In a triangle ABC,the coordinates of A are (1 | vedclass

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(B) The median through $C$ is $x = 4$. Since $C$ lies on this median,let $C = (4, y)$.The midpoint $D$ of $AC$ is $D = (\frac{1+4}{2}, \frac{2+y}{2})

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$9$

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(B) The median through $C$ is $x = 4$. Since $C$ lies on this median,let $C = (4, y)$.The midpoint $D$ of $AC$ is $D = (\frac{1+4}{2}, \frac{2+y}{2}) = (2.5, \frac{2+y}{2})$.Since $D$ lies on the median through $B$ $(x + y = 5)$,we have $2.5 + \frac{2+y}{2} = 5$.$2.5 + 1 + \frac{y}{2} = 5$ $\Rightarrow \frac{y}{2} = 1.5$ $\Rightarrow y = 3$. Thus,$C = (4, 3)$.The centroid $G$ is the intersection of the medians $x = 4$ and $x + y = 5$. Substituting $x = 4$ into $x + y = 5$,we get $4 + y = 5$,so $y = 1$. Thus,$G = (4, 1)$.To find $B$,we use the centroid formula $G = (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$.$4 = \frac{1+x_B+4}{3}$ $\Rightarrow 12 = 5 + x_B$ $\Rightarrow x_B = 7$.$1 = \frac{2+y_B+3}{3}$ $\Rightarrow 3 = 5 + y_B$ $\Rightarrow y_B = -2$. Thus,$B = (7, -2)$.The area of $\Delta ABC$ with vertices $A(1, 2)$,$B(7, -2)$,and $C(4, 3)$ is given by:Area $= \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$Area $= \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 - (-2))|$Area $= \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$ sq. units.

In a triangle $ABC$,the 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 the area of $\Delta ABC$ (in sq. units) is

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