Let A ( a , 0), B ( b , 2 b +1) and C (0, b ), b ≠ 0,| b | ≠ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :

Question

Let A ( a , 0), B ( b , 2 b +1) and C (0, b ), b ≠ 0,| b | ≠ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is : (A) (-2 b/b+1) (B) (2 b / b +1) (C) (2 b2/b+1) (D) (-2 b2/b+1)

Tardigrade Admin · Accepted Answer

|(1/2) | a 0 1 b 2 b +1 1 0 b 1| |=1 ⇒ || a 0 1 b 2 b +1 1 0 b 1 ||=± 2 ⇒ a (2 b +1- b )-0+1( b 2-0)=± 2 ⇒ a =(± 2- b 2/ b +1) ∴ a =(2- b 2/ b +1) and a =(-2- b 2/ b +1) sum of possible values of 'a' is =(-2 b2/a+1)

Q. Let $A (a, 0), B (b, 2 b + 1)$ and $C (0, b), b \neq = 0, ∣ b ∣ \neq = 1$ , be points such that the area of triangle $A BC$ is $1 s q$ . unit, then the sum of all possible values of a is :

$\frac{- 2 b}{b + 1}$

$\frac{2 b}{b + 1}$

$\frac{2 b ^{2}}{b + 1}$

$\frac{- 2 b ^{2}}{b + 1}$

Q. Let A(a,0),B(b,2b+1) and C(0,b),b=0,∣b∣=1, be points such that the area of triangle ABC is 1sq. unit, then the sum of all possible values of a is :

b+1−2b​

b+12b​

b+12b2​

b+1−2b2​

∣∣​21​∣∣​ab0​02b+1b​111​∣∣​∣∣​=1 ⇒∣∣​∣∣​ab0​02b+1b​111​∣∣​∣∣​=±2 ⇒a(2b+1−b)−0+1(b2−0)=±2 ⇒a=b+1±2−b2​ ∴a=b+12−b2​ and a=b+1−2−b2​ sum of possible values of 'a' is =a+1−2b2​

Q. Let $A (a, 0), B (b, 2 b + 1)$ and $C (0, b), b \neq = 0, ∣ b ∣ \neq = 1$ , be points such that the area of triangle $A BC$ is $1 s q$ . unit, then the sum of all possible values of a is :

$\frac{- 2 b}{b + 1}$

$\frac{2 b}{b + 1}$

$\frac{2 b ^{2}}{b + 1}$

$\frac{- 2 b ^{2}}{b + 1}$