Let the coordinates of the triangle be $A(1,2),\quad B(x_1,y_1),\quad C(x_2,y_2).$

Median through $B$ joins $B$ to the midpoint of $AC$.

Midpoint of $AC$ is $M_{AC}=\left(\frac{1+x_2}{2},\frac{2+y_2}{2}\right).$

Since the median through $B$ has equation $x+y=5,$ the midpoint $M_{AC}$ lies on this line. Hence $\frac{1+x_2}{2}+\frac{2+y_2}{2}=5$ $\frac{x_2+y_2+3}{2}=5$ $x_2+y_2=7. \qquad (1)$

Median through $C$ joins $C$ to the midpoint of $AB$.

Midpoint of $AB$ is $M_{AB}=\left(\frac{1+x_1}{2},\frac{2+y_1}{2}\right).$

Since the median through $C$ has equation $x=4,$ the midpoint $M_{AB}$ lies on this line. Therefore $\frac{1+x_1}{2}=4$ $1+x_1=8$ $x_1=7. \qquad (2)$

Also, point $B$ lies on its median $x+y=5$. So $x_1+y_1=5.$ Using $x_1=7$ from (2), $7+y_1=5$ $y_1=-2.$ Thus $B=(7,-2).$

Point $C$ lies on its median $x=4$, so $x_2=4.$ Using (1): $x_2+y_2=7$ $4+y_2=7$ $y_2=3.$ Hence $C=(4,3).$

Now compute the area of triangle $ABC$ with $A(1,2),\quad B(7,-2),\quad C(4,3).$

Using the determinant formula, $\text{Area}=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ $=\frac12\left|1(-2-3)+7(3-2)+4(2-(-2))\right|$ $=\frac12\left|-5+7+16\right|$ $=\frac12(18)=9.$

Therefore, the area is $\boxed{9}.$

Comparing with the stored correct answer $D$:

• Derived answer: $D$ ($9$)
• Stored answer: $D$
• They agree.