Pair of straight lines through A(1,1) are drawn to intersect the line 2x+4y=5 at B and C . If angle between the pair
of straight line is π3 , then the locus of incentre of ΔABC is
11x2−y2−16xy−10x+10y−5=0
11y2−x2+16xy−10y+20=0
11x2+y2−16xy−10x+10y+5=0
11y2−x2+16xy−10x−30y+15=0
Let the co-ordinates of In-centre are I(h, k)
∴ IM=INIn △AMI sin 300=IMAI⇒ AIsin30∘=IN⇒ (h−1)2+(k−1)212=2h+4k−520⇒ 11k2−h2+16hk−10h−30k+15=0⇒ Locus is 11y2−x2+16xy−10x−30y+15=0