In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are $$x + y = 5$$ and $$x = 4$$ respectively. Then area of $$\triangle ABC$$ (in sq. units) is:

We have a triangle whose vertex $$A$$ is fixed at the point $$(1,\,2)$$. The median drawn from vertex $$B$$ is the straight line whose equation is $$x + y = 5$$, while the median drawn from vertex $$C$$ is the vertical line $$x = 4$$. Using the definition of a median, each of these lines must pass through the corresponding vertex and also through the midpoint of the opposite side. We therefore assign

$$B\,(b_1,\,b_2), \qquad C\,(c_1,\,c_2).$$

First we work with the median through $$B$$. The midpoint of side $$AC$$ is obtained from the midpoint formula:

$$M_1\left(\frac{1 + c_1}{2},\; \frac{2 + c_2}{2}\right).$$

Because both $$B$$ and $$M_1$$ lie on the line $$x + y = 5$$, we substitute their coordinates into that equation one by one.

For the vertex $$B$$ we get

$$b_1 + b_2 = 5 \qquad\qquad (1).$$

For the midpoint $$M_1$$ we have

$$\frac{1 + c_1}{2} + \frac{2 + c_2}{2} = 5.$$

Simplifying the left‐hand side:

$$\frac{(1 + c_1) + (2 + c_2)}{2} = 5 \;\Longrightarrow\; \frac{3 + c_1 + c_2}{2} = 5.$$

Multiplying by two we obtain

$$3 + c_1 + c_2 = 10 \;\Longrightarrow\; c_1 + c_2 = 7 \qquad\qquad (2).$$

Now we use the information about the median through $$C$$. Since that median is the line $$x = 4$$, any point on it has an $$x$$-coordinate equal to $$4$$. Therefore

$$c_1 = 4 \qquad\qquad (3).$$

The same line also contains the midpoint of side $$AB$$. The midpoint of $$AB$$ is

$$M_2\left(\frac{1 + b_1}{2},\; \frac{2 + b_2}{2}\right).$$

Its $$x$$-coordinate must satisfy $$\dfrac{1 + b_1}{2} = 4$$. Multiplying by two gives

$$1 + b_1 = 8 \;\Longrightarrow\; b_1 = 7 \qquad\qquad (4).$$

We substitute $$b_1 = 7$$ into equation (1):

$$7 + b_2 = 5 \;\Longrightarrow\; b_2 = -2.$$

Next we substitute $$c_1 = 4$$ into equation (2):

$$4 + c_2 = 7 \;\Longrightarrow\; c_2 = 3.$$

Thus the three vertices are now completely known:

$$A(1,\,2), \quad B(7,\,-2), \quad C(4,\,3).$$

To find the area of $$\triangle ABC$$ we apply the determinant formula for the area of a triangle formed by points $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$:

$$\text{Area} \;=\; \frac12 \,\left|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\right|.$$

Assigning $$A(1, 2) \to (x_1, y_1),\; B(7, -2) \to (x_2, y_2),\; C(4, 3) \to (x_3, y_3)$$ we calculate each term carefully:

$$y_2 - y_3 = -2 - 3 = -5,$$

$$x_1(y_2 - y_3) = 1 \times (-5) = -5,$$

$$y_3 - y_1 = 3 - 2 = 1,$$

$$x_2(y_3 - y_1) = 7 \times 1 = 7,$$

$$y_1 - y_2 = 2 - (-2) = 4,$$

$$x_3(y_1 - y_2) = 4 \times 4 = 16.$$

Adding these three products we obtain

$$-5 + 7 + 16 = 18.$$

Taking the absolute value (already positive) and dividing by two gives the required area:

$$\text{Area} = \frac12 \times 18 = 9 \text{ square units}.$$

Hence, the correct answer is Option B.