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