Proving that this equation goes through the points

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In summary, using a vector solution, it can be shown that the scalar equation for the line through points P1(x1,y1) and P2(x2,y2) is y-y1/x-x2=y2-y1/x2-x1. This can be achieved by finding a vector that is normal to the line and using the dot product of this vector and P1P. By setting the two equations (x1, y1) = k(x2, y2) equal to each other, the constant k can be eliminated, resulting in the final equation (x1, y1)/(k, k) = (x2, y2).
  • #1
unknown101
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Homework Statement


Use a vector solution to show that a scalar equation for the line through the points P1(x1,y1) and P2(x2,y2) is y-y1/x-x2=y2-y1/x2-x1

Homework Equations


Find a vector which is normal to the line and then use the dot product of this vector and P1P

The Attempt at a Solution


I tried using numbers but I got lost. I don't what exactly this question is asking. Would I have to find the slope and then come up with a solution from that?
 
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  • #2
Hi unknown101! :smile:
unknown101 said:
Use a vector solution to show that a scalar equation for the line through the points P1(x1,y1) and P2(x2,y2) is y-y1/x-x2=y2-y1/x2-x1

(is that the right answer?)

Hint: call a point on the line P.

What do you know about the vectors PP1 and P1P2 ? :wink:
 
  • #3
OK so I call a point P. I don't know what you mean about those two vectors. Would I have to find the m which is the slop and then n which is the normal point. Looking at my notes none of this makes sense:confused:
 
  • #4
vector maths

unknown101 said:
OK so I call a point P. I don't know what you mean about those two vectors. Would I have to find the m which is the slop and then n which is the normal point. Looking at my notes none of this makes sense:confused:

ah … doesn't look as if your teacher has introduced you to the joys of vector maths, as opposed to coordinate maths.

Vector maths tries to avoid using coordinates …

for example, (a + b)2 = a2 + 2a.b + b2 is proved by simple algebra, and is a lot easier than the coordinate proof! :wink:

In vector maths, you can only really use three combinations …

a.b, axb, and ka (where k is a constant) …

which one(s) do you think would help in proving that PP1 is parallel to P1P2 ? :smile:
 
  • #5
Would I use a.b to prove that is it parallel?
 
  • #6
It's easier to use a = kb
 
  • #7
So if I use a=kb will I have to find the constant. Looking at the equation I'm trying to prove, do I have to to find out what y and x( the x and y without the number)?
 
  • #8
Hint: write a = kb in coordinates, and see if you can eliminate k. :wink:
 
  • #9
I wrote a=kb in coordinates. I don't if I did it correctly.
x1y1=k(x2y2)
x1k1/x2y2=k

Is that all i need to do?
 
  • #10
unknown101 said:
x1y1=k(x2y2)

you're making it look like one equation …

it's (x1 , y1) = k(x2 , y2), which is two equations …

so write them out and eliminate k :smile:
 
  • #11
So I did that I got (x1, y1)/x2, y2)=k
 
  • #12
unknown101 said:
So I did that I got (x1, y1)/x2, y2)=k

So this is the final answer?
 
  • #13
unknown101 said:
So I did that I got (x1, y1)/x2, y2)=k

uhhh?

that doesn't even make sense …

you can't divide a vector by another vector

write two equations
 
  • #14
You said I should write out 2 equations:
(x1,y1)=k(x2,y2)
x1,y1=kx2,ky2
Is that right?
 
  • #15
unknown101 said:
x1,y1=kx2,ky2
Is that right?

Yes … but it would be clearer if you wrote it explicitly as two equations …

anyway, now eliminate k :smile:
 
  • #16
tiny-tim said:
Yes … but it would be clearer if you wrote it explicitly as two equations …

anyway, now eliminate k :smile:

By writing as two equations do you mean as in...
1.(x1,y1)
2.(kx2,ky2)

Eliminate k. The only way I can think for eliminating k is
kx2,ky2=0
kx2,ky2/k=o/k
x2,y2=0

I don't know if I'm doing this right.
 
  • #17
unknown101 said:
By writing as two equations do you mean as in...
1.(x1,y1)
2.(kx2,ky2)

They aren't equations …

an equation is something with an = sign in the middle
kx2,ky2=0
kx2,ky2/k=o/k
x2,y2=0

?? this doesn't make any sense at all

write (x1 , y1) = k(x2 , y2) as two equations …

that's two completely separate sentences, each with an = in the middle
 
  • #18
Ok I kind of understand it now.
This is what I have so far
x1,y1=kx2,ky2
(x1,y1)/(k,k)=(kx2,ky2)/(k,k)
(x1,y1)/(k,k)=(x2,y2)
 
  • #19
unknown101 said:
Ok I kind of understand it now.
This is what I have so far
x1,y1=kx2,ky2
(x1,y1)/(k,k)=(kx2,ky2)/(k,k)
(x1,y1)/(k,k)=(x2,y2)

Is that right?
 

1. How do you prove that an equation goes through specific points?

To prove that an equation goes through specific points, you need to substitute the coordinates of those points into the equation and check if the equation holds true. If the equation is satisfied for all given points, it can be concluded that the equation goes through those points.

2. What is the mathematical process for proving an equation goes through certain points?

The mathematical process for proving an equation goes through certain points is known as substitution. This involves substituting the x and y coordinates of the given points into the equation and solving for both sides to check if they are equal. If they are equal, the equation is proven to go through those points.

3. Can an equation go through more than two points?

Yes, an equation can go through more than two points. In fact, an equation can go through an infinite number of points, depending on the complexity and type of equation. However, it is common to use two points to prove that an equation goes through a line.

4. Is it necessary for an equation to go through all given points to be considered valid?

No, it is not necessary for an equation to go through all given points to be considered valid. If an equation goes through two or more points, it can be considered valid for those points. However, it is important to note that an equation may not represent all the points on a curve or line, but it can still be a valid equation.

5. How do you use the slope-intercept form to prove an equation goes through two points?

To prove an equation goes through two points using the slope-intercept form, you need to first find the slope of the line passing through those points using the formula (y2-y1)/(x2-x1). Then, substitute the slope and one of the given points into the slope-intercept form (y=mx+b) to find the y-intercept (b). If the equation holds true for both points, it can be concluded that the equation goes through those points.

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