1. Establish Coordinates

We can solve this problem using a 2D coordinate system. Let point Z be at the origin (0, 0).

• W is 1 km west of Z. Coordinates of W: (-1, 0).
• P is 1 km south of W. Coordinates of P: (-1, -1).
• Q is 1 km east of P. Coordinates of Q: (-1 + 1, -1) = (0, -1).

2. Determine Coordinates of Y and X

Directions like northeast and southeast imply movement at a 45° angle. We use trigonometry for these displacements.

• Y is 1 km southeast of Z (0, 0). The displacement vector is $(1 \cdot \cos(-45^\circ), 1 \cdot \sin(-45^\circ)) = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Coordinates of Y: $(0 + \frac{1}{\sqrt{2}}, 0 - \frac{1}{\sqrt{2}}) = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.
• X is 1 km northeast of Y. The displacement vector from Y is $(1 \cdot \cos(45^\circ), 1 \cdot \sin(45^\circ)) = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Coordinates of X = Coordinates of Y + Displacement vector X = $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$ X = $(\frac{2}{\sqrt{2}}, 0) = (\sqrt{2}, 0)$.

3. Calculate Distance Between X and Q

Now we find the distance between X $(\sqrt{2}, 0)$ and Q (0, -1) using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$d = \sqrt{(0 - \sqrt{2})^2 + (-1 - 0)^2}$

$d = \sqrt{(-\sqrt{2})^2 + (-1)^2}$

$d = \sqrt{2 + 1}$

$d = \sqrt{3}$ km.

The distance between X and Q is $\sqrt{3}$ km.