Question

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

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

Answer 1

Answer: (B)

(r – 1) (s – 1) (s – r)

Answer 2

a, c, e are in G.P with common ratio = r then

c = ar, e = ar² b, d, f are in G.P with common ratio = s then d = bs, f = bs^{2 .} Area of D formed by the points (a, b) (c, d) (e, f)

= $\frac { 1 } { 2 }$ = $\frac { 1 } { 2 }$

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