1. ## Coordinate geomtry

Hi, I have a problem that is giving me trouble solving. The problem reads The vertices of a $\displaystyle \triangle$ ABC are at the points A$\displaystyle (a, 0)$, B$\displaystyle (0, b)$, C$\displaystyle (c, d)$. If $\displaystyle \angle$B =$\displaystyle 90 degrees$. find a relaionship between a, b, c and d. The book answere is $\displaystyle b(d-b)=ac$. Can someone help me to fugure out this proble please thank you.

2. It is simple. You gotta tally lengths:
$\displaystyle |AB|=\sqrt{a^2 + b^2}$
$\displaystyle |BC|=\sqrt{c^2+ (d-b)^2}$
$\displaystyle |AC|= \sqrt{(c-1)^2 + d^2}$

From Pitagoras' theorem:
$\displaystyle |AB|^2 + |BC|^2 = |AC|^2$

After simple transformation you should get:
$\displaystyle b^2 - bd + ca=0$
And finally the answer you wrote.

3. Originally Posted by Oloria
It is simple. You gotta tally lengths:
$\displaystyle |AB|=\sqrt{a^2 + b^2}$
$\displaystyle |BC|=\sqrt{x^2+ (d-b)^2}$
$\displaystyle |AC|= \sqrt{(c-1)^2 + d^2}$

From Pitagoras' theorem:
$\displaystyle |AB|^2 = |BC|^2 = |AC|^2$

After simple transformation you should get:
$\displaystyle b^2 - bd + ca=0$
And finally the answer you wrote.
Thank you very much. When i add the results I am having trouble comming up with the answer. Are you suppose to just add the results?

4. Show me your solution, please. Then I will point you to error.

5. Originally Posted by Oloria
Show me your solution, please. Then I will point you to error.
Hi, this is what I got so far
$\displaystyle a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2$

$\displaystyle a^2-b^2-2b(-d)-(-d)^2+b^2+c^2+2c(-1)+(-1)^2+c^2+d^2$

$\displaystyle a^2+2c^2+2bd-2c+1$
I have a feeling that I am on the wrong track.

6. Originally Posted by scrible
Hi, this is what I got so far
$\displaystyle a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2$
On the left side you subtract (d-b)^2 and you should sum it. Correct:
$\displaystyle a^2 + b^2 +c^2+(d-b)^2=(c-1)^2+d^2$

I have a feeling that I am on the wrong track.
Wrong feeling. I did it to the end and my result is the same as in the book.