https://file.aiquickdraw.com/mathbot/file/1724961707058q5hzwhxo.jpg

The accompanying figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is

10 units long.

**a.** Express the​ y-coordinate of P in terms of x.​ (You might start by writing an equation for the line​ AB.)

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.

Learn how to express the y-coordinate of Point P in an isosceles right triangle inscribed with a rectangle. This problem involves the geometry of triangles, coordinate geometry, and the equation of lines. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation