The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k.
Let four vertices of quadrilateral be A (1, 2) and B (−5, 6) and C (7, −4) and D (k, −2)
Area of □ ABCD = Area of ∆ABC + Area of ∆ACD = 0 sq. unit
Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3)
= 
|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|
Area of ∆ABC
= 
|1(6 – (-4)) - 5(-4 -2) + 7(2 – 6)|
= 
|10 + 30 -28|
= 6 sq. units
Area of ∆ACD
= 
|1(-2 – (-4)) + k(-4 -2) + 7(2 – (-2))|
= 
|2 - 6k + 30|
= (3k -15) sq. units
Area of ∆ABC + Area of ∆ACD = 0 sq. unit
∴ 6 + 3k -15 =0
3k -9 =0
∴ k =3
Hence, the value of k is 3
3