Answer:
By cross multiplication method
x=
(a
1
b
2
−a
2
b
1
)
(b
1
c
2
−b
2
c
1
)
&
y=
(a
1
b
2
−a
2
b
1
)
(c
1
a
2
−c
2
a
1
)
Where a
1
x+b
1
y+c
1
=0 & a
2
x+b
2
y+c
2
=0
are 2 lines
⇒x=
2(−2)−(3×3)
3(−6)−(−2)(−17)
=
−13
−52
=4
y=
2(−2)−(3×3)
(−17)3−(−6)2
=
−13
−39
=3
⇒
x=4&y=3
