1. Functions and their graphs

Been stuck on this for a while:

The figure shown here shows a rectangle inscibed in two isoceles right triangle whose hypotenuse is two units long
a.) express the y-coordinate of "P" in terms of x (you might start by writing an equation of line "AB"
b.) express the area of the rectangle in terms of x

Heres the diagram
59795_473501434601_592919601_6725805_3723994_n | Flickr - Photo Sharing!

2. I see that you have over thirty postings.
By now you should understand that this is not a homework service
So you need to either post some of your work on a problem or explain what you do not understand about the question.

3. confusion

Where I am confused is what the line equation has to do with "P"(x,?)
I understand that part a is required for a to figure out part b just using LxW= Area of the rectangle

in regards to part A I do not understand how I am suppose to use the data given to calculate the slope specifically
m=(y2-y1)/(x2-x1)

I apologize for my previous actions, as I will now put more thought into a question before posting.

4. The point $\displaystyle B0,b)$ is the y-intercept.
This is an isosceles right triangle. So length of $\displaystyle \overline{AB}$ is $\displaystyle \sqrt{2}$.
So $\displaystyle b=?$

5. P(x, ?) is a point ON the line. Once you have the equation as, say, y= mx+ b, then the "y coordinate of P" is mx+ b (where you should have numbers for m and b.) By The Pythagorean theorem applied to right triangle BOA, $\displaystyle b^2+ 1= 2$. I don't know why Plato says "So length of $\displaystyle \overline(AB)$ is $\displaystyle \sqrt{2}$". You are given in the problem that "the hypotenuse is 2 units long" and $\displaystyle \overline(AB)$ is the hypotenuse. You already know that A= (1, 0). Once you know the point B= (0, b), you should be able to find the equation y= mx+ b for the line and so (x, mx+ b) for any point on that line.

And once you have that, the area of the rectangle is "length times width"= (2x)(y)= 2x(mx+ b).

Now, what are m and b?

6. Halls,
No indeed $\displaystyle \overline{AB}$ is most certainly not the hypotenuses.
Did you look at the diagram?
The hypotenuses is along the x-axis from (-1,0) to (1,0)

The student’s description does not match the diagram that goes with the question.

7. Plato, are you inferring that "AB" is a component of a special triangle in order for it to be equal to root 2

8. I tried working it out and got this so far
$\displaystyle \frac{B-0}{0-1}+B$

9. First of all, what is correct: your description of the problem or the supplied diagram?
In the diagram the angle B is a right angle so the hypotenuse is from (-1,0) to (1,0).
That makes B(0,1) so the line determined by A & B is $\displaystyle y=-x+1$.

10. The question is word for word from the book and the supplied diagram is directly out of the book

11. Originally Posted by nightrider456
The figure shown here shows a rectangle inscibed in two isoceles right triangle whose hypotenuse is two units long
Originally Posted by nightrider456
The question is word for word from the book and the supplied diagram is directly out of the book
Either the diagram is drawn incorrectly or the description is wrong.
In the diagram point A is (1,0). So in order for $\displaystyle \Delta OAB$ an isoceles right triangle the point B has to be (0,1). Therefore the hypotenuse has length $\displaystyle \sqrt2$ not two units as in the description.

So something with the question one way or the other.

12. The right triangle in which the rectangle is inscribed has vertices at (-1, 0), (1, 0), and point B. Its hypotenuse is, indeed, of length 2. I made the foolish mistake of thinking that the "right triangle" referred to in the picture was $\displaystyle \Delta OAB$. The hypotenuse of that triangle is a leg of the large right triangle. The right triangle $\displaystyle \Delta OAB$ has hypotenuse $\displaystyle \sqrt{2}$ so AB is $\displaystyle \sqrt{2}$ as Plato said originally. I should have known! It should be easy to determine the length of OB, so the coordinates of point B and then the equation of the line. The area of the rectangle is xy where (x, y) is the point where the rectangle touches that line. Knowing the equation of the line, you know y in terms of x and so A= xy in terms of x only.