Part 1: Express the y-coordinate of $P$ in terms of $x$.

Given that the rectangle is inscribed in an isosceles right triangle, we can start by determining the equation of the line $AB$.

1. Identify the points:

   • Point $A$ is at $(13, 0)$.
   • Point $B$ is at $(0, 13)$.

2. Find the equation of the line $AB$: The slope of line $AB$ can be calculated as: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 0}{0 - 13} = -1$ The equation of the line in slope-intercept form $y = mx + b$ is: $y = -x + 13$

   So, the equation of the line $AB$ is: $y = -x + 13$

3. Express the y-coordinate of point $P$ in terms of $x$: Since point $P(x, y)$ lies on line $AB$, substitute $x$ into the equation of the line: $y = -x + 13$

Part 2: Express the area of the rectangle in terms of $x$.

Now, let's express the area of the rectangle formed by the coordinates $P(x, y)$.

1. Base of the rectangle: The base of the rectangle is the distance from $A$ to $P$ on the $x$-axis, which is simply $x$.

2. Height of the rectangle: The height of the rectangle is the $y$-coordinate of point $P$, which we found to be $y = -x + 13$.

3. Area of the rectangle: The area $A$ of the rectangle is given by: $A(x) = \text{base} \times \text{height} = x \times (-x + 13)$ Simplifying this, we get: $A(x) = -x^2 + 13x$

Summary

• Part 1 Answer: The $y$-coordinate of point $P$ in terms of $x$ is $y = -x + 13$.
• Part 2 Answer: The area of the rectangle in terms of $x$ is $A(x) = -x^2 + 13x$.

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

Related Questions:

1. How do we find the maximum area of the rectangle?
2. What is the significance of the equation $y = -x + 13$ in this context?
3. How would the equation of the line change if the triangle was not isosceles?
4. Can you explain how to derive the area formula $A(x) = -x^2 + 13x$?
5. How would the problem change if the hypotenuse was a different length?

Tip: Always start by identifying key points and their coordinates when working with geometric figures in algebraic problems.