The vertices of a ΔABC are A(4,6), B(1,5), and C(7,2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the ΔADE and compare it with the area of ΔABC.

In ΔABC, A(4,6), B(1,5), and C(7,2)
Area of ΔABC = 1/2[x1(y2−y3) + x2(y3−y1) + x3(y1−y2)] = 1/2[4(5−2) + 1(2−6) + 7(6−5)] = 1/2[12 − 4 + 7] = 15/2 sq. unit
A line is drawn to intersect sides AB and AC at D and E. Then AD/AB = 1/4
The point D divides the line AB in AD and DB. The ratio of AD and DB = m1:m2 = 1:3
∴ Coordinates of D = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
Here, (x1, y1) = (4, 6) and (x2, y2) = (1, 5) = (1×1 + 4×3/1 + 3, 1×5 + 3×6/3 + 1) = (13/4, 23/4)
The point E divides the line AC in AE and EC. The ratio of AE and EC = m1:m2 = 1:3
∴ Coordinates of E = (m1x2 + m2x1/m1 + m2, m1y2 + m2y1/m1 + m2)
Here, (x1, y1) = (4, 6) and (x2, y2) = (7, 2) = (1×7 + 3×4/1 + 3, 1×2 + 3×6/3 + 1) = (19/4, 20/4)
Now the area of ΔADE = 1/2[4(23/4 − 20/4) + 13/4(20/4 − 6) + 19/4(6 − 23/4)] = 1/2[4(3/4) + 13/4( −4/4) + 19/4(1/4)] = 1/2[3 − 13/4 + 19/16] = 1/2[48 − 52 + 19/16] = 1/2 × 15/16 = 15/32 sq. unit
Hence, Area(ΔADE) : Area (ΔABC) = 15/32 : 15/2 = 1/16