Mathematics - Straight Lines Question with Solution | TestHub

Given lines are $2\mathrm{x}-\mathrm{y}+3=0$, $6\mathrm{x}+3\mathrm{y}+1=0$ and $\mathrm{\alpha x}+2\mathrm{y}-2=0$.

These lines will not form a triangle if $\mathrm{\alpha x}+2\mathrm{y}-2=0$ is concurrent with $2\mathrm{x}-\mathrm{y}+3=0$ and $6\mathrm{x}+3\mathrm{y}+1=0$ or parallel to either of them.

Case-$1$: Concurrent lines

$\Rightarrow \left|\begin{array}{ccc}2& -1& 3\\ 6& 3& 1\\ \mathrm{\alpha }& 2& -2\end{array}\right|=0$

$\Rightarrow 2\left(-6-2\right)+1\left(-12-\alpha \right)+3\left(12-3\alpha \right)=0$

$\Rightarrow -16-12-\alpha +36-9\alpha =0$

$\Rightarrow -10\alpha +8=0$

$\Rightarrow \mathrm{\alpha }=\frac{4}{5}$

Case-$2$: Parallel lines

$\frac{-\alpha }{6}=\frac{-2}{3}$, $-\frac{\alpha }{2}=\frac{-2}{-1}$

$\Rightarrow \alpha =4$, $\alpha =-4$

$\Rightarrow \mathrm{P}=16+16+\frac{16}{25}$

$\left[\mathrm{P}\right]=\left[32+\frac{16}{25}\right]=32$