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The problem asks to calculate the area and perimeter of the triangle formed by the coordinate axes and a line that passes through the point (10, -12), which is perpendicular to the line given by the equation \( y = \frac{5}{12}x + 10 \).

### Let's break this down:

1. **Find the equation of the perpendicular line**:
   The slope of the given line \( y = \frac{5}{12}x + 10 \) is \( \frac{5}{12} \).
   - The slope of a line perpendicular to this is the negative reciprocal of \( \frac{5}{12} \), which is \( -\frac{12}{5} \).
   - The equation of the line passing through \( (10, -12) \) with slope \( -\frac{12}{5} \) is:
     \[
     y = -\frac{12}{5}x + b
     \]
   - Substitute \( (10, -12) \) into the equation to find \( b \):
     \[
     -12 = -\frac{12}{5}(10) + b
     \]
     \[
     -12 = -24 + b \implies b = 12
     \]
   - So, the equation of the perpendicular line is:
     \[
     y = -\frac{12}{5}x + 12
     \]

2. **Find the intercepts of the line**:
   To calculate the area and perimeter, we need the x-intercept and y-intercept of the line.

   - **x-intercept**: Set \( y = 0 \) in the equation \( y = -\frac{12}{5}x + 12 \):
     \[
     0 = -\frac{12}{5}x + 12
     \]
     \[
     \frac{12}{5}x = 12 \implies x = 5
     \]
     So the x-intercept is \( (5, 0) \).

   - **y-intercept**: Set \( x = 0 \) in the equation \( y = -\frac{12}{5}x + 12 \):
     \[
     y = 12
     \]
     So the y-intercept is \( (0, 12) \).

3. **Calculate the area of the triangle**:
   The area of the triangle formed by the x-axis, y-axis, and the line is given by:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
   \]
   Here, the base is the x-intercept (5 units), and the height is the y-intercept (12 units):
   \[
   \text{Area} = \frac{1}{2} \times 5 \times 12 = 30 \, \text{square units}
   \]

4. **Calculate the perimeter of the triangle**:
   The perimeter is the sum of the lengths of the sides of the triangle. The sides are the distances between the intercepts and the origin.
   - The distance between \( (0, 0) \) and \( (5, 0) \) is 5 units.
   - The distance between \( (0, 0) \) and \( (0, 12) \) is 12 units.
   - The distance between \( (5, 0) \) and \( (0, 12) \) is calculated using the Pythagorean theorem:
     \[
     \text{Distance} = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
     \]
   - The perimeter is:
     \[
     \text{Perimeter} = 5 + 12 + 13 = 30 \, \text{units}
     \]

### Final Answer:
- **Area** = 30 square units
- **Perimeter** = 30 units

Would you like further details on any part of the solution? Here are some related questions:

1. What is the geometric interpretation of perpendicular slopes?
2. How can you find the distance between two points using coordinates?
3. How does the area formula for triangles relate to coordinate geometry?
4. Can we generalize this method for lines with different slopes?
5. What if the point was outside the first quadrant—how would it affect the calculation?

**Tip**: When dealing with triangles formed by axes and lines, finding intercepts can quickly lead to the necessary dimensions for calculating area and perimeter.

Calcule l'aire et le périmètre du triangle formé par les axes des coordonnées et la droite d qui passe par le point (10, -12) et qui est perpendiculaire à la droite d'équation y = 5/12 * x + 10.

Learn how to calculate the area and perimeter of a triangle formed by coordinate axes and a line passing through a point, perpendicular to a given line equation. This problem covers key concepts in coordinate geometry, including line slopes and the Pythagorean theorem, suitable for students in Grades 9-12.

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