$Pair of straight lines through A (1, 1) are drawn to intersect the line 2 x + 4 y = 5 at B and C . If angle between the pair$
$of straight line is \frac{π}{3}, then the locus of incentre of Δ A B C is$

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$11 x^{2} + y^{2} - 16 x y - 10 x + 10 y + 5 = 0$

$11 x^{2} - y^{2} - 16 x y - 10 x + 10 y - 5 = 0$

$11 y^{2} - x^{2} + 16 x y - 10 x - 30 y + 15 = 0$

$11 y^{2} - x^{2} + 16 x y - 10 y + 20 = 0$

answer is D.

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Detailed Solution

Let the co-ordinates of In-centre are I(h, k)

$\begin{array}{l} ∴ I M = I N \\ I n △ A M I \sin 30^{0} = \frac{I M}{A I} \\ \Rightarrow A I \sin 30^{\circ} = I N \\ \Rightarrow \sqrt{(h - 1)^{2} + (k - 1)^{2}} \frac{1}{2} = |\frac{2 h + 4 k - 5}{\sqrt{20}}| \\ \Rightarrow 11 k^{2} - h^{2} + 16 h k - 10 h - 30 k + 15 = 0 \\ \Rightarrow Locus is 11 y^{2} - x^{2} + 16 x y - 10 x - 30 y + 15 = 0 \end{array}$