Analytical Geometry

[Jasonbot]

I'm writing a maths exam tomorrow and came across this problem I can't do...

P(-3;10) and Q(3;4) are vertices of triangle PQR. If the angle at Q= 90* determine the possible coordinates or R if QR= 4sqrt2 (sqrt32).

I've tried the distance formula for point on a cartesian plane but then I get an equation with x's and y's which I cant do. I understand that there must be two answers because the line could go up or down on the angle with the distance being equal.

The answer in the book is R1=(1;2) and R2=(5;7)

Oh and this will be my last exam for the mid year exam period

 

[sammich]

High school geometry...those were the days!

What you can simply do is find the equation for the line that is perpendicular to the line QP and passes through Q. It should be x + y = 1.

Then all you do with that is use the distance formula, use the relationship above to eliminate x or y, then solve!

 

[Jasonbot]

High school geometry...those were the days!

What you can simply do is find the equation for the line that is perpendicular to the line QP and passes through Q. It should be x + y = 1.

Then all you do with that is use the distance formula, use the relationship above to eliminate x or y, then solve!

wow! SO simple!

So...

[distance formula]=x+1. Should work. It seems logical!

EDIT: nope
heres what I did:

sqrt[(x-3)^2+(y-4)^2]=x+1
(x^2-6x+9)-(y^2-8y+16)=x^2+2x+1
-y^2+8y-8-6x=0

Its not working

 

[siurpeeman]

*edit*
thought about it some more...

the slope of PQ = -1, which means that the slope of QR = +1. the line that contains both Q and R is y=x+1 (using +1 for slope going through (3,4)). so, R is to have coordinates (x, x+1), which you use distance formula with Q to equal root 32, or use distance formula with P to equal root 104. incidentally, my answers are:

R = (-1,0) and (7,8). i checked the first one. too lazy/tired to check the second.


*edit #2*
psh, no gratitude.

 

[Jasonbot]

*edit #2*
psh, no gratitude.

k, I got it. thnx so much!

forget already?

 

[elfin buddy]

I'm writing a maths exam tomorrow and came across this problem I can't do...

P(-3;10) and Q(3;4) are vertices of triangle PQR. If the angle at Q= 90* determine the possible coordinates or R if QR= 4sqrt2 (sqrt32).

I've tried the distance formula for point on a cartesian plane but then I get an equation with x's and y's which I cant do. I understand that there must be two answers because the line could go up or down on the angle with the distance being equal.

The answer in the book is R1=(1;2) and R2=(5;7)

Oh and this will be my last exam for the mid year exam period

Just from checking the answers your book gives, I guarantee your book is wrong. The lengths of the QR sides are off by a factor of 2 from the problem requirements.

My calculations indicate that siurpeeman's answer is correct. I solved the problem from scratch and got the same answers as him.

 

[Penryn]

God how I hate maths...

 

[elfin buddy]

I wrote up a nice solution for you, with a diagram and all. Hope it helps

http://www.box.net/shared/oul37uague

 

[ErikCLDR]

I used to know how to do this last year or the year before. But its summer, so all gained knowledge has gone into deep storage.

 

[Bow Rekk]

I got R=(-1,0) and (7,8).

We can't all be wrong.

 

[Jasonbot]

I wrote up a nice solution for you, with a diagram and all. Hope it helps

http://www.box.net/shared/oul37uague

Wow! That was pretty higher grade. But yeah, I got it. Now wish me luck for my exam, I'm writing in an hour and 20 minutes!

 

[Mac One]

Well I got the same answer as you guys... not that it's of any use now.

Good luck for your exam Jasonbot, hope it goes OK
What year are you in? I did that sort of stuff last year. (I'm in my final year now.)

Enjoy not having any more exams!

 

[elfin buddy]

Wow! That was pretty higher grade. But yeah, I got it. Now wish me luck for my exam, I'm writing in an hour and 20 minutes!

You may not get this in time, but good luck! If you understood the math I wrote, I'm sure you'll do fine

 

[Jasonbot]

Well I got the same answer as you guys... not that it's of any use now.

Good luck for your exam Jasonbot, hope it goes OK
What year are you in? I did that sort of stuff last year. (I'm in my final year now.)

Enjoy not having any more exams!

Yeh, I'm in form 4 now. We have a total of five forms and the school year is from january to december.

Well I'm on holiday now but I get my exam results next wednesday *holds breath*