The equation of the straight line passing through the point (4, 5) and making equal angles with the two straight lines given by the equations 3x – 4y – 7 = 0 and 12x – 5y + 6 = 0, is

The equation of the bisectors of the angles between the given lines are3x-4y-732+(-4)2=±12x-5x+6122+-52⇒ (39x – 52y – 91) = ± (60x – 25y + 30).Taking the negative sign,we get 99x – 77y – 61 = 0                               (1)Taking the positive sign,we get 21x + 27y + 121 = 0                             (2)The required lines are the lines through (4, 5) and parallel to the bisectors of the angles between the given lines.The equation of a line parallel to (1), is99x – 77y + k1 = 0                                             (3)This passes through (4, 5),∴ 99 (4) – 77 (5) + k1 = 0 ⇒ k1 = –11.Putting k1 = –11 in (3), we get99x – 77y – 11 = 0 or 9x – 7y – 1 = 0.This is one of the required lines.The equation of a line parallel to (2), is21x + 27y + k2 = 0                                             (4)This passes through (4, 5),∴ 84 + 135 + k2 = 0 ⇒ k2 = –219.Putting k2 = –219 in (4), we get21x + 27y – 219 = 0 or 7x + 9y – 73 = 0,which is the other required line.