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(C) Let the vertices of the triangle be $A (4, -6), B (3, -2),$ and $C (5, 2)$.Let $D$ be the mid-point of side $BC$ of $	riangle ABC$. Therefore,$AD

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$3$

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(C) Let the vertices of the triangle be $A (4, -6), B (3, -2),$ and $C (5, 2)$.Let $D$ be the mid-point of side $BC$ of $	riangle ABC$. Therefore,$AD$ is the median of $	riangle ABC$.Coordinates of point $D = \left(\frac{3+5}{2}, \frac{-2+2}{2}ight) = (4, 0)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area of $	riangle ABD = \frac{1}{2} |4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))|$$= \frac{1}{2} |4(-2) + 3(6) + 4(-4)|$$= \frac{1}{2} |-8 + 18 - 16| = \frac{1}{2} |-6| = 3$ square units.Area of $	riangle ADC = \frac{1}{2} |4(0 - 2) + 4(2 - (-6)) + 5(-6 - 0)|$$= \frac{1}{2} |4(-2) + 4(8) + 5(-6)|$$= \frac{1}{2} |-8 + 32 - 30| = \frac{1}{2} |-6| = 3$ square units.Since the area of $	riangle ABD = 	ext{Area of } 	riangle ADC = 3$ square units,the median $AD$ divides $	riangle ABC$ into two triangles of equal areas.

$A$ median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A (4, -6), B (3, -2)$ and $C (5, 2)$. Find the area of each triangle (in square units).

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