Question 1081704
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(1)  3x-5ky+5 = 0   —>   y = (-3/(5k))x - (1/k)

(2)  4x+3y=2  —> y = (-4/3)x + (2/3)


In order for the lines to be perpendicular:   m1 = -1/m2

m1 = -3/(5k)
m2 = (-4/3)

   -3/(5k) = (-1)/(-4/3)    —>   {{{ highlight(k = -4/5) }}}