Proving that this equation goes through the points

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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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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:
 
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:
 
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:
 
Would I use a.b to prove that is it parallel?
 
It's easier to use a = kb
 
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)?
 
Hint: write a = kb in coordinates, and see if you can eliminate k. :wink:
 
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?
 
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:
 
So I did that I got (x1, y1)/x2, y2)=k
 
unknown101 said:
So I did that I got (x1, y1)/x2, y2)=k

So this is the final answer?
 
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
 
You said I should write out 2 equations:
(x1,y1)=k(x2,y2)
x1,y1=kx2,ky2
Is that right?
 
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:
 
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.
 
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
 
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)
 
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?