Question

If line $x+\alpha y+1=0$ is perpendicular to the line $2x-\beta y+1=0$ and parallel to the line $x-(\beta -3)y-1=0$, then the value of $|\alpha -\beta |$ is

Answer 1

Let the lines be
$l_1:x+\alpha y+1=0l_2:2x-\beta y+1=0l_3:x-(\beta -3)y-1=0$
Slope of the lines are
$m_1=-\frac{1}{\alpha },m_2=\frac{2}{\beta },m_3=\frac{1}{\beta -3}$
From the given conditions,
$m_1\times m_2=-1\Rightarrow -\frac{2}{\alpha \beta }=-1\Rightarrow \alpha \beta =2\cdots \left(1\right)m_1=m_3\Rightarrow -\frac{1}{\alpha }=\frac{1}{\beta -3}\Rightarrow \alpha +\beta =3\cdots \left(2\right)$
Now,
$|\alpha -\beta |=\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }\Rightarrow |\alpha -\beta |=\sqrt{9-8}=1$