Question

If (a, b), (c, d), (e, ƒ) are the vertices of a triangle such that a, c, e are in G.P. with common ratio r and b, d, ƒ are in G.P. with common ratio s, then the area of the triangle is-

• A. $\frac{ab}{2}$ (r + 1) (s + 2) (s + r)
• B. $\frac{ab}{2}$ (r – 1) (s – r) (s – 1)
• C. $\frac{ab}{2}$ (r – 1) (s + 1) (s – r)
• D. $\frac{ab}{2}$ (r + 1) (s + 1) (s – r)

Answer 1

Answer: (B)

$\frac{ab}{2}$ (r – 1) (s – r) (s – 1)

Answer 2

Here, c = ar, e = ar², d = bs, ƒ = bs²

\ D (Area of triangle) = $\frac { 1 } { 2 }$ $\left| \begin{matrix} a & b & 1 \\ c & d & 1 \\ e & ƒ & 1 \end{matrix} \right|$

= $\frac{1}{2}$ab (s – r) (s – 1) (r – 1).