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We are tasked with determining two things for this problem:

**Given:**
- Town \( T \) has coordinates \( (80, 140) \).
- A road \( R \) follows the equation \( y = x - 80 \), which is a straight line.
- We are to determine the location of a new airport \( S \) along the road \( R \), such that the distance from the town \( T \) to \( S \) is the shortest possible.
- All coordinates are in kilometers.

### a) Determine the coordinates of \( S \), the new airport.
The shortest distance from point \( T \) to the line \( R \) is the perpendicular distance. To solve this, we need the equation of the line perpendicular to \( y = x - 80 \) that passes through \( T(80, 140) \).

1. **Find the slope of the line \( y = x - 80 \):**
   The slope of this line is \( 1 \), since the equation is in the form \( y = mx + c \), where \( m = 1 \).

2. **Find the slope of the perpendicular line:**
   The slope of a line perpendicular to another is the negative reciprocal of the original slope. Since the slope of \( y = x - 80 \) is \( 1 \), the slope of the perpendicular line is \( -1 \).

3. **Find the equation of the perpendicular line:**
   Using the point-slope form of a line, the equation of the line passing through \( (80, 140) \) with slope \( -1 \) is:
   \[
   y - 140 = -1(x - 80)
   \]
   Simplifying:
   \[
   y = -x + 220
   \]

4. **Find the intersection of this line with \( y = x - 80 \):**
   Set the two equations equal to each other:
   \[
   -x + 220 = x - 80
   \]
   Solve for \( x \):
   \[
   220 + 80 = 2x \quad \Rightarrow \quad 300 = 2x \quad \Rightarrow \quad x = 150
   \]
   Substitute \( x = 150 \) into \( y = x - 80 \) to find \( y \):
   \[
   y = 150 - 80 = 70
   \]
   Therefore, the coordinates of \( S \) are \( (150, 70) \).

### b) Find the distance between the town and the new airport.
The distance between \( T(80, 140) \) and \( S(150, 70) \) can be found using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substitute the values:
\[
d = \sqrt{(150 - 80)^2 + (70 - 140)^2} = \sqrt{70^2 + (-70)^2} = \sqrt{4900 + 4900} = \sqrt{9800} = 99 \, \text{km}
\]

### Answers:
- **a)** The coordinates of \( S \) are \( (150, 70) \).
- **b)** The distance between the town and the new airport is \( 99 \) km.

Would you like more details or have any questions?

### Related Questions:
1. How do you derive the formula for the perpendicular distance between a point and a line?
2. What are other methods to find the shortest distance from a point to a line in 2D geometry?
3. How would the answer change if the line had a different slope or intercept?
4. What is the geometric significance of the distance formula used here?
5. How does vector calculus approach the same problem of finding shortest distance?

**Tip:** When solving geometry problems involving distance, always visualize the problem with a diagram to ensure accuracy in reasoning.

A new airport, S, is to be constructed along a straight road R, such that its distance from a nearby town, T, is the shortest possible. The town T has coordinates (80, 140) and the road R has the equation y = x - 80. Determine the coordinates of the new airport S and the shortest distance between the town and the new airport.

Learn how to determine the coordinates of a new airport and calculate the shortest distance from a town to a road using coordinate geometry. This problem involves perpendicular distance and line equations, suitable for students in Grades 10-12.

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