Given that
12x2+25xy+12y2+10x+11y+2=0 . . . (i)
First we take homogeneous part of Eq. (i),
i.e.
12x2+25xy+12y2=0⇒(3x+4y)(4x+3y)=0So, let the lines represented by Eq. (i) be
3x+4y+c1=0 . . . (ii)
4x+3y+c2=0 . . . (iii)
The combined Eqs. of (ii) and (iii), we get
(3x+4y+c1)(4x+3y+c2)=0⇒(3x+4y)(4x+3y)+c1(4x+3y)+c2(3x+4y)+c1c2=0⇒12x2+25xy+12y2+(4c1+3c2)x+(3c1+4c2)y+c1c2=0On comparing the equation with Eq. (i), we get
4c1+3c2=10 . . . (iv)
3c1+4c2=11 . . . (v)
On solving equations (iv) and (v), we get
4c1+3c2=103c1+4c2=11On solving equations (iv) and (v), we get
c1=1 c2=2Separate equation of lines are
3x+4y+1=0 . . . (vi)
and
4x+3y+2=0 . . . (vii)
The perpendicular distance from origin to the equations (vi) and (vii) are
p1== and p2==∴p1⋅p2=⋅=