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 1 sq. unit, then the sum of all possible values of a is:

Correct answer: C. - 2 b ^ 2 b + 1 Given coordinates of triangle ABC , A ( a, 0 ) , B ( b, 2b + 1 ) , C ( 0, b ) Area = (1)/(2)| cases a & 0 & 1 b & 2b + 1 & 1 0 & b & 1 cases | (1)/(2)[a ( b + 1 ) + b ^ 2 ] = ± 1 ab + a + b ^ 2 = ± 2 If 2 is positive a = 2 - b ^ 2 b + 1 If 2 is negative a = - 2 - b ^ 2 b + 1 Sum of possible values of a = 2 - b ^ 2 b + 1 + - 2 - b ^ 2 b + 1 = - 2 b ^ 2 b + 1

JEE Main 202127 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2021 · 27 August · Evening Shift · Mathematics · Straight Lines

Let A ( a, 0 ) , B ( b, 2b + 1 ) and C ( 0, b ) , b ≠ 0, | b | ≠ 1...

Let

A ( a, 0 ) , B ( b, 2b + 1 )

and

C ( 0, b ) , b \ne 0, | b | \ne 1

, be points such that the area of triangle

ABC

1

sq. unit, then the sum of all possible values of

a

is:

Options

A $\dfrac{- 2b}{b + 1}$
B $\dfrac{2{b}^{2}}{b + 1}$
C $\dfrac{- 2{b}^{2}}{b + 1}$
D $\dfrac{2b}{b + 1}$

Correct answer

C. $\dfrac{- 2{b}^{2}}{b + 1}$

Step-by-step solution

Given coordinates of triangle $ABC$ , $A ( a, 0 ) , B ( b, 2b + 1 ) , C ( 0, b )$

Area $= \dfrac{1}{2}\left|$ \begin{cases}a & 0 & 1 \\ b & 2b + 1 & 1 \\ 0 & b & 1\end{cases} $\right|$

$\dfrac{1}{2}\left[a ( b + 1 ) + {b}^{2}\right] = \pm 1$

$ab + a + {b}^{2} = \pm 2$

If $2$ is positive

$a = \dfrac{2 - {b}^{2}}{b + 1}$

If $2$ is negative

$a = \dfrac{- 2 - {b}^{2}}{b + 1}$

Sum of possible values of $a = \dfrac{2 - {b}^{2}}{b + 1} + \dfrac{- 2 - {b}^{2}}{b + 1}$

$= \dfrac{- 2{b}^{2}}{b + 1}$

Practice Straight Lines on Quantrex Academy →

More from Straight Lines