Distance Calculation for R and T

This problem involves determining the shortest distance between two points, R and T, based on a series of directional movements. We need to carefully plot the positions of each point relative to the others.

Mapping Point Locations

Let's establish a reference point. We can assume Point R is at the origin (0,0) of a coordinate plane, where East is the positive x-axis and South is the positive y-axis (or negative y-axis, depending on convention; let's use North as positive y and East as positive x for standard mapping).

Here's how we can determine the relative positions:

• Point R: Position (0, 0)
• Point Q: 4m towards East of R. Position: (4, 0)
• Point P: 3m towards South of Q. Position: (4, -3)
• Point S: 6m towards South of R. Position: (0, -6)
• Point T: 8m towards West of S. Position: (-8, -6)
• Point U: 3m North of T. Position: (-8, -3)
• Point V: 7m towards West of U. Position: (-15, -3)

Finding Shortest Distance Between R and T

The shortest distance between two points in a plane is a straight line. We have the coordinates for R and T:

• R = (0, 0)
• T = (-8, -6)

To find the distance, we can use the distance formula derived from the Pythagorean theorem:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Here, $(x_1, y_1) = (0, 0)$ for point R, and $(x_2, y_2) = (-8, -6)$ for point T.

Applying the Pythagorean Theorem

The difference in the x-coordinates is $\Delta x = -8 - 0 = -8$ meters.

The difference in the y-coordinates is $\Delta y = -6 - 0 = -6$ meters.

The horizontal distance is 8m (West) and the vertical distance is 6m (South). These form the two legs of a right-angled triangle, with the shortest distance between R and T being the hypotenuse.

Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

Distance$^2 = (\Delta x)^2 + (\Delta y)^2$

Distance$^2 = (-8)^2 + (-6)^2$

Distance$^2 = 64 + 36$

Distance$^2 = 100$

Distance $= \sqrt{100}$

Distance $= 10$ meters

Therefore, the shortest distance between point R and point T is 10 meters.