
d1:y−2=−1−1−2−2(x−1)
y−2=2(x−1)
y−2x=0
d2⊥d1⇒2y+x=k, (−1,−2) Lies on it ∴k=−5
⇒2y+x+5=0
Now point of intersection of d2 and L1
x−y+1=0......(1)
x+2y+5=0.......(2)
on solving (1)&(2)
−3y−4=0
y=−34
And point of intersection of d2 and L2
x+2y+5=0......(3)
14x−2y−10=0........(4)
On solving (3)&(4)
And y=−38
15x−5=0
⇒ x=31