Coordinate geometry

• Mar 16th 2008, 10:08 AM
Coach
Coordinate geometry
Dear forum members,

Find all possible values of a and b given that y=ax+14 is the perpendicular bisector of the line joining (1,2) to(b,6) .

I have tried everything(except for the right method of course), but I always end up with something illogical.

Thank you in advance!
• Mar 16th 2008, 10:18 AM
Moo
Hello,

First of all, find the equation of the line between A(1,2) and B(b,6) (by solving y=mx+n)

Then, let H be the intersection of the two lines.

H is the middle of [AB]. This may give you a relation between a and b.

For the second one, you can take a random point C on y=ax+14 (in function of a) and say that CA=CB
• Mar 16th 2008, 10:31 AM
Coach
Thank you, Moo, for answering to my post again.

I found the coordinates of the midpoint

X=(1+b)/2 and Y= 4

the equation for line which passes through points (1,2) and (b,6) is

$\displaystyle Y= \frac{4}{b-1}x-(\frac{4}{b-1}+\frac{2b-2}{b-1})$

then wouldn't $\displaystyle a$be the inverse negative number of the gradient of the first equation?

I have tried plugging x and y in the equation of the first line and then solving for b, but I have never gotten a logical answer.
• Mar 16th 2008, 10:39 AM
Moo
Perfect for the midpoint

For the equation of the line, i find something else :

n=2-m=2-4/(b-1)

$\displaystyle Y=\frac{4}{b-1} x + (\frac{2b-2}{b-1} - \frac{4}{b-1})$

Then, take the point at random of abscissa (b-1) (so that it will simplify for its y). This point is at equal distance of A and B.

Can you continue the exercise ? It's apparently not sufficient, but i come back in 30 minutes ;)

You could also act with direction vectors, but i don't know if you've studied it...
• Mar 16th 2008, 10:47 AM
Coach
Quote:

Originally Posted by Moo
Perfect for the midpoint

For the equation of the line, i find something else :

n=2-m=2-4/(b-1)

$\displaystyle Y=\frac{4}{b-1} x + (\frac{2b-2}{b-1} - \frac{4}{b-1})$

Then, take the point at random of abscissa (b-1) (so that it will simplify for its y). This point is at equal distance of A and B.

Can you continue the exercise ? It's apparently not sufficient, but i come back in 30 minutes ;)

You could also act with direction vectors, but i don't know if you've studied it...

Thank you Moo for replying. But can you tell me why I just can't plug the coordinates of the midpoint into that equation, so 4=\frac{4}{b-1} *\frac{1+b}{2}+ (\frac{2b-2}{b-1} - \frac{4}{b-1})

I mean the coordinates of the midpoint have to satisfy the equation of a line, or am I totally wrong?

And I have never studied direction vectors, so I have no idea what they are.
• Mar 16th 2008, 11:01 AM
Moo
Well, with my formula (which is a bit different from yours), i find it exact. When you factorised the - sign, you made a little mistake. Your idea of verifying is very good ;)
• Mar 16th 2008, 11:05 AM
Coach
Quote:

Originally Posted by Moo
Well, with my formula (which is a bit different from yours), i find it exact. When you factorised the - sign, you made a little mistake. Your idea of verifying is very good ;)

Thank you, and feel very stupid for saying this, I still don't get how to do it?

what did you get for your answer. According to my book the answers should be a=-2, b=9 or a= 5/2 and b=-9

Thank you so much for giving up your time to help me.
• Mar 16th 2008, 11:09 AM
Moo
I'm here to help :-)

I don't have the results, but i know the steps to go to it ^^

You wrote $\displaystyle Y= \frac{4}{b-1}x-(\frac{4}{b-1}+\frac{2b-2}{b-1})$ , which is [Math]Y=\frac{4}{b-1}x-\frac{4}{b-1}-\frac{2b-2}{b-1}[/tex]

But it's $\displaystyle +\frac{2b-2}{b-1}$, not -

And you can simplify : (2b-2)-4=2b-6
• Mar 16th 2008, 11:23 AM
Moo
Ok, let's take it again from the beginning :

- you got the coordinates of H((b+1)/2 ; 4), which is on the two lines, including y=ax+14. If you replace x and y in this equation, you will have a with an expression including b -> tell me what you have

- i'm really sorry, but i think finding the last equation was unnecessary (Headbang) (at least, you've been training (Giggle))

- choose a random point on the line y=ax+14, for example C(0;14) (simple, because it deletes a). Then, write the equation : CA=CB (C is on the perpendicular bisector of [AB])

$\displaystyle AC=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}$

$\displaystyle BC=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}$

This will give you two possible values for b, and you have a with a previous relation.

I've checked, you'll get the values given by the correction
• Mar 16th 2008, 11:24 AM
Coach
this is what I get when trying to solve for b, and it all cancels out :(

$\displaystyle 8b^2-16b+8=8b^2-16b+8$
• Mar 16th 2008, 11:28 AM
Moo
Read my last message :-)

I'm really sorry to have put you on a wrong way
• Mar 16th 2008, 11:35 AM
Coach
but if I substitute the coordinates into $\displaystyle Y=ax+14$, I will get $\displaystyle -20=a+b$
• Mar 16th 2008, 11:38 AM
Moo
Not really ;-)

H((b+1)/2 ; 4)

So in y=ax+14 :

4=a(b+1)/2 + 14

=> -20=a(b+1)

=> a=-20/(b+1) :-)

H is on this line, because it's the midpoint of [AB] and y=ax+14 is the perpendicular bisector.
• Mar 16th 2008, 11:50 AM
Coach
Quote:

Originally Posted by Moo
Not really ;-)

H((b+1)/2 ; 4)

So in y=ax+14 :

4=a(b+1)/2 + 14

=> -20=a(b+1)

=> a=-20/(b+1) :-)

H is on this line, because it's the midpoint of [AB] and y=ax+14 is the perpendicular bisector.

WOW!

You're math skills are amazing, I got the right answer now!

Thank you so much for helping me, I cannot say how much it means to me that someone has devoted so much of their time to voluntarily help someone else.

Though I think I am going to drop math next year, I just can't do stuff like this, and if I'm correct this is not so advanced for high school.

Have a good night, and thank you for everything!
• Mar 16th 2008, 11:52 AM
Moo
You're welcome, it was a pleasure ^^