Question

find the value of k for which the lines kx-5y+4=0 and 4x-3y+5=0 are perpendicular to each other

Answer 1

kx+4=5y
(kx+4)/5=y
4x+5=3y
(4x+5)/3=5

For lines to be perpendicular,products of slope=-1; m1m2=-1
(k/5)*(4/3)=-1
k/5=-3/4
k=-15/4

Answer 2

kx-5y+4=0

kx+4=5y

$\frac{kx+4}{5} =y\\\frac{kx}{5} +\frac{4}{5} =y$

slope of line is k/5

also 4x-3y+5=0

4x+5=3y

$\frac{4x}{3} +\frac{5}{3} =y$

slope of line is 4/3

given lines are perpendicular to each other

$(\frac{k}{5})(\frac{4}{3})\\\frac{k}{5} =-1*\frac{3}{4} \\k=-\frac{3}{4} *5\\=-\frac{15}{4}$