Intersection of 2x+y=0&x-y=3 is A(1,−2)
Perpendicular bisector of AB is y−3=21(x−2)⇒x−2y+4=0
Take image of A along the perpendicular bisector of AB, we get
1x−1=−2y+2=5−2(9)=−518
i.e. x=−513,y=526
B(−513,526)
This point lies on the line x+py=39
i.e. −513+p(526)=39⇒−1+2p=15
⇒p=8
Solving x+8y=39 and x−y=3, we get C(7,4)
Now (AC)2=(1−7)2+(−2−4)2=72=9p
and AC2+p2=72+64=136
Also ΔABC=21∣175−13−24526111∣
=21[18+18×513]
=9[518]=5162=32.4∈/(34,38)