Equation of Line: 3x + 2y + 1 = 0 | Coordinates of Third Point On Line

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In summary, the conversation discusses finding the equation and coordinates of a third point on a line given two points on that line. The formula for the equation of a line through two points is provided and the steps to find the equation of the line are explained. The correct equation of the line is determined to be -3x -4y + 7 = 0.
  • #1
lolimcool
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Homework Statement


assume the existence of a fixed coordinatization of the plane in which 0 is the origin. Find an equation and give the coordinates of a third point that is on the line a:(-3,4) b:(1,1)


Homework Equations





The Attempt at a Solution


ok so i get x = -3 + t(-1 3)
y = 4 + t(1 -4)
then i solve for t and make the two equation equal each other
and i get 3x +2y + 1 = 0
is that right?


and could someone explain what the existence of a fixed coordinatization of the plane is?

thanks
 
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  • #2
Hi lolimcool! :smile:

lolimcool said:
ok so i get x = -3 + t(-1 3)
y = 4 + t(1 -4)

What do you mean with (-1 3)?

then i solve for t and make the two equation equal each other
and i get 3x +2y + 1 = 0
is that right?

I'm sorry, that equation is not correct :frown: If I substitute (1,1) in there, I get 3.1+2+1+1=6 and this had to be 0. (-3,4) does lie on the line, though...

and could someone explain what the existence of a fixed coordinatization of the plane is?

I guess it's just a fancy way of saying that every point of the plane has a set of coordinates. So for example, every point can be given a coordinate (3,4). They make it sound harder than it is...
thanks[/QUOTE]
 
  • #3
micromass said:
Hi lolimcool! :smile:


What do you mean with (-1 3)?



I'm sorry, that equation is not correct :frown: If I substitute (1,1) in there, I get 3.1+2+1+1=6 and this had to be 0. (-3,4) does lie on the line, though...



I guess it's just a fancy way of saying that every point of the plane has a set of coordinates. So for example, every point can be given a coordinate (3,4). They make it sound harder than it is...
thanks
[/QUOTE]

sorry i mean x = -3 +t(-1,3)
y = 4 + t(1, -4)
which gives me x = -3 + 2t and y = 4 + -3t
i then isolate for t, make the equations equal to each other and i get 3x + 2y + 1 = 0
 
  • #4
lolimcool said:
sorry i mean x = -3 +t(-1,3)

But that makes no sense. How would you define -3+(-1,3)?? You can't add up a real number and a pair!

What did you really mean?
 
  • #5
micromass said:
But that makes no sense. How would you define -3+(-1,3)?? You can't add up a real number and a pair!

What did you really mean?

oh i don't know then, i was just kinda following my notes(i obv copied them down wrong)
how would i go about doing the first step then?
 
  • #6
The equation of a line through (a,b) and (c,d) is given by

[tex]\left\{\begin{array}{c} x=a+t(c-a)\\ y=b+t(d-b)\\ \end{array}\right.[/tex]

So this is the formula you will have to use...
 
  • #7
Given two points on a line,
1) find the slope of the line.
2) find the equation of the line by using the point-slope equation.

For example, if your points are (1, 1) and (2, 3), the slope of the segment joining these points is m = (3 - 1)/(2 - 1) = 2

Using the slope and the point (1, 1), we have
y - 1 = 2(x - 1)
==> y = 2x - 2 + 1
==> y = 2x - 1
 
  • #8
micromass said:
The equation of a line through (a,b) and (c,d) is given by

[tex]\left\{\begin{array}{c} x=a+t(c-a)\\ y=b+t(d-b)\\ \end{array}\right.[/tex]

So this is the formula you will have to use...


OH K, so that's what i initially did then
i had x = -3 +t(3-1) => x = -3 +2t
y = 4 + t(1-4) => y = 4 -3t

then i got
t= (x+3)/2 and t = (y - 4)/ -3
making both equations equal each other i get
-3x -9 = 2y -8
-3x -2y -1 = 0
3x +2y + 1 = 0
 
  • #9
lolimcool said:
OH K, so that's what i initially did then
i had x = -3 +t(3-1) => x = -3 +2t
y = 4 + t(1-4) => y = 4 -3t

No, because I don't see how you found (3-1)...
Your c=1 and a=-3 here...
 
  • #10
micromass said:
No, because I don't see how you found (3-1)...
Your c=1 and a=-3 here...

wow i screwd up
so x = -3 + 4t
and y = 4 - 3t

solved it and got -3x -4y + 7 = 0
which i believe is right
thanks for your help :P
 
  • #11
Yes, that's good! Well done! :wink:
 
  • #12
lolimcool said:
wow i screwd up
so x = -3 + 4t
and y = 4 - 3t
This seems like the long way around to me, writing parametric equations for the line, and then eliminating the parameter to write the equation in standard form. It's not wrong, just extra work that I don't believe is needed.

The suggestion I gave in post #7 is a lot more direct.
lolimcool said:
solved it and got -3x -4y + 7 = 0
which i believe is right
thanks for your help :P
 

FAQ: Equation of Line: 3x + 2y + 1 = 0 | Coordinates of Third Point On Line

What is the equation of the line with the given coordinates?

The equation of the line is 3x + 2y + 1 = 0.

How do you find the third point on this line?

To find the third point, you can choose any value for either x or y in the equation and solve for the other variable. For example, if we choose x = 1, then we can solve for y by substituting x = 1 into the equation: 3(1) + 2y + 1 = 0. This gives us y = -2. Therefore, the third point on the line is (1, -2).

Can you graph this equation?

Yes, we can graph this equation by plotting the points (0, -1) and (1, -2) and drawing a straight line through them. This line will represent the equation 3x + 2y + 1 = 0.

What do the coefficients in the equation represent?

The coefficient of x (3) represents the slope of the line, and the coefficient of y (2) represents the y-intercept of the line. The constant term (1) represents the value of y when x = 0.

How can this equation be used in real-world applications?

This equation can be used to model linear relationships in various fields such as physics, engineering, and economics. For example, the equation could represent the relationship between distance and time for an object moving at a constant speed, or the relationship between production cost and number of units produced in a business.

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