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)=0 So, 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=0 On 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=10 ‌3c1+4c2=11 On solving equations (iv) and (v), we get c1=1‌‌c2=2 Separate 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=‌