Find values of K, if the area of a triangle is 4 square units whose vertices are

(– 2,0), (0, 4) and (0, k)

Given: – Vertices of triangle are (– 2,0), (0, 4) and (0, k) and the area of the triangle is 4 sq. units.

Tip: – If vertices of a triangle are (x₁,y₁), (x₂,y₂) and (x₃,y₃), then the area of the triangle is given by:

Now,

Substituting given value in above formula

⇒

Removing modulus

⇒

Expanding along R₁

⇒

⇒ [ – 2(4 – k) – 0(0) + 1(0 – 0)] = ±8

⇒ – 8 + 2k = ± 8

Taking + ve sign, we get

⇒ 8 = – 8 + 2k

⇒ 2k = 16

⇒ k = 8

Taking – ve sign, we get

⇒ – 8 = 2x – 8

⇒ 2k = 0

⇒ k = 0

Thus k = 0, 8