Two lines are intersecting at the point $\left(1,1\right)$ . If  $2\mathrm{x}+3\mathrm{y}-5=0$ is the equation of one angular bisector of the two lines and the equation of the other bisector is $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$  then  $\mathrm{l}+\mathrm{m}+\mathrm{n}$ is 

detailed solution

Correct option is D

The angular bisector of any pair of lines are passing through the point of intersection of those two lines and the angular bisectors are mutually perpendicular to each other. The other angular bisector is a line passing through 1,1 and perpendicular to the line 2x+3y−5=0  The equation of any line passing through x1,y1 and perpendicular to the line ax+by+c=0 is  bx−x1−ay−y1=0Therefore, the equation of the other angular bisector is  3x−1−2y−1=0Rewrite the equation as  3x−2y−1=0Therefore, l+m+n=3−2−1=0