The product of the perpendicular from origin to the pair of lines 5x2+12xy+6y2+x+y−7=0is

The product of the perpendiculars drawn from the point P0,0to the pair of lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0 is ca−b2+4h2For the equation 5x2+12xy+6y2+x+y−7=0the product of the perpendiculars from the point 0,0 is dd=−75−62+122   =7145Therefore, the required distance is 7145