In triangle ABC,if A=(1,2) and the equations | vedclass

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(B) Let the equations of the medians through $B$ and $C$ be $L_1: x+y=5$ and $L_2: x=4$ respectively. The point of intersection of these medians is th

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$9$

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(B) Let the equations of the medians through $B$ and $C$ be $L_1: x+y=5$ and $L_2: x=4$ respectively. The point of intersection of these medians is the centroid $G$. Solving $x+y=5$ and $x=4$,we get $G=(4, 1)$.Let $C$ lie on $x=4$,so $C=(4, y_C)$. Let $B$ lie on $x+y=5$,so $B=(x_B, 5-x_B)$.Using the centroid formula $G = (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$:$4 = \frac{1+x_B+4}{3}$ $\Rightarrow 12 = 5+x_B$ $\Rightarrow x_B=7$. Thus $B=(7, 5-7) = (7, -2)$.$1 = \frac{2+y_B+y_C}{3}$ $\Rightarrow 3 = 2+(-2)+y_C$ $\Rightarrow y_C=3$. Thus $C=(4, 3)$.The vertices are $A(1, 2)$,$B(7, -2)$,and $C(4, 3)$.The area of $	riangle ABC = \frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$$= \frac{1}{2} |1(-2-3) + 7(3-2) + 4(2-(-2))|$$= \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$.

In $\triangle ABC$,if $A=(1,2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively,then the area of $\triangle ABC$ is

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