Geoff's Equation of a line given two points on it question!

J

jazzygeoff

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Jan 4, 2006
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Hi again all,

Please help me get going with this question?

Point P(x,y) lies on a line joining A(3,0) and B(5,6). Find expressions for gradients of lines AP and PB. Hence show that y=3x-9.

I have done:-
AP gradient is (y-0)/(x-3) = y/(x-3)
PB gradient is (6-y)/(5-x)

I come to a grinding halt at "Hence show that y=3x-9".

I have found that gradient of AP is y/(x-3) and gradient of PB is (6-y)/(5-x), and I guess that these two expressions equal eachother as AP and PB are parallel (They are on the same line.

Could someone please hint as to what the next stage I need to do is to come upwith an answer y=3x-9?

I would be very grateful, and would be happy to just receive a hint so that I can try to complete the problem myself.

Kind regards,
Geoff
 
I guess that these two expressions equal eachother as AP and PB are parallel (They are on the same line.
Click to expand...
Correct. So equate and solve for y.
 
I'm afraid I've hit rock bottom.

Are you saying that I need to write y/(x-3) = (6-y)/(5-x) and re-arrange it to a form y=?????

And it will pop out to be y=3x-9

If not, could you please venture the next line of the calculation so that I might be able to pick it up from there?

Geoff
 
y/(x-3) = (6-y)/(5-x)

Multiply both sides by (x-3)(5-x); expand both sides.

Too much? You can always alter path and use the slope of AB.
 
I have discovered after a couple of sides of A4 paper that I don't know how to multiply both sides by (x-3)(5-x)

I'm suddenly very aware of my inability to re-arrange algebra when there is division like that in the equation. I'll go and practice it before carrying on - do you have any web-site URL's, etc, with things similar to this that I can practice with?

Any chance of seeing the next line of the working? I realise it's rolling over and admitting defeat, but I'm using up paper at a rate of knots here. ;-)
 
\(\displaystyle \L \mbox{\frac{y}{x - 3} = \frac{6 - y}{5 - x}}\)

Multiply through by the denominators
(note that x cannot equal 3 or 5 because those would cause division by zero)

\(\displaystyle \L \mbox{y(5 - x) = (6 - y)(x - 3)}\)

Expand (FOIL the right-hand side):

\(\displaystyle \L \mbox{5y - xy = 6x - 18 - xy + 3y}\)

\(\displaystyle \mbox{-xy}\)'s cancel and simplify.
 
That's great, thanks!

y/(x-3) = (6-y)/(5-x)

y(5-x) = (6-y)(x-3)

5y-xy = 6x -xy + 3y -18

5y = 6x + 3y - 18

2y = 6x - 18

y = (6x - 18) / 2

y = (6x / 2) - (18 / 2)

y = 3x - 9

Brilliant! All done!
 
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