You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A(4,-6), B(3,-2), and C(5,2).

Let AD be the median of ΔABC. Hence, D is the midpoint of BC. We know that the coordinates of the midpoint of the line segment joining (x1, y1) and (x2, y2) are: P(x, y) = (x1+x2/2, y1+y2/2)
∴ Coordinates of D = (3+5/2, -2+2/2) = (4, 0)
Median AD divides the ΔABC into two triangles.
∴ Area of ΔABD = |1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]|
Here, (x1, y1) = (4, -6), (x2, y2) = (3, -2) and (x3, y3) = (4, 0)
Hence, area of ΔABD = 1/2[4(-2-0) + 3(0+6) + 4(-6+2)] = 1/2[-8 + 18 -16] = |-6/2| = 3
Also, area of ΔADC = |1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]|
Here, (x1, y1) = (4, -6), (x2, y2) = (4, 0) and (x3, y3) = (5, 2)
Hence, area of ΔADC = 1/2[4(0+2) + 4(2+6) + 5(-6-0)] = 1/2[8 + 32 - 30] = |-6/2| = 3
∴ Area of ΔABD = Area of ΔADC = 3
Hence, it is proved that a median of a triangle divides it into two triangles of equal areas.