If the line `x-3y+5+ k(x+y-3)=0`, is perpendicular to the line `x+y=1`, and `k`.

The value of k such that the line 2x+3y+4+k(6x-y+12)=0 is perpendicular to the line 7x+5y=c is

If one of the lines given by kx^(2)-5xy-3y^(2)=0 is perpendicular to the line x-2y+3=0 , then k=

If one of the lines given by kx^(2)-5xy-3y^(2)=0 is perpendicular to the line x-2y+3=0 , then k=

If one of the lines given by kx^(2)-5xy-3y^(2)=0 is perpendicular to the line x-2y+3=0 ,then the value of k is

Find the value of k if the straight line 2x+3y+4+k(6x-y+12)=0 is perpendicular to the line 7x+5y-4=0.

If one of the lines given by 3x^(2)-kxy+5y^(2)=0 is perpendicular to the line 5x+3y=0 , then k=