Mathematics - Point Question with Solution | TestHub

$∆=\frac{1}{2}\left|\begin{array}{ccc}0& 2& 1\\ 1& -1& 1\\ x\text{'}& y\text{'}& 1\end{array}\hspace{0.17em}\right|$
$\Rightarrow -2\left(1-x\text{'}\right)+\left(y\text{'}+x\text{'}\right)=\pm 10$
$\Rightarrow -2+2x\text{'}+y\text{'}+x\text{'}=\pm 10$
$\Rightarrow 3x\text{'}+y\text{'}=12$ Or $3x\text{'}+y\text{'}=-8 ...\left(i\right)$

$\left(x\text{'}, y\text{'}\right)$ lies on the line $3x+y-4\lambda =0$,

So, $3x\text{'}+y\text{'}-4\lambda =0...\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, we get,
$\lambda =3,\hspace{0.17em}-2$