A point P is in equilibrium under the action of four forces. Three of the forces are 2 N at 45 degrees to the left of the vertical, 1 N at 45 degrees to the right of the vertical, and F1 N horizontall

The point P is in equilibrium, meaning the net force in both the horizontal (x) and vertical (y) directions is zero.
Let's define the positive x-direction as to the right and the positive y-direction as upwards.

The forces acting on point P are:

• A 2 N force acting at $45^\circ$ to the left of the vertical.
• A 1 N force acting at $45^\circ$ to the right of the vertical.
• Force $F_1$ acting horizontally to the right.
• Force $F_2$ acting vertically downwards.

We use $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\cos(45^\circ) = \frac{1}{\sqrt{2}}$.

The sum of forces in the x-direction must be zero ($\sum F_x = 0$).

• x-component of 2 N force: $-2 \sin(45^\circ)$ (negative because it's to the left).
• x-component of 1 N force: $+1 \sin(45^\circ)$ (positive because it's to the right).
• x-component of $F_1$: $+F_1$ (positive because it's to the right).
• x-component of $F_2$: 0.

$$\sum F_x = -2 \sin(45^\circ) + 1 \sin(45^\circ) + F_1 = 0$$ $$-2 \left(\frac{1}{\sqrt{2}}\right) + 1 \left(\frac{1}{\sqrt{2}}\right) + F_1 = 0$$ $$-\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} + F_1 = 0$$ $$-\frac{1}{\sqrt{2}} + F_1 = 0$$ $$F_1 = \frac{1}{\sqrt{2}} \text{ N}$$

The sum of forces in the y-direction must be zero ($\sum F_y = 0$).

• y-component of 2 N force: $+2 \cos(45^\circ)$ (positive because it's upwards).
• y-component of 1 N force: $+1 \cos(45^\circ)$ (positive because it's upwards).
• y-component of $F_1$: 0.
• y-component of $F_2$: $-F_2$ (negative because it's downwards).

$$\sum F_y = 2 \cos(45^\circ) + 1 \cos(45^\circ) - F_2 = 0$$ $$2 \left(\frac{1}{\sqrt{2}}\right) + 1 \left(\frac{1}{\sqrt{2}}\right) - F_2 = 0$$ $$\frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} - F_2 = 0$$ $$\frac{3}{\sqrt{2}} - F_2 = 0$$ $$F_2 = \frac{3}{\sqrt{2}} \text{ N}$$

We found $F_1 = \frac{1}{\sqrt{2}}$ N and $F_2 = \frac{3}{\sqrt{2}}$ N.
This matches option (a).

The final answer is $\text{(a)}$.