Question

The real value of $\lambda$ for which the image of the point $(\lambda ,\lambda -1)$ with respect to the line $3x+y=6\lambda$ is the point $({\lambda }^2+1,\lambda )$ is

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

Let $(\lambda ,\lambda -1)$ be $P$,
$({\lambda }^2+1,\lambda )$ be $Q$
and $R$ be the point of intersrection of $PQ$ and $L$

Here $Q$ is the image of the point $P$
So $PQ\perp L$ and $PR=QR$
$\Rightarrow$ slope of $PQ$ $\times$ slope of $L=-1$
$\Rightarrow \left(\frac{\lambda -(\lambda -1)}{{\lambda }^2+1-\lambda }\right)\times (-3)=-1$
$\Rightarrow {\lambda }^2-\lambda +1=3$
$\Rightarrow \lambda =-1,2$
But this is not sufficient, we need $PR=QR$
For $\lambda =2$
$P\equiv (2,1),Q\equiv (5,2)$
$R\equiv (\frac{7}{2},\frac{3}{2})$
Putting value of $R$ in equation of $L$
$R$ satisfies the equation of $L$

For $\lambda =-1$
$P\equiv (-1,-2),Q\equiv (2,-1)$
$R\equiv (\frac{1}{2},-\frac{3}{2})$
Putting value of $R$ in equation of $L$
$R$ does not satisfies the equation of $L$

So $\lambda =2$