The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. k.

Question

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.

Answer 1

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)

= \(\frac{1}2\) |x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)| 

Area of ∆ABC

= \(\frac{1}2\) |1(6 – (-4)) - 5(-4 -2) + 7(2 – 6)| 

= \(\frac{1}2\) |10 + 30 -28| 

= 6 sq. units 

Area of ∆ACD 

= \(\frac{1}2\) |1(-2 – (-4)) + k(-4 -2) + 7(2 – (-2))| 

= \(\frac{1}2\) |2 - 6k + 30| 

= \(\frac{1}2\) (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