The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0.The equation of the other line is

The family of line through the given lines is L≡x−7y+5+λ(x+3y−2)=0 =x(1+λ)+y(-7+3λ)+5-2λ-----(1)let  (2, 0) be a point on x+3y-2=0 which is equidistant from  x - 7y + 5 = 0 and the line L = 0Therefore,2-0+512+72=2(1+λ)+0(-7+3λ)+5-2λ1+λ2+(-7+3λ)22+550=2+2λ+5−2λ(1+λ)2+(3λ−7)2 or 10λ2−40λ=0 i.e., λ=4 or 0 Hence, L=0,λ=4. Therefore, the required line is 5x+5y−3=0 .