To solve the problem of expressing the $y$-coordinate of point $P(x, y)$ in terms of $x$, we can proceed step by step by analyzing the geometry of the given figure. Let's summarize the key details from the image and then derive the required equation.

Key Observations:

1. Isosceles Right Triangle: The triangle $\triangle AOB$ is isosceles and right-angled at $O$, with the hypotenuse $AB$ having a length of 10 units.
2. Rectangle: There is a rectangle inscribed in the triangle with one side on the base and the top corners touching the hypotenuse of the triangle.
3. Coordinate System: The coordinate system is centered at $O(0, 0)$, with $A$ and $B$ being symmetric about the y-axis.

Step 1: Equation of the Hypotenuse $AB$

Given that $\triangle AOB$ is isosceles and right-angled with hypotenuse $AB = 10$, the endpoints $A$ and $B$ lie on the x-axis symmetrically. The equation of a line can be written in slope-intercept form $y = mx + c$.

• Coordinates of $A$ and $B$: Since the triangle is isosceles, both $A$ and $B$ are equidistant from the origin. Let’s denote the coordinates of $A$ as $(a, 0)$ and $B$ as $(-a, 0)$.

• Slope of the Line $AB$: The slope of the line $AB$ is negative because the triangle is oriented with $B$ above $A$ on the y-axis. Since the triangle is symmetric, the slope is $m = -\frac{1}{1} = -1$.

Thus, the equation of the line $AB$ (which is the hypotenuse of the triangle) can be written as: $y = -x + c$

To find $c$, use the distance formula considering $AB = 10$. The midpoint of $AB$ will be $(0, \frac{c}{2})$. But the distance from $A$ to $B$ must satisfy the equation $10 = \sqrt{(2a)^2 + c^2}$.

Given that $a^2 + a^2 = 25$, because the sum of squares of the legs equals the square of the hypotenuse, we get: $c = 5\sqrt{2}$

The equation becomes: $y = -x + 5\sqrt{2}$

Step 2: Expressing the y-coordinate of $P(x, y)$

Given $P(x, y)$ lies on line $AB$, substitute $x$ into the equation: $y = -x + 5\sqrt{2}$

So, the $y$-coordinate of point $P(x, y)$ is: $y = 5\sqrt{2} - x$

This expression gives you the y-coordinate of point $P$ in terms of $x$.

Summary:

• Equation of line $AB$: $y = -x + 5\sqrt{2}$
• y-coordinate of $P(x, y)$: $y = 5\sqrt{2} - x$

Would you like more details, or do you have any questions?

5 Related Questions:

1. How would the equation of the line change if the triangle were not isosceles?
2. Can you derive the area of the rectangle as a function of $x$?
3. What happens to the y-coordinate of $P$ as $x$ approaches the x-coordinate of $A$?
4. How does the symmetry of the triangle affect the location of point $P$?
5. What is the maximum area of the rectangle that can be inscribed in the triangle?

Tip:

When solving geometric problems involving coordinates, always start by identifying the key geometric properties (like symmetry) to simplify your equations.