Mathematics - Straight Lines Question with Solution | TestHub

As line passes through the point of intersection of $x-2y-2=0$ and $2x-$ by $-6=0$

It can be represented as $\lambda (x-2y-2)+(2x-$ by -6$)=0$

As it passes through the origin

$-2\lambda -6=0$

$\lambda =-3$

∴ equation of the line is $-\mathrm{x}+(6-\mathrm{b})\mathrm{y}=0$

Its slope is $\frac{1}{6-b}$

As its angle with $y=0$ is less than $\frac{\pi }{4}$

$\therefore -1<\frac{1}{6-\mathrm{b}}<1$

$\Rightarrow 6-\mathrm{b}>1$ or $<-1\Rightarrow \mathrm{b}<5$ or $\mathrm{b}>7$

But $\mathrm{b}\ne 4$ (as the lines intersect)

$\therefore \mathrm{b}\in (-\infty ,4)\cup (4,5)\cup (7,\infty )$