Prove that the line `k^2 x+ ky+1 =0` is perpendicular to the line `x-ky=1` for all real values of `k(!=0)`.

Show that the line a^2x+a y+1=0 is perpendicular to the line x-a y=1 for all non zero real values of adot

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

Find k so that the line 2x + ky – 9 = 0 may be perpendicular to 2x + 3y – 1 = 0

If the straight lines 2x+ 3y -3=0 and x+ky +7 =0 are perpendicular, then the value of k is

If Lines 2x+3y=18 and 4x+ky=39 are perpendicular line then find k?.

If the line 2x - ky + 6 = 0 passes through the point (2,-8) , then find the value of k.

The line (K+1)^(2)x+Ky-2K^(2)-2=0 passes through the point (m, n) for all real values of K, then

Plane passing through the intersection of the planes x+2y+z-1=0 and 2x+y+3z-2=0 and perpendicular to the plane x+y+z-1=0 and x+ ky+3z-1=0 .Then the value of k is

The two lines 3x + 4y - 6 = 0 and 6x + ky - 7 = 0 are such that any line which is perpendicular to the first line is also perpendicular to the second line . Then , k = "_______" .

If one of the lines represents by 3x^(2)-4xy+y^(2)=0 is perpendicular to one of the line 2x^(2)-5xy+ky^(2)=0 then k=