Geoff's gradient of a line proof question

[jazzygeoff]

Hi again all,
I wonder of someone could please help point me in the right direction for this question? It would be much appreciated!

"Let P, with coordinates (p,q), be a fixed point on the 'curve' with equation 'y=mx + c', and Q, with coordinated (r,s), be any other point on 'y=mx + c'. Use the fact the the coordinates of P and Q satisfy the equation 'y=mx + c' to show that the gradient of PQ is m for all positions of Q.

I understand that the gradient m is (q-s)/(p-r), and that will always be the same because the two points are always on the same line 'y=mx + c', but I can't work out what to do first in order to "show" that this is always the case.

Many thanks in advance for any help!
Geoff

 

[galactus]

I suppose they're just asking you to derive the slope-intercept form of a line.

You know the slope m is given by \(\displaystyle m=\frac{y-y_{1}}{x-x_{1}}\)

Using some algebra, we can rewrite this as:

\(\displaystyle y-y_{1}=m(x-x_{1})\)

\(\displaystyle y=mx-mx_{1}+y_{1}\)

Let \(\displaystyle c=-mx_{1}+y_{1}\)

And you have \(\displaystyle y=mx+c\)

BTW, b is normally used as the y-intercept, not c.

Is this what you was looking for?.

 

[jazzygeoff]

Thanks for that Galactus.

Unfortunately, my maths books doesn't have the answer to this question, so I can't check to see if that is what they want from the question.

I was expecting the answer to require some working with the given coordinates (p,q), and (r,s), otherwise there doesn't seem much point in them being defined???

Anyway, I think that a basic understanding of the equations you mention is what they are after, and that is what I have been doing for the 20 questions before this one that I got stuck on.

Thanks again for your help!
Geoff

 

[galactus]

p,q,r,s are the x and y coordinates. You could say, "let p,q,r,s be the respective x and y coordinates" and proceed from there.

\(\displaystyle q-s=y-y_{1}\)

\(\displaystyle p-r=x-x_{1}\)

 

[Mrspi]

Hi again all,
I wonder of someone could please help point me in the right direction for this question? It would be much appreciated!

"Let P, with coordinates (p,q), be a fixed point on the 'curve' with equation 'y=mx + c', and Q, with coordinated (r,s), be any other point on 'y=mx + c'. Use the fact the the coordinates of P and Q satisfy the equation 'y=mx + c' to show that the gradient of PQ is m for all positions of Q.

I understand that the gradient m is (q-s)/(p-r), and that will always be the same because the two points are always on the same line 'y=mx + c', but I can't work out what to do first in order to "show" that this is always the case.

Many thanks in advance for any help!
Geoff

Here's how I would approach this:

Since P with coordinates (p, q) is on the 'curve' with equation
y = mx + c
we know that the equation will be true when we substitute p for x and q for y:
1) q = mp + c

And since Q with coordinates (r, s) is also on this curve, the equation will also be true when we substitute r for x and s for y:
2) s = mr + c

Solve each of these equations for c:
1) q = mp + c
q - mp = c

2) s = mr + c
s - mr = c

Now, we have two expressions which are each equal to "c", so they must be equal to each other:
q - mp = s - mr
And let's solve this for m:
q - s = -mr + mp
q - s = m(p - r)
(q - s)/(p - r) = m

Is that the kind of proof you are looking for?

 

[jazzygeoff]

That's fantastic! I think that I just what I as looking for!

Thankyou very much indeed!