(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)    —>    

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
 

Find the value of parameter k so that the line 3x - 5ky + 5 = 0 will be perpendiculat to 4x + 3y = 2.

Correct answer:  

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