EAMCET Engineering 2006 Solved Paper

The equation of lines are
x−3y+2=0 . . . (i)
and 2x+5y−7=0 . . . (ii)
On solving Eqs. (i) and (ii), we get
x=1,y=1
∴ The co-ordinates of a point of intersection of given lines are (1,1).
The equation of line perpendicular to 3x+2y+5=0 is 2x−3y+λ=0.
Which passes through (1,1).
∴‌2−3+λ=0
⇒λ=1
∴ Required equation of line is
2x−3y+1=0