Find values of k if area of triangle is 4 sq. units and vertices are

(–2, 0), (0, 4), (0, k)

Given vertices of the triangle are (–2, 0), (0, 4), (0, k)

Let the vertices of the triangle be given by (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area of triangle is given by Δ =

Given, Area of triangle = Δ = 4 sq. units

= 4

⇒ 4 = 1/2 |[(-2) × (4 × 1 – k × 1) – 0 × (0 × 1 – 0 × 1) + 1 × (0 × k – 0 × 4)]|

⇒ 4 = 1/2 × |[(-2) × (4 – k) – 0 + 1 × (0 – 0)]|

⇒ 4 × 2 = |(-8 + 2k)

⇒ � 8= 2k - 8

⇒ 8 = 2k - 8 and ⇒ -8 = 2k - 8

⇒ 8 + 8 = 2k and ⇒ 8 - 8 = 2k

⇒ k = 16/2 and ⇒ k = 0/2

⇒ k = 8 and ⇒ k = 0