2x+4y+3=0.....(i)4x+3y+3=0.....(ii)x+1=0........(iii)
Solving (ii) and (iii) we get A(−1,13)
solving (iii) and (i) we get B(−1,−14)
solving (i) and (ii) we get C(−310,−35)
Distance BC=a
a=√(−310+1)2+(−35+14)2=7√520
Distance CA=b
b=√(−310+1)2+(−35−13)2=76
Distance AB=c
c=√(−1+1)2+(13+14)2=712
Coordiantes of incentre are
(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)
⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(7√520×−1)+(76×−1)+(712×−310)7√220+76+712⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭,⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(7√520×13)+(76×−14)+(712×−35)7√220+76+712⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
(−85−7√5120,21√5−65120)
Radius is equal to distance from x+1=0
=∣∣
∣
∣
∣∣−85−7√5120+1√12+02∣∣
∣
∣
∣∣
=35−7√5120