Summary MAT1503 ASSIGNMENT 3 2024

MAT1503
ASSIGNMENT 3
2024

, QUESTION 1




Solution:



1.1).

Y
B(a, b)




C(c, d)
(x, y)A




0 F E D X




Area of triangle ABC = area of trapezium ABEF + area of trapezium BCDE − area of trapezium ACDF



1
area of trapezium ABEF = (y + b)(a − x)
2
1
= (ay − xy + ab − bx)
2


1
area of trapezium BCDE = (b + d)(c − a)
2
1
= (bc − ab + cd − ad)
2

, 1
area of trapezium ACDF = (y + d)(c − x)
2
1
= (cy − xy + cd − xd)
2


1 1 1
Area of triangle ABC = (ay − xy + ab − bx) + (bc − ab + cd − ad) − (cy − xy + cd − xd)
2 2 2
1
Area of triangle ABC = [ay − xy + ab − bx + bc − ab + cd − ad − cy + xy − cd + xd]
2
1
Area of triangle ABC = [ay − bx + bc − ad − cy + xd]
2
1
Area of triangle ABC = [xd − xb + ay − cy + bc − ad]
2
1
Area of triangle ABC = − [−xd + xb − ay + cy − bc + ad]
2
1
Area of triangle ABC = − [xb − xd + cy − ay + ad − bc]
2


If the three point are collinear then , Area of triangle is zero


1
Area of triangle ABC = − [xb − xd + cy − ay + ad − bc] = 0
2
1
− [xb − xd + cy − ay + ad − bc] = 0
2
xb − xd + cy − ay + ad − bc = 0



We can see that
b 1 a 1 a b
xb − xd + cy − ay + ad − bc = x | | − y| |+1⌈ ⌉
d 1 c 1 c d


xb − xd + cy − ay + ad − bc = 0
b 1 a 1 a b
x| | − y| | + 1⌈ ⌉=0
d 1 c 1 c d
x y 1
|a b 1| = 0 Shown
c d 1