y greater than or equals x
x squared plus y squared less than or equals 169
List all points of intersection.

To find the points of intersection between the two given inequalities:

1. \( y \geq x \)
2. \( x^2 + y^2 \leq 169 \)

First, we need to solve for the boundary lines of these inequalities.

1. \( y = x \)
2. \( x^2 + y^2 = 169 \)

To find the points of intersection, we substitute \( y = x \) into the circle equation \( x^2 + y^2 = 169 \):

\[ x^2 + x^2 = 169 \]
\[ 2x^2 = 169 \]
\[ x^2 = \frac{169}{2} \]
\[ x = \pm \sqrt{\frac{169}{2}} \]
\[ x = \pm \sqrt{84.5} \]
\[ x \approx \pm 9.19 \]

Thus, the points of intersection are:
\[ ( \sqrt{84.5}, \sqrt{84.5} ) \approx ( 9.19, 9.19 ) \]
\[ ( -\sqrt{84.5}, -\sqrt{84.5} ) \approx ( -9.19, -9.19 ) \]

Now, let’s verify these points:

- For \( (9.19, 9.19) \):
  - \( y \geq x \) is satisfied because \( 9.19 \geq 9.19 \).
  - \( x^2 + y^2 = 169 \) is satisfied because \( 9.19^2 + 9.19^2 = 169 \).

- For \( (-9.19, -9.19) \):
  - \( y \geq x \) is satisfied because \( -9.19 \geq -9.19 \).
  - \( x^2 + y^2 = 169 \) is satisfied because \( (-9.19)^2 + (-9.19)^2 = 169 \).

Therefore, the points of intersection are:

\[ ( 9.19, 9.19 ) \]
\[ ( -9.19, -9.19 ) \]

Do you want details or have any questions?

Here are some more questions you might want to ask:
1. How do we solve systems of inequalities graphically?
2. What is the significance of the unit circle in trigonometry?
3. How can we find the area of the region bounded by inequalities?
4. What are some methods for solving quadratic equations?
5. How do we determine the distance between two points in a plane?

**Tip:** When solving systems of inequalities, it's often helpful to graph each inequality to visualize the solution set and points of intersection.

Explore how to find the intersection points of the inequalities y ≥ x and x² + y² ≤ 169. This detailed solution includes steps on solving circle equations and verifying points of intersection. Suitable for high school students.

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