Shortest Distance Between M and N: A Step-by-Step Solution

To determine the shortest distance between M and N at the end of their travel, we can effectively use a coordinate system. Let the common starting location for both M and N be the origin (0,0) of a Cartesian coordinate plane. This approach allows us to represent their movements as vectors and easily calculate their final positions.

We will calculate the final coordinates for M and N separately based on their given paths and then use the distance formula to find the shortest distance between their final positions.

M's Travel Path Analysis

M starts at the origin (0,0) and completes two distinct displacements:

1. First Displacement: 10 km East. This movement is directed purely along the positive x-axis.
2. Second Displacement: 10 km North-East. This movement implies a direction at an angle of 45 degrees from the East (positive x-axis). To find the coordinates, we need to decompose this displacement into its horizontal (x) and vertical (y) components.

Let's track M's coordinates throughout the journey:

Therefore, M's final coordinates, denoted as $M_f$, are $(10 + 5\sqrt{2}, 5\sqrt{2})$.

Using the approximate value $\sqrt{2} \approx 1.4142$ for calculations:

• $x_M = 10 + 5(1.4142) = 10 + 7.071 = 17.071$
• $y_M = 5(1.4142) = 7.071$

So, $M_f \approx (17.071, 7.071)$.

N's Travel Path Analysis

N also starts at the origin (0,0) and completes two displacements:

1. First Displacement: 5 km South. This movement is purely along the negative y-axis.
2. Second Displacement: 4 km South-East. This movement implies a direction at an angle of -45 degrees (or 315 degrees counter-clockwise from the positive x-axis). Similar to M's second displacement, we resolve this into x and y components.

Let's track N's coordinates throughout the journey:

Therefore, N's final coordinates, denoted as $N_f$, are $(2\sqrt{2}, -5 - 2\sqrt{2})$.

Using the approximate value $\sqrt{2} \approx 1.4142$ for calculations:

• $x_N = 2(1.4142) = 2.828$
• $y_N = -5 - 2(1.4142) = -5 - 2.828 = -7.828$

So, $N_f \approx (2.828, -7.828)$.

Calculating Shortest Distance Between M and N

Now that we have the final coordinates for both M and N, we can calculate the shortest distance between their final positions using the distance formula. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Using M's final position $(x_M, y_M) = (10 + 5\sqrt{2}, 5\sqrt{2})$ and N's final position $(x_N, y_N) = (2\sqrt{2}, -5 - 2\sqrt{2})$:

\[ d = \sqrt{((2\sqrt{2}) - (10 + 5\sqrt{2}))^2 + ((-5 - 2\sqrt{2}) - (5\sqrt{2}))^2} \]

\[ d = \sqrt{(2\sqrt{2} - 10 - 5\sqrt{2})^2 + (-5 - 2\sqrt{2} - 5\sqrt{2})^2} \]

\[ d = \sqrt{(-10 - 3\sqrt{2})^2 + (-5 - 7\sqrt{2})^2} \]

Since the square of a negative number is positive, $(-a)^2 = a^2$, we can simplify:

\[ d = \sqrt{(10 + 3\sqrt{2})^2 + (5 + 7\sqrt{2})^2} \]

Expand each squared term using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

• $(10 + 3\sqrt{2})^2 = 10^2 + 2(10)(3\sqrt{2}) + (3\sqrt{2})^2 = 100 + 60\sqrt{2} + (9 \cdot 2) = 100 + 60\sqrt{2} + 18 = 118 + 60\sqrt{2}$
• $(5 + 7\sqrt{2})^2 = 5^2 + 2(5)(7\sqrt{2}) + (7\sqrt{2})^2 = 25 + 70\sqrt{2} + (49 \cdot 2) = 25 + 70\sqrt{2} + 98 = 123 + 70\sqrt{2}$

Substitute these expanded values back into the distance formula:

\[ d = \sqrt{(118 + 60\sqrt{2}) + (123 + 70\sqrt{2})} \]

\[ d = \sqrt{118 + 123 + 60\sqrt{2} + 70\sqrt{2}} \]

\[ d = \sqrt{241 + 130\sqrt{2}} \]

Now, substitute the approximate value of $\sqrt{2} \approx 1.41421356$ for precision:

\[ d \approx \sqrt{241 + 130 \times 1.41421356} \]

\[ d \approx \sqrt{241 + 183.8477628} \]

\[ d \approx \sqrt{424.8477628} \]

\[ d \approx 20.611835 \]

Final Distance Result

Rounding the result to two decimal places, the shortest distance between M and N at the end of their travel is approximately 20.61 km.