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  4. A straight line L: 4 x-4 y+3=0 is rotated in clockwise about the point where the line cuts the y-axis and a circle S 1 whose centre is (λ, (3/4)) touches both the line L and L 1 ( L1 is the line obtained after rotation) and the x-axis. If area of the triangle formed by the lines L1, angle bisector between L L1 and the x-axis is (p/q), p q ∈ R then least value of ( p + q ) equals

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Q. A straight line $L: 4 x-4 y+3=0$ is rotated in clockwise about the point where the line cuts the $y$-axis and a circle $S _{1}$ whose centre is $\left(\lambda, \frac{3}{4}\right)$ touches both the line $L$ and $L _{1}$ ( $L_{1}$ is the line obtained after rotation) and the $x$-axis.
If area of the triangle formed by the lines $L_{1}$, angle bisector between $L \,\& \,L_{1}$ and the $x$-axis is $\frac{p}{q}, p q \in R$ then least value of $( p + q )$ equals

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Solution:

Using $p = r$
(i)$\frac{4 \lambda-3+3}{4 \sqrt{2}}=r=\frac{3}{4} $
$\Rightarrow \frac{\lambda}{\sqrt{2}}=\frac{3}{4} $
$\therefore \lambda=\frac{3}{2 \sqrt{2}} $
$\Rightarrow [\lambda]=1$
image
(ii) $\Delta OAB =\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4}$
$=\frac{9}{32} \equiv \frac{ p }{ q }$
$\therefore (p+q)=41$