Find the values of k so that area of triangle with vertices (1, – 1) (– 4, 2k) (– k, – 5) is 24 Sq units.

Question

Philoid · Accepted Answer

When the coordinates of vertices of triangle are

(a₁,b₁);(a₂,b₂);(a₃,b₃) then ;

Area = [a₁(b₂ – b₃) – a₂(b₃ – b₁) + a₃(b₁ – b₂)]

hence,

1(2k + 5) + 4( – 5 + 1) – k( – 1 – 2k) = 24

⇒ 2k + 5 – 16 + k + 2k² = 24

⇒ 2k² + 3k – 35 = 0

⇒ 2k² + 10k – 7k – 35 = 0

⇒ 2k(k + 5) – 7(k + 5) = 0

⇒ (2k – 7)(k + 5) = 0

So, k = 7/2 or k = – 5