2k$, and thus $y=2x$ is the equation of this line (or $2x-y=0$).

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

user7857462user7857462

1238

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

user7857462user7857462

1238

answered 2 hours ago

user7857462user7857462

1238

answered 2 hours ago

user7857462user7857462

1238

2

$begingroup$

Assuming that $(1,2)$ is a vector parallel to the line, the line has slope $frac21 = 2$.

From this, we can deduce that the Cartesian equation of the line is

$$y = 2x + c$$

for some constant $c$. We cannot determine $c$ unless we are given more information (such as a particular point on the line).

share|cite|improve this answer

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

$endgroup$

add a comment | 

2

$begingroup$

Assuming that $(1,2)$ is a vector parallel to the line, the line has slope $frac21 = 2$.

From this, we can deduce that the Cartesian equation of the line is

$$y = 2x + c$$

for some constant $c$. We cannot determine $c$ unless we are given more information (such as a particular point on the line).

share|cite|improve this answer

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

$endgroup$

add a comment | 

$begingroup$

Assuming that $(1,2)$ is a vector parallel to the line, the line has slope $frac21 = 2$.

From this, we can deduce that the Cartesian equation of the line is

$$y = 2x + c$$

for some constant $c$. We cannot determine $c$ unless we are given more information (such as a particular point on the line).

share|cite|improve this answer

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

$endgroup$

share|cite|improve this answer

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

answered 3 hours ago

雨が好きな人雨が好きな人

2,022316

2

$begingroup$

I take it you want the line through the origin and $(1,2)$. The equation is $y=2x$. This is in slope-intercept form (slope $2$, $y$-intercept $0$). The slope is $m=dfrac Delta yDelta x=dfrac2-01-0=2$.

share|cite|improve this answer

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

$endgroup$

add a comment | 

2

$begingroup$

I take it you want the line through the origin and $(1,2)$. The equation is $y=2x$. This is in slope-intercept form (slope $2$, $y$-intercept $0$). The slope is $m=dfrac Delta yDelta x=dfrac2-01-0=2$.

share|cite|improve this answer

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

$endgroup$

add a comment | 

$begingroup$

I take it you want the line through the origin and $(1,2)$. The equation is $y=2x$. This is in slope-intercept form (slope $2$, $y$-intercept $0$). The slope is $m=dfrac Delta yDelta x=dfrac2-01-0=2$.

share|cite|improve this answer

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

answered 2 hours ago

Chris CusterChris Custer

15.3k3827

answered 2 hours ago

Chris CusterChris Custer

15.3k3827