If area of triangle is 35 sq units with vertices (2, –6), (5, 4) and (k, 4). Then k is

Given vertices of the triangle are (2, – 6), (5, 4) and (k, 4).

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 = 35 sq. units

= 35

⇒ � 35 = 1/2 × [2 × (4 × 1 – 4 × 1) – (-6) × (5 × 1 – k × 1) + 1 × (5 × 4 – k × 4)]

⇒ � 35 = 1/2 × [2 × (4 – 4) + 6 × (5 – k) + 1 × (20 – 4k)]

⇒ � 35 × 2 = (2 × 0 + 30 – 6k + 20 – 4k)

⇒ � 70 = 30 – 6k + 20 – 4k

⇒ � 70 = 50 – 10k

⇒ 70 – 50 = -10k and ⇒ -70 – 50 = -10k

⇒ 20 = -10k and ⇒ -120 = -10k

⇒ k = -20/10 and ⇒ k = 120/10

⇒ k = -2 and ⇒ k = 12