Find the area of the triangle ABC, coordinates of whose vertices are A(4,1), B(6,6) and C(8,4).

A triangle contains three line segments AB, BC and CA, so let's find the equations for these line segments.
Let us first find for A(4,1) and B(6,6)
Equation of line AB is given by
y - 1/x - 4 = 6 - 1/6 - 4 = 5/2
=> 2(y - 1) = 5(x - 4)
=> 2y - 2 = 5x - 20
=> 2y = 5x - 18
=> y = (5x - 18)/2 (i)
Let's find for B(6,6), C(8,4)
Equation of line BC is given by
y - 6/x - 6 = 4 - 6/8 - 6 = -2/2 = -1
=> y - 6 = -1(x - 6)
=> y - 6 = -x + 6
=> y = -x + 12 (ii)
Let's find for C(8,4), A(4,1)
Equation of line CA is given by
y - 4/x - 8 = 1 - 4/4 - 8 = -3/-4 = 3/4
=> 4(y - 4) = 3(x - 8)
=> 4y - 16 = 3x - 24
=> 4y = 3x - 8
=> y = (3x - 8)/4 (iii)
Area of triangle ABC is given by
A(ΔABC) = ∫64 [(5x - 18)/2] dx + ∫86 (-x + 12) dx - ∫84 [(3x - 8)/4] dx
= 1/2 ∫64 (5x - 18) dx + ∫86 (-x + 12) dx - 1/4 ∫84 (3x - 8) dx
= 1/2 [5x²/2 - 18x]64 + [-x²/2 + 12x]86 - 1/4 [3x²/2 - 8x]84
= 1/2 [5(36 - 16)/2 - 18(6 - 4)] + [- (64 - 36)/2 + 12(8 - 6)] - 1/4 [3(64 - 16)/2 - 8(8 - 4)]
= 1/2 [5(20)/2 - 18(2)] + [-(28)/2 + 12(2)] - 1/4 [3(48)/2 - 8(4)]
= 1/2 [50 - 36] + [-14 + 24] - 1/4 [72 - 32]
= 1/2 (14) + 10 - 1/4 (40)
= 7 + 10 - 10
= 7
Hence, A(ΔABC) = 7 units